Understanding Periodic Motion and SHM

So what exactly is periodic motion? Periodic motion is motion that repeats itself or that oscillates back and forth. Some good examples of periodic motion and simple harmonic motion is the mass spring system and also the simple pendulum. Let's talk about the mass spring system. Let's say this is the wall and we have a spring connected to a mass horizontally. And let's say that this is the equilibrium position of the spring. So the spring has been stretched towards the right. So that means we're applying a force to pull it towards the right to stretch it from its equilibrium position. As we pull it towards the right, There is another force that wants to push it back towards its equilibrium position. This is known as the restoring force. Now, once you let go, the spring is going to move back towards the left. Now, it's not going to stop at the equilibrium position. It's going to move past it. And once that happens, it's going to bounce back the other direction. So it's going to move back and forth, towards the left and towards the right. And as it oscillates back and forth, you have simple harmonic motion, or periodic motion. Now according to Hooke's Law, the restoring force...... equal to negative times K times X. X is the distance measured from the equilibrium position. So let's say this is the wall and we have a spring and here's the block of mass M. The distance between where the spring is located and the equilibrium position is x. At the middle, at the equilibrium position, x is 0. So the restoring force will always be 0 at the equilibrium position. K is known as the spring constant, and K has the units newtons per meter. So let's say if you have a spring constant of 100 newtons per meter. What that means is that it requires 100 newtons of force to either stretch or compress the spring by 1 meter from its equivalent position. So if you wish to stretch it by 2 meters, it requires 200 newtons of force. So if you have a very stiff spring that's hard to compress, the K constant is very high. Let's say if K is 1000. So for this spring, it requires 1000 N of force just to stretch it by 1 meter. So this spring is going to be very stiff. It's hard to stretch or compress. This particular spring is going to be easy to stretch or compress compared to the other one. So this spring is relatively loose. By the way, before we work on some practice problems, I do want to mention one thing. In the equation in Hooke's Law, why is there a negative symbol for the restoring force? Why do we have this negative sign? What's the reason for that? So notice that if you apply a force towards the right to stretch the spring, the displacement vector will be towards the right. And the restoring force is towards the left. Because the restoring force is opposite to the displacement vector, FR is negative. Because it's always opposite to the direction of motion. the restoring force will always act in such a way to bring the object back to its equaling position. So if the equaling position is here, the restoring force wants to bring it to the left. Likewise, let's say if the spring is compressed. The restoring force wants to move it towards the right, back to its equal position. And remember, at the equal position, the restoring force is zero. since x is equal to 0 and FR is negative kx so negative k times 0 is 0. Let's work on some practice problems. How much force is required to stretch a 300 Newton per meter spring by 25 centimeters? So let's use the equation F is equal to kx. And let's not worry about the negative sign. If you see the unit Newton's Perimeter, that tells you that that value represents the spring constant, which is... 300 newtons per meter and since it has the unit meters in it that tells us that we need to convert centimeters into meters it turns out that 100 centimeters is equal to a single meter so to convert it needs to divide by 100 Or, move the decimal two units to the left. So this is equal to 0.25 meters, which is the value of x. So what is 300 times 0.25? Or, what is 25% of 300? This is going to be about 75. So that is the force required to stretch the spring by 25 centimeters. How far can you compress the spring with a force of 150 newtons? Now let's think about this. It required 75 newtons to stretch the spring by 25 centimeters. It also requires 75 newtons to compress the spring by 25 centimeters. So if we double the force to 150, the amount that we can stretch or compress the spring is going to be twice the value. It's going to be 50 centimeters. F equals kx. If you double the distance, if you stretch it by 2 times the amount, the force will increase by a factor of 2. Likewise, if you double the force, you can double the distance that you stretch or compress the spring by. Now, for those of you who prefer to use an equation, you can use this. Let's divide F2 by F1. F1 is equal to K times X1, and F2 is K times X2. Since we're dealing with the same spring, K is the same. so as you can see the force is proportional to x let's say f1 is 75 minutes and we're looking for f2 X1, the distance that corresponds to 75, is 25 centimeters. Actually, we're not looking for F2, we have F2. F2 is 150. We're looking for X2. 150 divided by 75 is 2. And at this point, we can cross multiply. So, x2 is equal to 2 times 25, which is 50. Here's another problem. It takes 200 newtons to stretch a spring by 40 centimeters. How much force is required to stretch a spring by 120 centimeters? So 40 centimeters corresponds to a force of 200 newtons. What is the force that's required to triple the distance by a factor of 3? If you triple the distance by a factor of 3, the force is going to increase by a factor of 3. So the force that's required is 600 N. Again, you can use this equation, F2 over F1 is equal to X2 divided by X1. So we're looking for F2 this time. F1 is 200 N. X1 is 40 centimeters, and X2 is 120 centimeters. You don't need to convert the units to meters, because the unit centimeters will cancel. So 120 divided by 40 is 3. So therefore, F2 divided by 200 is equal to 3. And if we multiply both sides by 200... We can see that F2 is going to be 3 times 200, which is 600 newtons. So that's how much force is required to stretch a spring by 120 centimeters. Now what about the second part to the question? What is the value of the spring constant? So using the formula F is equal to kx. Solving for k, k is equal to the force divided by the spring constant. So we can use the force of 200 newtons, and instead of using 40 centimeters, we need to convert it to meters. So that's about 0.4 meters. 200. divided by 0.4 is equal to 500. So the spring constant is 500 newtons per meter. So to stretch the spring by a distance of one meter requires a force of 500 newtons. Now what about the last part of the problem? How much work is required to stretch a spring by 120 centimeters? Work is equal to force times displacement, which displacement could be represented as x. Now this is if you have a constant force. Now what about if the force is a variable force? For example, in this case, work is equal to the integral of f times dx, where f is a function of x. And we know f is equal to k times x, ignoring the negative sign. So work is equal to the antiderivative of k times x dx. To find the antiderivative of kx, add 1 to the exponent and divide by that number. The antiderivative of x to the first power is x squared divided by 2. So the work required to stretch a spring is going to be basically 1 half kx squared. This is also the potential energy that is stored in a spring. In some textbooks you'll see this U symbol. This is the elastic potential energy of a spring. It's equal to 1 half k x squared. Which is the work required to stretch it or compress it. So k, we know it's 500 newtons per meter. I'm going to write it like this. And x... It's going to be 120 centimeters, but we need to convert it to meters, so let's divide it by 100, which is 1.2 meters squared. And the unit meters will cancel, which leaves behind newtons times meters. There's 2 meters here, since it's squared. Newtons times meters is equal to the unit joules. Half of 500 is 250 times 1.2 squared is equal to 360. So it takes 360 joules of work to stretch this particular spring by 120 centimeters. Now let's go back to our mass spring system slash oscillator. So let's say if we have a mass of M attached to a spring, and let's say this is the equilibrium position. Now, as we mentioned before, the force required to stretch the spring is the applied force. And whenever you stretch it, there's going to be a restoring force that pulls it back towards the left. Now, you also need to know that at this point, the net force is zero. Now, this is only true... while you're holding it. As soon as you release the mass from rest, the applied force will disappear. At that point the net force is equal to the restoring force. And at this instant, as soon as you release it, this object has maximum acceleration. And initially the velocity is zero. The velocity reaches its maximum value. once it reaches equilibrium. At equilibrium the restoring force is zero and because the restoring force is zero the acceleration is equal to zero at the equilibrium position. However because the velocity is at a maximum it's moving towards the left. So because it has momentum because it's moving due to inertia it's going to continue to move to the left compressing the spring. As it begins to compress the spring a restoring force is going to build up, decelerating the object. Once the object reaches a velocity of zero, the restoring force is at its maximum value, which means acceleration is at its maximum value at this point as well. So make sure you understand this, that at the equilibrium position, the velocity is at its maximum, and the acceleration is zero. When it's fully stretched or compressed, the acceleration is at its maximum value, and the velocity is zero. Now, if no friction is present in the system, this oscillator will continue to oscillate indefinitely. If there's friction, friction is going to oppose motion. So, in the first picture, the object is moving towards the left. So, the force of friction will be directed towards the right. Now, in the middle, it's still moving towards the left, so friction is still directed towards the right. And, on the last picture, because the velocity is 0, friction will be 0. since it's not moving instantaneously. But as it begins to move towards the right, friction will oppose it. So friction always opposes motion. By the way, once you release it, there's no friction until it begins moving towards the left. So as it accelerates, friction is going to oppose it. Whenever there's friction present, the oscillator will be dampened by it. The amplitude will... gradually decreased. Now let's talk about X and A. At the equilibrium position, let me draw a new picture. So let's say if we have an oscillator, and let's say this is the equilibrium position. X is equal to 0 at the equilibrium position. And let's say at this point x is equal to 1. So this is the maximum displacement. The maximum displacement is represented by the symbol capital A. So in this case, for this problem, A is 1. At this position, x is 0.5, since it's to the right of the equilibrium position. And at this position... x is negative 0.5. So let's say k is 100 newtons per meter. What is the force, the restoring force, at let's say a position of 0.5? Include the appropriate sign as well. So f is equal to kx. It's going to be negative times 100 times 0.5, which is negative 50. So at this point, we have a force of negative 50. The negative sign means that it's towards the left. Now what about at this position? If we plug in negative 0.5, okay, that does not look like a 5, it's going to be negative 100 times negative 0.5, which is positive 50. Since the object is moving at this location towards the left, the restoring force is towards the right. And notice that the force is positive 50 because it's directed towards the right. When the restoring force is directed towards the left, Notice that it's negative now. Let's talk about the energy of this particular oscillator when it's fully stretched and Once you release it The acceleration is at its maximum and the velocity is zero. The kinetic energy of any moving object is one half mv squared. So the kinetic energy is dependent on the speed. So when it's fully stretched or fully compressed, let's say it's fully compressed at this position here, the kinetic energy is at a minimum because the velocity is zero. So, Ke is going to equal 0 when it's fully stretched or compressed. Now, at the equivalent position, the velocity is at a max. So, therefore, Ke is at its maximum value at the equivalent position. As we mentioned before, the acceleration is 0 at the equivalent position, and the restoring force is equal to 0 since x is 0. And f equals kx, so k times 0 is 0. Now what about the potential energy? The potential energy is at its maximum when it's fully stretched or compressed. But at equilibrium the potential energy is zero. The potential energy is equal to one-half kx squared. Since x is 0 at the equilibrium position, the potential energy is 0 because 1 half k times 0 squared is 0. Now, there's another thing that you need to be familiar with, and that is mechanical energy. The mechanical energy is the sum of the kinetic energy and the potential energy. Keep in line. When the amplitude is 1, x can vary anywhere between negative 1 to 1. The mechanical energy depends on the amplitude, that is the maximum value of x, or the maximum displacement. So the mechanical energy is 1 half k a squared. The kinetic energy is 1 half mv squared. And the potential energy is 1 half k x squared. So, whenever x is equal to a, the potential energy is the same as the mechanical energy. At this point, the kinetic energy is 0. And whenever x is equal to 0, that is at the equilibrium position, potential energy is equal to 0. So let me organize this information. So let's say this is the equilibrium position where x is 0. At that point, the potential energy will be equal to 0. So Ke is at its maximum. Ke equals the mechanical energy. The mechanical energy of the system is constant if there's no friction. The mechanical energy is the total energy of the system. Now let's say if it's at the right, if it's fully stretched. At this point, the maximum displacement is equal to A. So when X is equal to A, the potential energy is at a maximum. It's equal to the mechanical energy, which means that Ke is equal to 0. And when it's fully compressed... PE is still equal to the mechanical energy, and KE is equal to 0. Now, sometimes you may need to calculate the maximum acceleration and the maximum velocity. What equations can we use to do that? So, let's start with maximum velocity. Now, we said that... It's going to be, the velocity is going to be at its maximum when x is equal to 0. So when x is equal to 0, the potential energy is 0, so this disappears. So therefore, 1 half kA squared is equal to 1 half mv squared. We can multiply both sides by 2 to get rid of the fraction. So kA squared is equal to mv squared. To isolate V squared, we need to divide both sides by m. So V is equal to the square root of k divided by m times the amplitude, a. So that's how you could find the maximum velocity if you ever need to. What about the acceleration? How can we find the maximum acceleration? We know that F is equal to kx. And according to Newton's second law, F is ma. So what we need to do is divide both sides by m. So k divided by m times x is equal to the acceleration at any point. So that's the acceleration as a function of x. Now, the acceleration is at its maximum value when x is the maximum, which is the amplitude. So the maximum acceleration is k times a over m. Now what if we don't want to find the maximum velocity? What if we need to find the velocity as a function of x? Let's say when x is 0.4 or 0.7. How can we do that? Well, let's go back to our conservation of energy equation. Mechanical energy is equal to kinetic plus potential energy. So, let's begin by multiplying everything by 2. 2 times 1 half is 1. This will cancel all of the fractions. So, we have k a squared is equal to m v squared plus k x squared. Now, let's subtract k x squared by both sides. k a squared minus k x squared is equal to m v squared. And now let's factor out the GCF, the greatest common factor, which is K. So this is going to be A squared minus X squared. Next, let's divide the expression by the mass. So V squared is equal to K divided by M times A squared minus X squared. Now, what's our next step? The next thing we need to do is take out a. I mean, a squared. If we take out a squared, a squared divided by a squared is simply 1. And this, divided by a squared, that's going to be negative x squared divided by a squared. So now, what we want to do is take the square root of both sides. So the velocity as a function of x is going to be the square root of k divided by m times a times the square root of 1 minus x squared divided by a squared. Now the velocity can be positive or negative. If the velocity is moving towards the right, it's positive. If it's moving towards the left, it's negative. Now if you recall, the quantity... Square root k over m times a is equal to the maximum velocity. So therefore, we can rewrite the equation like this. So the velocity as a function of x is equal to plus or minus the maximum velocity times the square root of 1 minus x squared divided by a squared. So make sure you write this equation down. It might be useful later. Now there's something else that we need to talk about, and that is the frequency and period of this spring mass oscillator. So let's say this is the equilibrium position. moves to the left and as it moves back to where it started from, that is equal to one cycle. The distance of one cycle is equal to basically four times the value of x. So from here, back to its equaling position, it travels a distance of x. And then when it's fully compressed, it travels another distance of x. And then when it goes back to its equaling position, that's another x value. And when it returns to where it started, that's x again. So the distance of one cycle... is four times its maximum displacement. So we should really say this is four times A, since A represents the maximum displacement. So this is A. Now what about the period and the frequency? The period is the time it takes to complete one cycle. The period, which is represented by capital T, is the number of, is the time, divided by the number of cycles. So let's say if it takes 30 seconds to complete 10 cycles, 30 divided by 10 is 3. That means that it takes 3 seconds to complete a single cycle. Therefore, the period is 3. So that's how you could find a period. It's the total time divided by the total number of cycles, or the time it takes to complete one single cycle. The frequency is the number of cycles per second. So if you take the total number of cycles and divide it by the total time, you can get the frequency. So for example, let's say if it takes, let me choose a nice number. Let's say if this system can cycle 200 times. in 10 seconds. So if it makes 200 cycles in 10 seconds, that means that it's doing 20 cycles in 1 second. So the frequency is 20 or 20 Hertz. Hertz is the unit of frequency. Now frequency and period are inversely related. The frequency is 1 divided by the period and the period is 1 divided by the frequency. So let's work on some problems. What happens to the following if the maximum displacement of the spring is doubled? So what happens to the total energy of the system? The total energy of the system is the mechanical energy. Now the first thing we want to do is write an equation that relates the mechanical energy with the maximum displacement. And that's this equation. mE is equal to 1 half kA squared, where A is the maximum displacement. So if we double the value of a, what happens to m8? For these types of problems, anything that's constant or remains the same, replace it with a 1. So 1 half k, we're just going to replace it with 1. And plug in the value of a. If you double the value of a, because the function is squared, the mechanical energy will increase by a factor of 4. If we were to triple the value of a, the mechanical energy will increase by 3 squared, which is 9. If we quadruple the value of A, the mechanical energy will increase by a factor of 16. Now what about part B, the maximum velocity? What equation can we write that relates maximum velocity to the maximum displacement? Now we came up with the equation earlier. We said that the maximum velocity is equal to the square root of k divided by m, where k is the spring constant, times the maximum displacement. So notice that it's raised to the first power. So if we double the value of a, and this is going to stay the same, so we're going to replace it with 1, the velocity will increase by a factor of 2. If you triple the displacement, the velocity will increase by a factor of 3. If you cut the displacement in half, the velocity will decrease by a factor of 2, or it's going to be 1 half of its original value. Now what about part C, the maximum acceleration? The equation that we wrote for that is this equation. The maximum acceleration is equal to k divided by m times the maximum displacement. So if you double the value of the maximum displacement, the acceleration will increase by a factor of 2. If you triple it, the acceleration will triple. If you reduce it by one half... the acceleration will reduce by a factor of two. Let's work on this problem. A massless spring is pulled by 30 centimeters from its natural length and then released. The spring is attached to a 0.4 kilogram block and oscillates across a horizontal frictionless surface. Let's begin by calculating the maximum velocity. So if you want to you can draw a picture. So here's the block and it rests across the horizontal surface. Let's say this is the equilibrium position. So right now it's pulled by 30 centimeters which is 0.3 meters. So let's make a list of what we know. A, the maximum displacement is 0.3. And K, the spring constant, is 300 newtons per meter. What equation do we need to calculate the maximum velocity? The maximum velocity is simply the square root of K divided by m times A. The mass... is 0.4 kilograms so we have everything we need to find the maximum velocity so this is going to be square root 300 divided by 0.4 times 0.3 300 divided by 0.4 is 750. The square root of 750 is 27.386 times 0.3. And you should get a maximum velocity of 8.216 meters per second. So now, what about part B? What's the maximum acceleration? So let's use this equation. It's equal to k times a divided by m. So it's going to be 300 times 0.3 divided by 0.4. 0.3 divided by 0.4 is basically 0.75 times 300. That's 225. So the maximum acceleration is 225 meters per second squared. Now what about part C? How can we find the velocity at a position of 20 centimeters from its natural length? So keep in mind, the maximum velocity was about 8.2 something. 8.216 We can use this equation. The velocity as a function of x is equal to the maximum velocity times the square root 1 minus x squared divided by a squared. That's the equation that we had earlier in this video. So we want to find the velocity at a position of 20 centimeters from its natural length. Now, because we have a ratio between x and a, it doesn't really matter if you use centimeters or meters. The units simply have to match, because they're both going to cancel. Centimeters over centimeters and meters over meters will cancel. So we can plug in 20 centimeters if we want to. So the maximum velocity is 8.216, and then this is going to be 1 minus 20 squared. divided by 30 squared. 20 squared divided by 30 squared is about 4 over 9, which is 0.4 repeating. And 1 minus that number is about 0.5 repeating. And the square root of that is 0.7454 times 8.216. So you should get a velocity of about... Let's make some space. 6.124. So your answer should be less than the maximum velocity. If for some reason it's larger, you know it's not correct. Now what about the next part? How can we find the restoring force 20 centimeters from its natural length? Now let's say 20 centimeters is located at this point. We know the restoring force is directed towards the left, which means that it should be negative. So the restoring force is equal to negative times kx. So this is going to be negative 300 newtons per meter times x, which is 0.2 meters. Now in this case, we need to convert centimeters to meters because we're using k, and k is in meters. 3 times 2 is 6, so 300 times 0.2 must be 60. So the restoring force is negative 60 newtons. It's negative because it's directed towards the left. Now what about the acceleration? Well, we know that the force is equal to mass times acceleration. So the acceleration is the restoring force divided by m. And the restoring force is negative 60 newtons. The mass is 0.4 kilograms. So negative 60 divided by 0.4 is negative 150 meters per second squared. Keep in mind, the maximum acceleration was 225. And this is less than 225, so that works out. Now, you can also use this equation. Acceleration is equal to k over m times x. It's similar to this equation. The maximum acceleration is k times a divided by m. But here, it's k times x over m. So, k is 300, x is 0.20, and m is 0.4. So, if you use this equation, you should get the same answer. which is 150. Now we know it should be negative 150, because the resultant, I mean the restoring force is towards the left. Acceleration and the force that causes that acceleration are usually in the same direction. So because the restoring force is negative, the acceleration is negative, which really means that this equation should have a negative sign. So let's go ahead and add it. So that is it for this problem. In this problem, we have a vertical spring. So let's say this is the spring without a mass attached to it. Now once we add a mass to it, it's going to stretch. Now it stretches by 0.4 meters. So this is the new equilibrium position of the mass. Now, if we stretch it even further by an additional 0.20 meters from its new equilibrium position, once we release it, it's going to bounce up and down. So with this information, determine the spring constant, the amplitude, and everything else. So let's focus on the spring constant. We can use this picture to find the value of the spring constant. The applied force that stretches the spring from its original position is basically the weight force of the object, which is mg. The spring wants to go back to its original length. So the restoring force is equal to the weight force, so that the net force is zero, creating that new equilibrium position. So FR is equal to MG. And the restoring force is negative KX. and G is negative 9.8 so you really don't need to worry about the negative sign they will cancel now let's solve for K so K is going to be equal to MG divided by X so the mass is 2 kilograms G is 9.8 and it stretches by distance of 0.4 meters So 2 times 9.8 is 19.6 divided by 0.4 is 49. So K is 49 newtons per meter. Now what about part B? What is the amplitude in this problem? What do you think the amplitude is equal to? Now remember, this is the new equilibrium position, and it stretched 0.20 meters from that equilibrium position. So therefore, the amplitude is 0.20. So once we release it from this position, it's going to bounce up and down.20 from this position. So it's going to go up.20 and down.20 meters. Now what about C? How can we calculate the maximum acceleration? What is the equation that we need? If you recall, the maximum acceleration is equal to the spring constant times the amplitude divided by the mass. So it's 49 times 0.2 divided by the mass of 2. So the maximum acceleration is 4.9 meters per second squared. Now what about d, the maximum velocity? The equation that we need for that is the square root of k divided by m times a. 49 divided by 2 is 24.5. And the square root of that... is about 4.95 times.2 that's about.99 meters per second. So now what about part E? What is the kinetic energy, potential energy, and mechanical energy when it's.10 meters from its equilibrium position? So let's find the potential energy first. The potential energy is 1 half kx squared, which is basically 1 half times 49 times an x value of 0.10. 49 times 0.10 squared, that's about 0.49 times half, that's 0.25. So the potential energy is 0.245 joules. Now what about the kinetic energy? The kinetic energy is equal to 1 half mv squared. Now we can't use the maximum velocity. The velocity will be 0.99, it will be at its max when x is 0 at the equilibrium position, but we're not at the equilibrium position, x is 0.10. So we've got to find the velocity first at 0.10. The velocity as a function of x is the maximum velocity times the square root of 1 minus x squared divided by a squared. So the maximum velocity is about 0.99, and x is 0.1, a, the amplitude, is 0.2. 0.1 squared divided by 0.2 squared is 1 fourth. 1 minus 1 fourth is 3 fourths, and the square root of 3 fourths is about 0.866 times 0.99. You should get 0.8574. meters per second. Now that we have the velocity, we can now use this equation to find the kinetic energy. So let's plug it into that equation. So Ke is equal to one half times the mass, which is two kilograms, times the speed, 0.8574 squared. So half of two is one. And 1 times 0.8574 squared is about 0.735 joules. So that's the kinetic energy, 0.735. Now let's calculate the mechanical energy. We're going to do it two ways. So first let's use the equation mechanical energy is equal to 1 half. K times the amplitude squared. So K is 49 and the amplitude is 0.20 meters or 0.2 meters. 0.2 squared is 1 over 25 or 0.04 times 49 which is 1.96. Half of that is 0.98. So the mechanical energy which represents the total energy of the system is 0.98 joules. Another way in which we could find it is by adding the potential energy and the kinetic energy. So if we add 0.245 plus 0.735, it also gives us the same answer of 0.98 joules. So let's say if you didn't know the equation to find the velocity at any point x. You can use this to get the mechanical energy. And if you subtract the mechanical energy by the potential, that's one way you can get the kinetic energy. So there's different ways of solving different things that you may need. So you're not faced with one option. There's many options that you can use. Try this problem. A spring is compressed 0.35 meters with a 0.25 kilogram block. by an applied force of 500 Newtons. What speed will the block have as soon as it's released from the spring across a horizontal, frictionless surface? So let's say this is the wall, and we have a compressed spring and the mass attached to it. And let's say this is the horizontal... frictionless surface. Now it's compressed by an applied force of 500 newtons and it's compressed by a distance let's say of 0.35 meters and the mass of the block is 0.25. Right now The applied force is balanced by a restoring force of 500 N. So as soon as we release it, the applied force will disappear, and it's going to accelerate towards the right due to this restoring force of 500. Now, this restoring force is not constant. As the spring expands, the restoring force will decrease eventually to zero. At that point, the mass will be released from the spring and it's going to move to the right at constant speed. So, when it's completely released from the spring, when it's no longer attached to the spring, what is the speed that the block will have? How can we calculate that value? Take a minute and work on that example. The first thing that we need to do is find the spring constant, K. So F is equal to Kx, therefore K is the force divided by x. So we have an applied force of 500 newtons, which compresses the spring by 0.35 meters. So if we divide these two, we'll see that the spring constant is about 1429. Round it to the nearest whole number, newtons per meter. So now that we have the spring constant, we can use conservation of energy to calculate how fast it's going to move. So the mechanical energy is equal to the kinetic energy. Or you could say the potential energy stored in the spring is equal to the kinetic. The mechanical energy is 1 half kA squared. If you use the potential energy equation, it's 1 half kX squared, but in this case, X is equal to A because it's stretched. I mean, it's compressed all the way. X is basically at its maximum displacement when it's 0.35. So we'll use the symbol A, though. 1 half KA squared is equal to 1 half MV squared. So if we multiply both sides by 2, we can get rid of the 1 half. So k is 1429, a is 0.35, m is 0.25, and we need to solve for v. 0.35 squared times 1429. that's about 175 and if we divide that by.25 we see that v squared is equal to about 700.21 and now if we take the square root of both sides the velocity is about 26.5 meters per second so that's how fast it's going to be moving By the way, let's say if we wanted to find the kinetic energy of the object as soon as it's released. Using this equation, it's 1 half times the mass, which is 0.25, times the speed, which is really 26.46 squared. The kinetic energy of the object is about 87.53 joules. Now another equation that you can use to get that answer is the work equation. So keep in mind, as soon as we release it, there's a restoring force that's going to accelerate it towards the right. Work is equal to force times distance. however we don't have a constant force we have a variable force now let's assume that this force decreases constantly well it actually does so we don't really have to assume the force will decrease at a constant rate from its maximum value of 500 to 0 as the block is displaced by 0.35 meters. So basically, the energy transferred by this force should be equal to the kinetic energy, which is 87.5. So how can we find the area of that triangle? Well area is basically 1 half base times height or 1 half times the height which is 500 that's basically the restoring force times the base which is the displacement of 0.35. Half of 500 is 250 and 250 times 0.35 gives you 87.5 joules. So that's another way in which you can calculate the kinetic energy gained by the block, is by using the work equation. But if you're going to use the restoring force, you need to add a 1 half to it, because the restoring force is not constant. Now there are some other topics that you need to know, and that is the equations for the period and the frequency as it is related to the mass and the spring constant of a spring. Now we need to compare circular motion with the simple harmonic motion of a spring. So let's draw a circle. Now, let's say if an object on a circle moves from, let's say, point A to point B. As it moves this way, if you can view the circle... horizontally from this perspective rather than viewing it from top to bottom if you can just have like a side view all you will see is that this particle is moving in the negative x direction you're only going to see the displacement along the x-axis Now, as it moves from, let's say, position B to position C, if you take a side view of it, you're going to see it's still moving towards the left. Now, as it moves from C to D, it's going to appear as if it's moving towards the right, which it is. And from D to A. it's moving this way. So if you can view it from the side, the circular motion would seem similar to the oscillation of a spring as it moves back and forth. So knowing that, we can say that the speed of the object around a circle is basically the distance that it travels divided by the time. And the distance around a circle is the circumference, which is 2 pi r. The time it takes to make one revolution or one cycle is the period, which is represented by capital T. Now, notice that the radius of the circle is basically the amplitude. So, we can replace r with a. Now, we no longer need this picture. So now at this point, let's multiply both sides by t. So vt is equal to 2 pi a, and now let's divide both sides by v. So the period is 2 pi times the amplitude divided by the velocity. And if you recall, the maximum velocity we said was equal to... the square root of k divided by m times a. So to solve for a, we can multiply both sides by the square root of m divided by k so that m will cancel on the right and k will cancel on the right as well. So v times the square root of m over k is equal to a. So now let's replace this a with this expression. So the period T is 2 pi divided by V times V max times the square root of m over k. So we can cancel V. So therefore, the period... of a spring is 2 pi times the square root of m divided by k. So now let's talk about this equation. If we increase the mass of the spring, what's going to happen to the period? Well the time it takes to complete one cycle will it increase or decrease? Because the mass is an enumerate of the fraction, the period will increase as well. So let's say if you double the mass of the spring, what effect will it have on the period? So for these questions, plug in the value that is being changed, everything else plug in the 1. So if we double the mass, the period will increase by the square root of 2. Now what if we quadruple the mass? The period will increase by the square root of 4, which is 2. So if you quadruple the mass, the period will double. If you increase the mass by a factor of 9, the period will triple in value. What if you reduce the mass by 1 half? The square root of 1 half, which is 1 over root 2, that's root 2 over 2. So the period will increase. Well, actually, it's going to... decreased by root 2 over 2. That's about 0.707. So it's going to be 70% of its value. Or you could say it's reduced by a factor of square root 2, because you're dividing it by root 2. So if you decrease the mass to half of its value, the period is going to be reduced by a factor of square root 2. It's going to be 70% of its original value. or 70.7. Now what about the spring constant? Let's say if we increase the value of k, what happens to the period? The period will decrease since k is on the bottom. So if you double the value of k, the period will decrease by the square root of 2. It's going to be 70% or 70.7% of its original value. If you quadruple the value of k, the period is going to be 1 half of its original value. Now what if you reduce the spring constant by... 1 4th. What effect will that have on the period? 1 divided by 1 4th is 4. So if you decrease k to 1 4th its value, the period will double. Now let's think about what this means. So we could see from the equation why the period will increase if we increase the mass, but let's understand it. conceptually as you increase the mass of the block is going to have more inertia so it's going to be harder to move as a result it's going to have less acceleration f is equal to ma if the force is constant whenever you increase the mass the acceleration will decrease so if it has a lower acceleration it's going to move more slowly And therefore, an object that moves slowly takes a longer time to get to its destination. Therefore, as you increase the mass, the period increases. Since the object is heavier, it's going to take a longer time to move back and forth. And so that's why the period increases. Now what about the second situation, where if we increase the value of k, the period decreases? A large k value means that... The spring requires a greater force to compress or stretch it, so the spring is stiff. Because the spring is stiff, it doesn't really move very far. It doesn't really, it's not easily stretched or compressed, so it can vibrate faster. As a result, because it can vibrate faster, its period will be reduced. So keep in mind k is f over x. So a large k constant means that a greater force is required to stretch it with the same distance. And if the force is greater, the acceleration will be greater as well. And a higher acceleration means a fast moving object. It can quickly gain speed. And if the object is moving faster, then it's going to take a longer time. I mean a shorter time actually, to oscillate back and forth. And so that's why as you increase the value of k, it oscillates faster and the period is reduced. Now going back to this equation, there's one more equation that you need, and that is frequency. Frequency is 1 divided by period. So the frequency is going to be 1 divided by 2 pi. times the square root of k divided by m. So this time, if you increase the k value, if you increase the spring constant, a greater force is required to stretch or compress the spring, which means that there's going to be a greater acceleration, which means that it's oscillating faster. And because it's oscillating faster, the frequency will increase. Now if you increase the mass, the object requires a greater acceleration to move it back and forth. But if the force is constant, as you increase the mass, the acceleration will be reduced. And so it's going to be moving slower, so to speak. So therefore, the frequency will be reduced. Frequency and speed, they're related. An object that can oscillate back and forth at a faster rate will have a higher frequency. And an object that oscillates at a slower rate will have a low frequency. So as you increase the value of k, the speed of the oscillations will increase, and so the frequency will increase. If you increase the mass, heavy objects tend to move slower, and so the frequency will decrease. A spring attached to a 0.25 kilogram block is stretched horizontally across a frictionless floor 0.25 meters by an applied force of 200 N. Calculate the frequency and period of the oscillations of this mass spring system. So let's begin by drawing a picture. So here we have a spring attached to a mass. and here is the floor So as we stretch it with an applied force, once we release it, the spring will oscillate back and forth. So how can we find the frequency of the oscillations? Well, we can use this equation, 1 divided by 2 pi times the square root of k divided by m. Now, in this problem, we don't have the spring constant, k, so we need to find it. But we do have the applied force and the distance that it's stretched by. So, k is equal to the force divided by x. So, that's 200 newtons divided by 0.25 meters. So, let's go ahead and divide those two numbers. So, k is 800 newtons per meter. Now that we have the spring constant, we can calculate the frequency. So it's 1 divided by 2 pi times the square root of k, which is 800, divided by the mass, which is 0.25 kilograms. 800 divided by 0.25 is 3200, and the square root of 3200 is about 56.57. And let's divide that by 2 pi. And you should get a frequency of 9 Hz. Now that we have the frequency, we can easily calculate the period. The period is simply 1 over f, which is basically 1 divided by 9. 1 divided by 9 is about 0.1 repeated. So the period is about 0.11 seconds. So it takes 0.11 seconds to make a full cycle. That is, it takes 0.11 seconds for the block to travel towards the left and back to its original position. And in one second, the spring will oscillate nine times. That's the frequency, the number of cycles that occurs in one second. So in one second, it's going to go backwards and forwards nine times in that single second. Here's another problem you can try. When a 70 kilogram person gets on a 1200 kilogram car, the spring is compressed by 2 centimeters, what will be the frequency of vibration when the car hits a bump. So first, let's find K. So F is equal to Kx. The force that compresses the spring by 2 centimeters is the weight of the person, because once he gets on the car, X changes by 2 centimeters, so we only need to use the weight of that person, and not the weight of the car and the person, to find K. So K is equal to mg divided by X. So it's the mass of the person times 9.8 divided by 2 centimeters converted to meters. To convert centimeters to meters, divide by 100. So this is 0.02 meters. So it's 70 times 9.8 divided by that number and the spring constant is 34,300 newtons per meter. Now that we have the spring constant, we can calculate the frequency of vibration. Now when the car hits a bump, you have to consider the mass of the person and the car because the spring... is attached to both, or it's affected by the mass of the person and the car. It has to support the weight of both. So, using the equation F is equal to 1 over 2 pi times the square root of k divided by m. This is 1 over 2 pi times root 34,300 divided by the total mass. That's 70 plus 1,200, or 1,270. 34,300 divided by 1270 is about 27. And the square root of 27 is about 5.2. And if you divide that by 2 pi, you should get a frequency of 0.83 hertz. So that's it for this particular problem. So here's another problem. Here we have an insect caught in a spider web. We have the mass of the insect and the frequency and we need to find the spring constant k. So let's start with the equation f is equal to 1 over 2 pi times the square root of k divided by m. Now let's solve for k. So first let's multiply both sides by 2 pi. So 2 pi f is equal to the square root of k divided by m. Now at this point, we're going to square both sides, so we can get rid of the square root on the right. So 2 pi f squared is equal to k divided by m. Now the last thing we need to do is multiply both sides by m. So now we have the equation that we need. So the spring constant K is equal to the mass times 2 pi f squared. And let's assume that the mass of the spider web is negligible. So we're only concerned with the mass of the insect. Now, we need to plug in the mass in kilograms. To convert grams to kilograms, you can divide by 1,000, or instead of writing 0.25g, you can write 0.25 times 10 to the minus 3 kilograms. 1 gram is about 0.001 kilograms, or 1 times 10 to the minus 3 kilograms. And the frequency is 20 hertz. 2 pi times 20 is about 40 pi or 125.66. If you square it, you should get 15,791. And then if you multiply it by 0.25 times 10 to the minus 3, this should give you a spring constant of 3.95 newtons per meter. Now that we have the spring constant K, how can we answer the second part of the problem? So, if the mass is now 0.10 grams, what is the frequency? So, using the equation 1 over 2 pi times the square root of K over M, it's going to be different. So, K is about 3.9478. 3.95 is the right answer. and the mass is going to be 0.1 times 10 to the minus 3 kilograms. So 3.9478 divided by 0.1 times 10 to the minus 3, that's 39,478. If you take the square root of that, you should get 198.69, and then divide that by 2 pi, and this will be equal to 31.6 hertz. Now, you can get the same answer. Even if you don't have the value of K, and let me show you so let's say if we had a ratio F2 divided by F1 F2 is 1 over 2 pi square root K divided by M2 and we're assuming that K is the same since we're dealing with the same spider web and this is going to be same thing but m1. So we can cancel the 2 pi and we can cancel k. So we get the equation f2 divided by f1 is equal to the square root of 1 over m2 divided by the square root of 1 over m1. So how can we simplify the expression that we have on the right? What would you do to simplify it? Let's multiply the top and the bottom by the square root of m1. If we do that, m1 will cancel on the bottom. And so we're going to have f2 divided by f1. And if you multiply 1 over m2 times m1, this is just going to be the square root of m1 over m2. So this is the equation that we can use to get the answer to the second part of the problem if we don't have the spring constant K. So notice that a frequency of 20 Hz corresponds to a mass of 0.25 g. Because we have a ratio of the masses, we don't need to convert it to kilograms. So m1 is going to be 0.25 grams. Now we're looking for the frequency f2 when the mass is 0.10. 0.25 divided by 0.10 is 2.5. And the square root of 2.5 is about 1.581. So f2 divided by 20 is equal to... 1.581. To solve for F2, multiply both sides by 20. So 1.581 times 20 is 31.62 Hz, which is about the answer that we got a few minutes ago. A block of mass attached to a spring with a constant of 200 Nm vibrates at 15 Hz. What is the frequency of vibration if the same block is attached to a spring of 500 Nm? So the mass is constant here. The only thing that changes is the spring constant and f. So we know that f is equal to 1 over 2 pi times the square root of k divided by m. So we can write a ratio, f2 divided by f1 is 1 over 2 pi square root k2 over m. So we're going to write the subscript for k since k changes, but m is constant, so we don't need to write the subscript for m. So we can cancel the 2 pi and the mass. So we're going to get this equation, f2 divided by f1 is equal to the square root. of k2 divided by k1. So as you increase the spring constant, the force will increase. So, I mean not the force, but the frequency will increase. So the frequency should be higher than 15 hertz. So f1 is 15, and that corresponds to a k constant of 200. k2 is 500, and we're looking for f2. So if we divide 500 by 200, that's 2.5 and the square root of 2.5 is 1.581 so f2 divided by 15 is that number so to find f2 we need to multiply 1.581 by 15 and you should get a frequency of 23.7 Hertz The block undergoes SHM with amplitude 0.4, that simple harmonic motion. If you see SHO, that's a simple harmonic oscillator. What is the total distance that it travels in 8 periods? Now, keep this in mind. The distance that it travels in 1 period, or 1 cycle, is 4 times the amplitude. Let's say this is the equilibrium position. This is the amplitude, the amount that it's fully stretched by. So for this block to travel one complete cycle, it has to travel A towards the equilibrium position and another distance of A, and then back towards the equilibrium position and back towards where it started. So as you can see, The distance that it travels in one cycle or one period is 4 times 8. So 4 times 0.4 is 1.6 meters. So in one period, it's going to travel a distance of 1.6 meters. In 8 periods, it's going to be 1.6 times 8, which is about 12.8 meters. So that's the total distance that... the block will travel in eight periods. Now there's some other functions that can describe the motion of a mass spring system. So let's say if we have a vertical spring and currently it's compressed. Let's say this is the equilibrium position so it's capable of bouncing Up and down so therefore This is going to be equal to the amplitude and this will be negative A. Now, as it bounces up and down, if you plot this on a graph, let's say time is the x-axis and the displacement is on the y-axis. So this is going to be the maximum displacement, and here's negative a. So according to this graph, it's starting at its compressed position. So that's at the top. If you graph it over time, it's going to form a sinusoidal function. It's basically a sine wave. Now, because it started at the top, this is the cosine function. If it started at the middle, it would be sine. Now if we draw the circle that we had earlier in this video, this is X and the radius of the circle is A. On the inside, this is the angle theta, and this is the right triangle. According to Circulator, if you've taken trigonometry at this point, you know that cosine of the angle theta is equal to the adjacent side divided by the hypotenuse. That's the Ka part in SOHCAHTOA. The adjacent side is the side that's very close to the angle. So that's x. The hypotenuse is across the box. That's a. So cosine theta is x divided by a. So if you multiply both sides by a, you can get the value of x. So therefore x is equal to a cosine theta. Now we know that linear displacement is equal to velocity multiplied by time, d equals vt. Angular displacement is equal to omega, which is angular velocity multiplied by time. Now, perhaps you have seen this equation as you've studied rotational kinematics. If you haven't, you can look up my video that I've created on YouTube, you can check that out, and you'll see that equation. Omega which is the angle of velocity, it's equal to 2 pi times the frequency. So therefore, we could say x is equal to a cosine omega times t, since theta equals omega times t, and then we can replace omega with 2 pi f. So x, the position function with respect to time, is equal to the amplitude times cosine. 2 pi f times t. So make sure that you write this equation down. This is an equation that you need to know. Now if you've taken calculus, you know that the derivative of the position function x gives you the instantaneous velocity. So let's find the derivative. The derivative of cosine is equal to a negative sine. Now you have to keep the angle. the same, the 2 pi f t. And then you have to differentiate the nested function, or the inside part of sine, which is 2 pi f t. The derivative of 2 pi f t is simply 2 pi f. t is the variable that you're differentiating with respect to. The derivative of 5x is 5. The derivative of 9x is 9. The derivative of 8t is 8. So the derivative of 2 pi f t is 2 pi f. By the way, for those of you who haven't taken calculus and who want to understand how to find the derivative of a function to understand a step, you can find my video on derivatives in YouTube. You just gotta search for it. You may have to type in like power rule, product rule, quotient rule, but you only need the power rule for this function. And also, you need to know how to find the derivative of trigonometric functions as well, which you can also find out on YouTube. So now let's continue. So we need to replace a few things at this point. Let's replace f. We know that the frequency is 1 divided by 2 pi times the square root of k divided by m. In addition, earlier in this video we mentioned that the amplitude is equal to the maximum velocity. times the square root of m divided by k. This came from the equation v max is equal to square root k over m times a. So let's replace a with this expression, and let's replace f with that expression. So the velocity as a function of time is going to be a, which is v max, times the square root of m divided by k times negative sine 2 pi ft times 2 pi times f, which is 1 divided by 2 pi, square root k divided by m. So we can cancel 2 pi, we can cancel k, and we can cancel m. So now we have the velocity function with respect to time. So here it is. The instantaneous velocity is equal to the maximum velocity. Let's put the negative in front. So negative Vmax times sine 2 pi ft. And that's it. So make sure that you add this to the list of equations that you need to know. Now instead of writing Vmax, some textbooks will use the symbol VO, but it's the same thing, it's the maximum velocity. Now let's find the acceleration. The acceleration is the derivative of the velocity function. So Vmax is a constant, we don't have to change that. The derivative of sine is cosine, and the derivative of 2 pi ft is 2 pi f. Now let's replace Vmax with what we wrote before. We said it's equal to negative square root k over m times a. And the frequency, as we mentioned before, is 1 over 2 pi square root k over m. So we can cancel 2 pi. k times k is k squared. And m times m is m squared. This time they don't cancel. So we have k squared over m squared, a cosine 2 pi ft. The square root of k squared is simply k, and the square root of m squared is simply m. So the acceleration as a function of time is negative k over m times a. cosine 2 pi ft. Now if you recall, f is equal to kx, and the force is basically mass times acceleration. And let's replace x with the amplitude a. So the maximum acceleration is k times the amplitude divided by m, which is exactly what we have here. So therefore, we could say that the acceleration function with respect to time, is equal to the maximum acceleration, you can say a-naught, or simply just a-max, times cosine 2 pi ft. So t is the variable, f is a constant, and the maximum acceleration is a constant. So here's the third equation that you need. A 0.75 kilogram mass vibrates according to the equation. x is equal to 0.6 cosine 9.2t. x is in meters, t is in seconds. So, what is the amplitude? Now, we have the position function with respect to time, and we know the equation is a times cosine 2 pi ft. So, the amplitude is simply the number in front of cosine. Therefore, the amplitude is 0.6. That's the maximum displacement of the 0.75 kg mass. Now what about part B, the frequency? Notice that the 9.2, the number in front of T, is equal to 2 pi f. So let's set 2 pi f equal to 9.2 and let's solve for f. To solve for f, divide both sides by 2 pi. So f is 9.2 divided by 2 pi. So the frequency is about 1.464 hertz. Now that we have the frequency, we can find the period, which is 1 divided by the frequency, or 1 divided by 1.464, which is about 0.683 seconds. Now, how can we find the spring constant? What equation would you use to find it? You can use this equation. The spring constant is the mass. times 2 pi f squared. We've established this equation earlier in the video. So k is going to be the mass, which is 0.75 times 2 pi times the frequency 1.464 squared. 2 pi times 1.464 is 9.198, and if you square that, you should get about 84.61 times 0.75. So k is equal to 63.46 newtons per meter. So now, the next thing we need to do is find the total energy, which is really the mechanical energy of the system. And that's 1 half times k times the amplitude squared. So, one half times 63.46 times 0.6 squared. Half of 63.46 is about 31.73. And if you multiply that by 0.6 squared, you should get 11.42 joules. So, that is the total energy of the system. Now what is the potential and the kinetic energy at 0.2 and 0.6? So let's say if this is the spring, and let's say this is its compressed position, let's say this is the equilibrium position. At equilibrium, we know that x is 0. When it's fully stretched or compressed, x is equal to 8, which is about 0.6. When x is 0, we know that the velocity is at its maximum, which means kinetic energy is at its maximum. So at x equals 0, the kinetic energy is equal to the total energy, which is 11.42. And when x is 0, the potential energy is therefore equal to 0. So that's the answer for the first part. So when x is 0, Ke equals Me, that's 11.42 Joules, and Pe is equal to 0. Now what about the other extreme, when x is 0.6? When it's fully compressed or stretched, the kinetic energy will be 0 at this point. But the potential energy will not be zero. It's going to be 11.42 at a displacement of 0.6. So whenever it's fully stretched or compressed, it's not moving. The velocity is zero. And all of the energy is stored in the form of potential energy. But at the equilibrium position, it's moving at its greatest speed. There's no restoring force at that point. So all of the energy is in the form of kinetic energy and there's no potential energy. Now what about in the middle, somewhere between 0.6, which is at 0.2? In the middle, it's going to have kinetic and potential energy. So we need to use the equations to get that answer. Now there's two ways you can do this. You can do it the fast way or the long way. But it's good to understand both, so you can fully get the sense of this chapter. So let's do it the fast way first. Let's find the potential energy when it's 0.2 meters from the equilibrium position. So we can use the equation 1 half kx squared. So that's 1 half times the value of k, which... I forgot what that is since I erased the board. But let me recalculate it. So k was 63.46. So let me just write that here, just in case I need that again. And x is 0.2. So therefore... The potential energy at this point is about 1.27 joules. Now, if you know the mechanical energy and the potential energy, you can find the kinetic. The kinetic energy is simply the difference between a mechanical energy and a potential energy. So 11.42 minus 1.27 is about 10.15. So that is the kinetic energy. That's how you can find it the fast way. Now let's use the long way to get the kinetic energy. So before we can find the kinetic energy, we need to use this equation. We've got to find the velocity at point 2. So it's Vmax times the square root of 1 minus x squared divided by a squared. Now we've got to find Vmax first. So Vmax is equal to the square root of k divided by m times a. So that's going to be the square root of... 63.46 divided by the mass which is.75 times the amplitude which is.6. 63.46 divided by.75, that's like 84.61. If you take the square root of that, you should get 9.2 times.6. So the maximum velocity is 5.519. So now we could find the velocity at 0.2. So that's going to be 5.519 square root 1 minus, let's plug in the x value of 0.2. So 0.2 squared divided by 0.6 squared. 0.2 squared divided by 0.6 squared is basically 1 over 9, which is 0.1 repeated. 1 minus that is 0.8 repeated. Take the square root, and then that's 0.9428 multiplied by 5.519. So, V... at this position is about 5.203. It's slightly less than the maximum velocity. So now we can find the kinetic energy. It's 1 half mv squared, so 1 half times the mass, times the velocity squared. 5.203 squared is about 27.07 times 0.75 times 0.5. And you should get... 10.15 joules. So you get the same answer. But as you can see, taking the difference between a mechanical energy and a potential energy was a lot easier. So make sure you know how to find the kinetic energy using both techniques. A 0.55 kilogram block vibrates according to the equation x is equal to 1.5 cosine 12.14. Determine the equations of the velocity and acceleration as a function of time. So let's write the generic equations. So we know the position function is a cosine 2 pi f t. Now, by the way, you can describe the position function in terms of sine as well. It all depends on where the block starts at t equals 0. For this function... If you're using positive cosine, the block has to start from the top if it's a vertical spring. If it's horizontal, it has to start towards the right of the equilibrium position. If it's starting at the equilibrium position, you need to use the sine function. Now, the velocity function, we said... It was negative VO, which is Vmax, times sine 2 pi FT. So notice for the velocity function, we don't need to change the inside part of the sine and cosine function. So 2 pi f t is 12.14. So that's going to be the same for the velocity function. All we need to find is the maximum velocity. So let's go ahead and do that. Now v max is equal to the square root of k divided by m times a. We know that the amplitude is the number in front of cosine, which is 1.5. We know the mass, it's 0.55. The only thing that we're missing is the spring constant K, which we can use this equation, 2 pi f squared times the mass. So we've got to find the frequency. 12.4t is equal to 2 pi ft, which means that 2 pi f is 12.4. So therefore, the frequency... is 12.4 divided by 2 pi. Well, actually, we really don't need to find a frequency. What we could do is simply replace 2 pi f with 12.4, since we have 2 pi f in the equation. So 2 pi f is going to be 12.4, and then we're going to square it and multiply it by 0.55. So the spring constant... is 84.568. So with that information, we can now find Vmax. So that's going to be the square root of K, which is 84.568, divided by M, which is 0.55, times the amplitude of 1.5. So 84.568 divided by 0.55, that's 153.76. And the square root of that is 12.4 times 1.5 therefore the maximum velocity is 18.6 meters per second so the equation of the velocity as a function of time it's negative 18.6 sine and this part is just 12.4 so sine 12.4t Now let's find the acceleration function. So we gotta find the maximum acceleration first. And the maximum acceleration is equal to k times a divided by m. So k is 84.568. The amplitude is 1.5 and the mass is 0.55. Therefore, the maximum acceleration is 230.64. The equation for the acceleration as a function of time is equal to the negative Maximum acceleration times cosine 2 pi ft. So that's going to be negative 230.64 cosine 12.4t. So if you have the position function, it's not that difficult to write the velocity function and the acceleration function. You need to know the generic equation. This part is going to stay the same. And you need to know the equations to find the maximum velocity and the maximum acceleration. So now let's work on the next part. Let's find the velocity 0.5 meters from its equilibrium position. Using the equation that we have, V of t is equal to negative 18.6 sine 12.4t. So we're looking for V of 0.5. So that's negative 18.6 sine 12.4 times 0.5. That's going to be 6.2. So make sure the calculator is in radian mode for this to work. sine of 6.2 in radian mode is about negative 0.083 and if you multiply that by negative 18.6 The velocity at this point is 1.545 meters per second. So now let's find the acceleration. So let's write the function first. It's negative 230.64 cosine 12.4t. So this is going to be cosine 6.2, just like... sine 6.2. Cosine 6.2 is about 0.9965. Multiply by negative 230.64. The acceleration at this point is negative 229.84 meters per second squared. Now that we have the acceleration, we can calculate the restoring force. So the restoring force is simply going to be the mass times the acceleration. So the mass is 0.55 and the acceleration negative 229.84. So the restoring force at this point is about negative 126.4 newtons. So that is it for this problem. A spring of force constant 300 newtons per meter vibrates with an amplitude of 45 centimeters when a 0.35 kilogram mass hangs from it. What is the equation describing this motion? If the mass passes through the equilibrium point with positive velocity at t equals 0. So we have a vertical spring, we have a mass hanging from it. Now, there's four points at which... There's four ways that the system can start. It can start above the equilibrium position, in which case it has no choice but to go downward. So at t equals 0, for this particular shape, or this particular situation, it has negative velocity because it's going in the negative y direction. So therefore, to graph it, it is starting at the top. with an amplitude of positive a. So this would be the cosine graph, positive cosine to be exact. So x would be equal to positive a cosine 2 pi ft. But we don't have that situation. This is if it's starting above the equilibrium point. and if it's going down with negative velocity at t equals 0. The situation that we have for part A is that it's starting at the equilibrium position, and it's moving with positive velocity, that means it's going this way, in the positive y direction. So to graph it, because it's at the equilibrium point, it starts at the center. And it's going in the positive y direction, so it's going to go up. This graph is positive sine, so the position function is going to equal positive a sine 2 pi ft. So that's what we have for part a. But let's say if part A was different. Let's say if everything was the same, the mass passes through the equilibrium point, but with negative velocity, meaning that it's going in a downward direction. I'm going to graph it in blue. So, in that case, it starts at the equilibrium position that is the origin, but it's going down, and then up. So, the blue sine wave would represent this function. Instead of being positive sine, it's going to be negative. A sine 2 pi FT, if it was going in the negative y direction. So anytime it starts at the equilibrium position, use the sine function. If it goes up, use positive sine. If it's going down from the equilibrium point, use negative sine. So let's get back to the stuff that we're trying to figure out, which is positive A sine 2 pi FT. So we have the amplitude. It's 45 centimeters, which is 0.45 meters. Now what we need to figure out is 2 pi f. We know that k is equal to 2 pi f squared times m. So k divided by m is 2 pi f squared. So 2 pi f is the square root of k over m. K is 300, the mass is 0.35. 300 divided by 0.35 is 857.1. If you take the square root of that, you should get 29.28, or 29.3. Therefore, the position function is going to be A, which is 0.45, times sine. 2 pi f, which is 29.28 times t, and this is positive 0.45. Now, what about the second part of the question? What if the mass starts at the lowest point below the equilibrium point? So let's say this is the equilibrium point. If it starts below it, It has no choice but to go up. It can't go down if it's at the lowest point. So it has to be moving up with positive velocity. So if we graph in, we're starting at negative a. So if you don't start at the center, it's going to be the cosine function. Sine always starts at the center. Since we're starting at the bottom, this is negative cosine. So x is negative a cosine 2 pi ft. So the equation that describes this motion is negative 0.45 cosine 29.28 times t. So that's the answer for part b. So let's review. There's four functions that you need to know. So it can start at the equilibrium position and go up with positive velocity. Or. It can go down with negative velocity, starting from the equilibrium position. Or it can start from the top, or it can start from the bottom. Anytime it starts from the center, you have the positive sine function if it's going up. If it starts from the center and goes down, it's negative sine. If it starts at the top, it's positive cosine. If it starts at the bottom, negative cosine. And remember, the origin represents the equilibrium position. Now let's talk about damp harmonic motion. This form of motion occurs when friction is present. But let's draw two graphs, when friction is not present and when it is present. So let's say this is positive A and negative A. And let's say the graph starts at the top. The mass is pulled to the right and then released from rest. So if there's no friction, the amplitude will be constant. The energy in the system remains the same. Notice that the amplitude is not decreasing. That is the maximum amplitude, or the maximum displacement. Now what if friction is present? How will that change? the appearance of the graph. So if friction is present, it's going to lose energy. And so the amplitude is going to decrease over time until eventually it reaches zero. In this case, dampen has occurred. So dampen is caused by Friction. It could be internal friction or friction due to air molecules, but it reduces the energy of the system, and this is seen by a decrease in amplitude over time. Now, there's three different types of dampen that you need to be familiar with. There's over dampen, under dampen, and critical dampen. When the system is under damp, The object will make a few swings before it comes to rest. So it's going to make a few cycles, and eventually it's going to become zero. Now, if it's over damp, it won't get the chance to make multiple cycles. Within its first cycle, it's going to lose all of its energy and go to zero. But it's going to take a long time. for it to do that. And then there's critical dampen, where it reaches zero faster, and it doesn't make multiple cycles. Critical dampen is useful for the shock absorbers in a car. You don't want your car bouncing up and down every time it hits a bump. You want it to absorb that energy fast, and so you want critical dampen. In a situation where you have an underdamp system, let's say like a door that's old, it can bounce back and forth before it closes. And you don't want that. So whenever you want to reduce bouncing in a device, you want critical damping. You want to remove the energy as quickly as possible from the system. Now there's one other topic that we need to discuss and that is resonant frequency. When you apply a force to an oscillator, you can cause forced vibrations to occur. And as you force the oscillator to move, you can increase the amplitude. However, if you apply the force at the natural frequency of the object, it's going to be most effective when increasing the amplitude. So if the applied frequency matches the natural frequency or if it's very close to it, the amplitude can increase greatly. When the applied frequency matches the natural frequency, it's known as the resonant frequency. A good example to illustrate this, is a swing. Imagine if you have a child on a swing and to increase the amplitude of the swing you need to push the child at the right position and that is while the child is swinging back towards you. That's when you apply the push and at that point the child can go higher and higher. However, let's say if you try to push the child at the middle at the equal position. you probably will not get the greatest effect in increasing the amplitude. You want to apply the force at a rate that's equal to the natural frequency. If you do that, then the applied force is at the resonant frequency and that's when you can achieve maximum amplitude. That's when you can increase the amplitude of the oscillator most effectively. So just make sure that... You keep this fact in mind. Whenever the applied force matches the natural frequency of the system, that is the resonant frequency, the applied force will be most effective in increasing the amplitude of the oscillator.

Let's talk about the mass spring system. Let's say this is the wall and we have a spring connected to a mass horizontally. And let's say that this is the equilibrium position of the spring. So the spring has been stretched towards the right.

So that means we're applying a force to pull it towards the right to stretch it from its equilibrium position. As we pull it towards the right, There is another force that wants to push it back towards its equilibrium position. This is known as the restoring force.

Now, once you let go, the spring is going to move back towards the left. Now, it's not going to stop at the equilibrium position. It's going to move past it.

And once that happens, it's going to bounce back the other direction. So it's going to move back and forth, towards the left and towards the right. And as it oscillates back and forth, you have simple harmonic motion, or periodic motion.

Now according to Hooke's Law, the restoring force...... equal to negative times K times X. X is the distance measured from the equilibrium position. So let's say this is the wall and we have a spring and here's the block of mass M. The distance between where the spring is located and the equilibrium position is x.

At the middle, at the equilibrium position, x is 0. So the restoring force will always be 0 at the equilibrium position. K is known as the spring constant, and K has the units newtons per meter. So let's say if you have a spring constant of 100 newtons per meter. What that means is that it requires 100 newtons of force to either stretch or compress the spring by 1 meter from its equivalent position.

So if you wish to stretch it by 2 meters, it requires 200 newtons of force. So if you have a very stiff spring that's hard to compress, the K constant is very high. Let's say if K is 1000. So for this spring, it requires 1000 N of force just to stretch it by 1 meter. So this spring is going to be very stiff. It's hard to stretch or compress.

This particular spring is going to be easy to stretch or compress compared to the other one. So this spring is relatively loose. By the way, before we work on some practice problems, I do want to mention one thing.

In the equation in Hooke's Law, why is there a negative symbol for the restoring force? Why do we have this negative sign? What's the reason for that?

So notice that if you apply a force towards the right to stretch the spring, the displacement vector will be towards the right. And the restoring force is towards the left. Because the restoring force is opposite to the displacement vector, FR is negative.

Because it's always opposite to the direction of motion. the restoring force will always act in such a way to bring the object back to its equaling position. So if the equaling position is here, the restoring force wants to bring it to the left.

Likewise, let's say if the spring is compressed. The restoring force wants to move it towards the right, back to its equal position. And remember, at the equal position, the restoring force is zero.

since x is equal to 0 and FR is negative kx so negative k times 0 is 0. Let's work on some practice problems. How much force is required to stretch a 300 Newton per meter spring by 25 centimeters? So let's use the equation F is equal to kx.

And let's not worry about the negative sign. If you see the unit Newton's Perimeter, that tells you that that value represents the spring constant, which is... 300 newtons per meter and since it has the unit meters in it that tells us that we need to convert centimeters into meters it turns out that 100 centimeters is equal to a single meter so to convert it needs to divide by 100 Or, move the decimal two units to the left. So this is equal to 0.25 meters, which is the value of x.

So what is 300 times 0.25? Or, what is 25% of 300? This is going to be about 75. So that is the force required to stretch the spring by 25 centimeters.

How far can you compress the spring with a force of 150 newtons? Now let's think about this. It required 75 newtons to stretch the spring by 25 centimeters. It also requires 75 newtons to compress the spring by 25 centimeters. So if we double the force to 150, the amount that we can stretch or compress the spring is going to be twice the value.

It's going to be 50 centimeters. F equals kx. If you double the distance, if you stretch it by 2 times the amount, the force will increase by a factor of 2. Likewise, if you double the force, you can double the distance that you stretch or compress the spring by. Now, for those of you who prefer to use an equation, you can use this. Let's divide F2 by F1.

F1 is equal to K times X1, and F2 is K times X2. Since we're dealing with the same spring, K is the same. so as you can see the force is proportional to x let's say f1 is 75 minutes and we're looking for f2 X1, the distance that corresponds to 75, is 25 centimeters. Actually, we're not looking for F2, we have F2.

F2 is 150. We're looking for X2. 150 divided by 75 is 2. And at this point, we can cross multiply. So, x2 is equal to 2 times 25, which is 50. Here's another problem.

It takes 200 newtons to stretch a spring by 40 centimeters. How much force is required to stretch a spring by 120 centimeters? So 40 centimeters corresponds to a force of 200 newtons.

What is the force that's required to triple the distance by a factor of 3? If you triple the distance by a factor of 3, the force is going to increase by a factor of 3. So the force that's required is 600 N. Again, you can use this equation, F2 over F1 is equal to X2 divided by X1.

So we're looking for F2 this time. F1 is 200 N. X1 is 40 centimeters, and X2 is 120 centimeters.

You don't need to convert the units to meters, because the unit centimeters will cancel. So 120 divided by 40 is 3. So therefore, F2 divided by 200 is equal to 3. And if we multiply both sides by 200... We can see that F2 is going to be 3 times 200, which is 600 newtons. So that's how much force is required to stretch a spring by 120 centimeters. Now what about the second part to the question?

What is the value of the spring constant? So using the formula F is equal to kx. Solving for k, k is equal to the force divided by the spring constant. So we can use the force of 200 newtons, and instead of using 40 centimeters, we need to convert it to meters. So that's about 0.4 meters.

200. divided by 0.4 is equal to 500. So the spring constant is 500 newtons per meter. So to stretch the spring by a distance of one meter requires a force of 500 newtons. Now what about the last part of the problem?

How much work is required to stretch a spring by 120 centimeters? Work is equal to force times displacement, which displacement could be represented as x. Now this is if you have a constant force. Now what about if the force is a variable force?

For example, in this case, work is equal to the integral of f times dx, where f is a function of x. And we know f is equal to k times x, ignoring the negative sign. So work is equal to the antiderivative of k times x dx. To find the antiderivative of kx, add 1 to the exponent and divide by that number. The antiderivative of x to the first power is x squared divided by 2. So the work required to stretch a spring is going to be basically 1 half kx squared.

This is also the potential energy that is stored in a spring. In some textbooks you'll see this U symbol. This is the elastic potential energy of a spring. It's equal to 1 half k x squared.

Which is the work required to stretch it or compress it. So k, we know it's 500 newtons per meter. I'm going to write it like this.

And x... It's going to be 120 centimeters, but we need to convert it to meters, so let's divide it by 100, which is 1.2 meters squared. And the unit meters will cancel, which leaves behind newtons times meters. There's 2 meters here, since it's squared. Newtons times meters is equal to the unit joules.

Half of 500 is 250 times 1.2 squared is equal to 360. So it takes 360 joules of work to stretch this particular spring by 120 centimeters. Now let's go back to our mass spring system slash oscillator. So let's say if we have a mass of M attached to a spring, and let's say this is the equilibrium position.

Now, as we mentioned before, the force required to stretch the spring is the applied force. And whenever you stretch it, there's going to be a restoring force that pulls it back towards the left. Now, you also need to know that at this point, the net force is zero. Now, this is only true... while you're holding it.

As soon as you release the mass from rest, the applied force will disappear. At that point the net force is equal to the restoring force. And at this instant, as soon as you release it, this object has maximum acceleration.

And initially the velocity is zero. The velocity reaches its maximum value. once it reaches equilibrium. At equilibrium the restoring force is zero and because the restoring force is zero the acceleration is equal to zero at the equilibrium position. However because the velocity is at a maximum it's moving towards the left.

So because it has momentum because it's moving due to inertia it's going to continue to move to the left compressing the spring. As it begins to compress the spring a restoring force is going to build up, decelerating the object. Once the object reaches a velocity of zero, the restoring force is at its maximum value, which means acceleration is at its maximum value at this point as well.

So make sure you understand this, that at the equilibrium position, the velocity is at its maximum, and the acceleration is zero. When it's fully stretched or compressed, the acceleration is at its maximum value, and the velocity is zero. Now, if no friction is present in the system, this oscillator will continue to oscillate indefinitely.

If there's friction, friction is going to oppose motion. So, in the first picture, the object is moving towards the left. So, the force of friction will be directed towards the right.

Now, in the middle, it's still moving towards the left, so friction is still directed towards the right. And, on the last picture, because the velocity is 0, friction will be 0. since it's not moving instantaneously. But as it begins to move towards the right, friction will oppose it. So friction always opposes motion. By the way, once you release it, there's no friction until it begins moving towards the left.

So as it accelerates, friction is going to oppose it. Whenever there's friction present, the oscillator will be dampened by it. The amplitude will... gradually decreased. Now let's talk about X and A.

At the equilibrium position, let me draw a new picture. So let's say if we have an oscillator, and let's say this is the equilibrium position. X is equal to 0 at the equilibrium position.

And let's say at this point x is equal to 1. So this is the maximum displacement. The maximum displacement is represented by the symbol capital A. So in this case, for this problem, A is 1. At this position, x is 0.5, since it's to the right of the equilibrium position.

And at this position... x is negative 0.5. So let's say k is 100 newtons per meter.

What is the force, the restoring force, at let's say a position of 0.5? Include the appropriate sign as well. So f is equal to kx. It's going to be negative times 100 times 0.5, which is negative 50. So at this point, we have a force of negative 50. The negative sign means that it's towards the left. Now what about at this position?

If we plug in negative 0.5, okay, that does not look like a 5, it's going to be negative 100 times negative 0.5, which is positive 50. Since the object is moving at this location towards the left, the restoring force is towards the right. And notice that the force is positive 50 because it's directed towards the right. When the restoring force is directed towards the left, Notice that it's negative now.

Let's talk about the energy of this particular oscillator when it's fully stretched and Once you release it The acceleration is at its maximum and the velocity is zero. The kinetic energy of any moving object is one half mv squared. So the kinetic energy is dependent on the speed. So when it's fully stretched or fully compressed, let's say it's fully compressed at this position here, the kinetic energy is at a minimum because the velocity is zero.

So, Ke is going to equal 0 when it's fully stretched or compressed. Now, at the equivalent position, the velocity is at a max. So, therefore, Ke is at its maximum value at the equivalent position.

As we mentioned before, the acceleration is 0 at the equivalent position, and the restoring force is equal to 0 since x is 0. And f equals kx, so k times 0 is 0. Now what about the potential energy? The potential energy is at its maximum when it's fully stretched or compressed. But at equilibrium the potential energy is zero. The potential energy is equal to one-half kx squared. Since x is 0 at the equilibrium position, the potential energy is 0 because 1 half k times 0 squared is 0. Now, there's another thing that you need to be familiar with, and that is mechanical energy.

The mechanical energy is the sum of the kinetic energy and the potential energy. Keep in line. When the amplitude is 1, x can vary anywhere between negative 1 to 1. The mechanical energy depends on the amplitude, that is the maximum value of x, or the maximum displacement.

So the mechanical energy is 1 half k a squared. The kinetic energy is 1 half mv squared. And the potential energy is 1 half k x squared.

So, whenever x is equal to a, the potential energy is the same as the mechanical energy. At this point, the kinetic energy is 0. And whenever x is equal to 0, that is at the equilibrium position, potential energy is equal to 0. So let me organize this information. So let's say this is the equilibrium position where x is 0. At that point, the potential energy will be equal to 0. So Ke is at its maximum.

Ke equals the mechanical energy. The mechanical energy of the system is constant if there's no friction. The mechanical energy is the total energy of the system.

Now let's say if it's at the right, if it's fully stretched. At this point, the maximum displacement is equal to A. So when X is equal to A, the potential energy is at a maximum. It's equal to the mechanical energy, which means that Ke is equal to 0. And when it's fully compressed... PE is still equal to the mechanical energy, and KE is equal to 0. Now, sometimes you may need to calculate the maximum acceleration and the maximum velocity.

What equations can we use to do that? So, let's start with maximum velocity. Now, we said that...

It's going to be, the velocity is going to be at its maximum when x is equal to 0. So when x is equal to 0, the potential energy is 0, so this disappears. So therefore, 1 half kA squared is equal to 1 half mv squared. We can multiply both sides by 2 to get rid of the fraction.

So kA squared is equal to mv squared. To isolate V squared, we need to divide both sides by m. So V is equal to the square root of k divided by m times the amplitude, a.

So that's how you could find the maximum velocity if you ever need to. What about the acceleration? How can we find the maximum acceleration?

We know that F is equal to kx. And according to Newton's second law, F is ma. So what we need to do is divide both sides by m.

So k divided by m times x is equal to the acceleration at any point. So that's the acceleration as a function of x. Now, the acceleration is at its maximum value when x is the maximum, which is the amplitude. So the maximum acceleration is k times a over m. Now what if we don't want to find the maximum velocity?

What if we need to find the velocity as a function of x? Let's say when x is 0.4 or 0.7. How can we do that? Well, let's go back to our conservation of energy equation.

Mechanical energy is equal to kinetic plus potential energy. So, let's begin by multiplying everything by 2. 2 times 1 half is 1. This will cancel all of the fractions. So, we have k a squared is equal to m v squared plus k x squared. Now, let's subtract k x squared by both sides. k a squared minus k x squared is equal to m v squared.

And now let's factor out the GCF, the greatest common factor, which is K. So this is going to be A squared minus X squared. Next, let's divide the expression by the mass. So V squared is equal to K divided by M times A squared minus X squared.

Now, what's our next step? The next thing we need to do is take out a. I mean, a squared. If we take out a squared, a squared divided by a squared is simply 1. And this, divided by a squared, that's going to be negative x squared divided by a squared.

So now, what we want to do is take the square root of both sides. So the velocity as a function of x is going to be the square root of k divided by m times a times the square root of 1 minus x squared divided by a squared. Now the velocity can be positive or negative.

If the velocity is moving towards the right, it's positive. If it's moving towards the left, it's negative. Now if you recall, the quantity...

Square root k over m times a is equal to the maximum velocity. So therefore, we can rewrite the equation like this. So the velocity as a function of x is equal to plus or minus the maximum velocity times the square root of 1 minus x squared divided by a squared.

So make sure you write this equation down. It might be useful later. Now there's something else that we need to talk about, and that is the frequency and period of this spring mass oscillator.

So let's say this is the equilibrium position. moves to the left and as it moves back to where it started from, that is equal to one cycle. The distance of one cycle is equal to basically four times the value of x.

So from here, back to its equaling position, it travels a distance of x. And then when it's fully compressed, it travels another distance of x. And then when it goes back to its equaling position, that's another x value.

And when it returns to where it started, that's x again. So the distance of one cycle... is four times its maximum displacement. So we should really say this is four times A, since A represents the maximum displacement. So this is A.

Now what about the period and the frequency? The period is the time it takes to complete one cycle. The period, which is represented by capital T, is the number of, is the time, divided by the number of cycles.

So let's say if it takes 30 seconds to complete 10 cycles, 30 divided by 10 is 3. That means that it takes 3 seconds to complete a single cycle. Therefore, the period is 3. So that's how you could find a period. It's the total time divided by the total number of cycles, or the time it takes to complete one single cycle. The frequency is the number of cycles per second.

So if you take the total number of cycles and divide it by the total time, you can get the frequency. So for example, let's say if it takes, let me choose a nice number. Let's say if this system can cycle 200 times.

in 10 seconds. So if it makes 200 cycles in 10 seconds, that means that it's doing 20 cycles in 1 second. So the frequency is 20 or 20 Hertz. Hertz is the unit of frequency.

Now frequency and period are inversely related. The frequency is 1 divided by the period and the period is 1 divided by the frequency. So let's work on some problems.

What happens to the following if the maximum displacement of the spring is doubled? So what happens to the total energy of the system? The total energy of the system is the mechanical energy. Now the first thing we want to do is write an equation that relates the mechanical energy with the maximum displacement.

And that's this equation. mE is equal to 1 half kA squared, where A is the maximum displacement. So if we double the value of a, what happens to m8?

For these types of problems, anything that's constant or remains the same, replace it with a 1. So 1 half k, we're just going to replace it with 1. And plug in the value of a. If you double the value of a, because the function is squared, the mechanical energy will increase by a factor of 4. If we were to triple the value of a, the mechanical energy will increase by 3 squared, which is 9. If we quadruple the value of A, the mechanical energy will increase by a factor of 16. Now what about part B, the maximum velocity? What equation can we write that relates maximum velocity to the maximum displacement? Now we came up with the equation earlier. We said that the maximum velocity is equal to the square root of k divided by m, where k is the spring constant, times the maximum displacement.

So notice that it's raised to the first power. So if we double the value of a, and this is going to stay the same, so we're going to replace it with 1, the velocity will increase by a factor of 2. If you triple the displacement, the velocity will increase by a factor of 3. If you cut the displacement in half, the velocity will decrease by a factor of 2, or it's going to be 1 half of its original value. Now what about part C, the maximum acceleration? The equation that we wrote for that is this equation. The maximum acceleration is equal to k divided by m times the maximum displacement.

So if you double the value of the maximum displacement, the acceleration will increase by a factor of 2. If you triple it, the acceleration will triple. If you reduce it by one half... the acceleration will reduce by a factor of two.

Let's work on this problem. A massless spring is pulled by 30 centimeters from its natural length and then released. The spring is attached to a 0.4 kilogram block and oscillates across a horizontal frictionless surface. Let's begin by calculating the maximum velocity. So if you want to you can draw a picture.

So here's the block and it rests across the horizontal surface. Let's say this is the equilibrium position. So right now it's pulled by 30 centimeters which is 0.3 meters. So let's make a list of what we know.

A, the maximum displacement is 0.3. And K, the spring constant, is 300 newtons per meter. What equation do we need to calculate the maximum velocity? The maximum velocity is simply the square root of K divided by m times A. The mass...

is 0.4 kilograms so we have everything we need to find the maximum velocity so this is going to be square root 300 divided by 0.4 times 0.3 300 divided by 0.4 is 750. The square root of 750 is 27.386 times 0.3. And you should get a maximum velocity of 8.216 meters per second. So now, what about part B? What's the maximum acceleration? So let's use this equation.

It's equal to k times a divided by m. So it's going to be 300 times 0.3 divided by 0.4. 0.3 divided by 0.4 is basically 0.75 times 300. That's 225. So the maximum acceleration is 225 meters per second squared.

Now what about part C? How can we find the velocity at a position of 20 centimeters from its natural length? So keep in mind, the maximum velocity was about 8.2 something.

8.216 We can use this equation. The velocity as a function of x is equal to the maximum velocity times the square root 1 minus x squared divided by a squared. That's the equation that we had earlier in this video. So we want to find the velocity at a position of 20 centimeters from its natural length.

Now, because we have a ratio between x and a, it doesn't really matter if you use centimeters or meters. The units simply have to match, because they're both going to cancel. Centimeters over centimeters and meters over meters will cancel.

So we can plug in 20 centimeters if we want to. So the maximum velocity is 8.216, and then this is going to be 1 minus 20 squared. divided by 30 squared.

20 squared divided by 30 squared is about 4 over 9, which is 0.4 repeating. And 1 minus that number is about 0.5 repeating. And the square root of that is 0.7454 times 8.216. So you should get a velocity of about... Let's make some space.

6.124. So your answer should be less than the maximum velocity. If for some reason it's larger, you know it's not correct.

Now what about the next part? How can we find the restoring force 20 centimeters from its natural length? Now let's say 20 centimeters is located at this point.

We know the restoring force is directed towards the left, which means that it should be negative. So the restoring force is equal to negative times kx. So this is going to be negative 300 newtons per meter times x, which is 0.2 meters.

Now in this case, we need to convert centimeters to meters because we're using k, and k is in meters. 3 times 2 is 6, so 300 times 0.2 must be 60. So the restoring force is negative 60 newtons. It's negative because it's directed towards the left.

Now what about the acceleration? Well, we know that the force is equal to mass times acceleration. So the acceleration is the restoring force divided by m. And the restoring force is negative 60 newtons. The mass is 0.4 kilograms.

So negative 60 divided by 0.4 is negative 150 meters per second squared. Keep in mind, the maximum acceleration was 225. And this is less than 225, so that works out. Now, you can also use this equation.

Acceleration is equal to k over m times x. It's similar to this equation. The maximum acceleration is k times a divided by m.

But here, it's k times x over m. So, k is 300, x is 0.20, and m is 0.4. So, if you use this equation, you should get the same answer. which is 150. Now we know it should be negative 150, because the resultant, I mean the restoring force is towards the left.

Acceleration and the force that causes that acceleration are usually in the same direction. So because the restoring force is negative, the acceleration is negative, which really means that this equation should have a negative sign. So let's go ahead and add it.

So that is it for this problem. In this problem, we have a vertical spring. So let's say this is the spring without a mass attached to it. Now once we add a mass to it, it's going to stretch. Now it stretches by 0.4 meters.

So this is the new equilibrium position of the mass. Now, if we stretch it even further by an additional 0.20 meters from its new equilibrium position, once we release it, it's going to bounce up and down. So with this information, determine the spring constant, the amplitude, and everything else.

So let's focus on the spring constant. We can use this picture to find the value of the spring constant. The applied force that stretches the spring from its original position is basically the weight force of the object, which is mg.

The spring wants to go back to its original length. So the restoring force is equal to the weight force, so that the net force is zero, creating that new equilibrium position. So FR is equal to MG.

And the restoring force is negative KX. and G is negative 9.8 so you really don't need to worry about the negative sign they will cancel now let's solve for K so K is going to be equal to MG divided by X so the mass is 2 kilograms G is 9.8 and it stretches by distance of 0.4 meters So 2 times 9.8 is 19.6 divided by 0.4 is 49. So K is 49 newtons per meter. Now what about part B?

What is the amplitude in this problem? What do you think the amplitude is equal to? Now remember, this is the new equilibrium position, and it stretched 0.20 meters from that equilibrium position.

So therefore, the amplitude is 0.20. So once we release it from this position, it's going to bounce up and down.20 from this position. So it's going to go up.20 and down.20 meters. Now what about C? How can we calculate the maximum acceleration?

What is the equation that we need? If you recall, the maximum acceleration is equal to the spring constant times the amplitude divided by the mass. So it's 49 times 0.2 divided by the mass of 2. So the maximum acceleration is 4.9 meters per second squared. Now what about d, the maximum velocity? The equation that we need for that is the square root of k divided by m times a.

49 divided by 2 is 24.5. And the square root of that... is about 4.95 times.2 that's about.99 meters per second.

So now what about part E? What is the kinetic energy, potential energy, and mechanical energy when it's.10 meters from its equilibrium position? So let's find the potential energy first.

The potential energy is 1 half kx squared, which is basically 1 half times 49 times an x value of 0.10. 49 times 0.10 squared, that's about 0.49 times half, that's 0.25. So the potential energy is 0.245 joules. Now what about the kinetic energy? The kinetic energy is equal to 1 half mv squared.

Now we can't use the maximum velocity. The velocity will be 0.99, it will be at its max when x is 0 at the equilibrium position, but we're not at the equilibrium position, x is 0.10. So we've got to find the velocity first at 0.10. The velocity as a function of x is the maximum velocity times the square root of 1 minus x squared divided by a squared.

So the maximum velocity is about 0.99, and x is 0.1, a, the amplitude, is 0.2. 0.1 squared divided by 0.2 squared is 1 fourth. 1 minus 1 fourth is 3 fourths, and the square root of 3 fourths is about 0.866 times 0.99.

You should get 0.8574. meters per second. Now that we have the velocity, we can now use this equation to find the kinetic energy. So let's plug it into that equation.

So Ke is equal to one half times the mass, which is two kilograms, times the speed, 0.8574 squared. So half of two is one. And 1 times 0.8574 squared is about 0.735 joules.

So that's the kinetic energy, 0.735. Now let's calculate the mechanical energy. We're going to do it two ways. So first let's use the equation mechanical energy is equal to 1 half.

K times the amplitude squared. So K is 49 and the amplitude is 0.20 meters or 0.2 meters. 0.2 squared is 1 over 25 or 0.04 times 49 which is 1.96. Half of that is 0.98.

So the mechanical energy which represents the total energy of the system is 0.98 joules. Another way in which we could find it is by adding the potential energy and the kinetic energy. So if we add 0.245 plus 0.735, it also gives us the same answer of 0.98 joules. So let's say if you didn't know the equation to find the velocity at any point x. You can use this to get the mechanical energy.

And if you subtract the mechanical energy by the potential, that's one way you can get the kinetic energy. So there's different ways of solving different things that you may need. So you're not faced with one option.

There's many options that you can use. Try this problem. A spring is compressed 0.35 meters with a 0.25 kilogram block.

by an applied force of 500 Newtons. What speed will the block have as soon as it's released from the spring across a horizontal, frictionless surface? So let's say this is the wall, and we have a compressed spring and the mass attached to it.

And let's say this is the horizontal... frictionless surface. Now it's compressed by an applied force of 500 newtons and it's compressed by a distance let's say of 0.35 meters and the mass of the block is 0.25.

Right now The applied force is balanced by a restoring force of 500 N. So as soon as we release it, the applied force will disappear, and it's going to accelerate towards the right due to this restoring force of 500. Now, this restoring force is not constant. As the spring expands, the restoring force will decrease eventually to zero. At that point, the mass will be released from the spring and it's going to move to the right at constant speed.

So, when it's completely released from the spring, when it's no longer attached to the spring, what is the speed that the block will have? How can we calculate that value? Take a minute and work on that example.

The first thing that we need to do is find the spring constant, K. So F is equal to Kx, therefore K is the force divided by x. So we have an applied force of 500 newtons, which compresses the spring by 0.35 meters.

So if we divide these two, we'll see that the spring constant is about 1429. Round it to the nearest whole number, newtons per meter. So now that we have the spring constant, we can use conservation of energy to calculate how fast it's going to move. So the mechanical energy is equal to the kinetic energy.

Or you could say the potential energy stored in the spring is equal to the kinetic. The mechanical energy is 1 half kA squared. If you use the potential energy equation, it's 1 half kX squared, but in this case, X is equal to A because it's stretched.

I mean, it's compressed all the way. X is basically at its maximum displacement when it's 0.35. So we'll use the symbol A, though.

1 half KA squared is equal to 1 half MV squared. So if we multiply both sides by 2, we can get rid of the 1 half. So k is 1429, a is 0.35, m is 0.25, and we need to solve for v. 0.35 squared times 1429. that's about 175 and if we divide that by.25 we see that v squared is equal to about 700.21 and now if we take the square root of both sides the velocity is about 26.5 meters per second so that's how fast it's going to be moving By the way, let's say if we wanted to find the kinetic energy of the object as soon as it's released.

Using this equation, it's 1 half times the mass, which is 0.25, times the speed, which is really 26.46 squared. The kinetic energy of the object is about 87.53 joules. Now another equation that you can use to get that answer is the work equation.

So keep in mind, as soon as we release it, there's a restoring force that's going to accelerate it towards the right. Work is equal to force times distance. however we don't have a constant force we have a variable force now let's assume that this force decreases constantly well it actually does so we don't really have to assume the force will decrease at a constant rate from its maximum value of 500 to 0 as the block is displaced by 0.35 meters. So basically, the energy transferred by this force should be equal to the kinetic energy, which is 87.5.

So how can we find the area of that triangle? Well area is basically 1 half base times height or 1 half times the height which is 500 that's basically the restoring force times the base which is the displacement of 0.35. Half of 500 is 250 and 250 times 0.35 gives you 87.5 joules.

So that's another way in which you can calculate the kinetic energy gained by the block, is by using the work equation. But if you're going to use the restoring force, you need to add a 1 half to it, because the restoring force is not constant. Now there are some other topics that you need to know, and that is the equations for the period and the frequency as it is related to the mass and the spring constant of a spring. Now we need to compare circular motion with the simple harmonic motion of a spring. So let's draw a circle.

Now, let's say if an object on a circle moves from, let's say, point A to point B. As it moves this way, if you can view the circle... horizontally from this perspective rather than viewing it from top to bottom if you can just have like a side view all you will see is that this particle is moving in the negative x direction you're only going to see the displacement along the x-axis Now, as it moves from, let's say, position B to position C, if you take a side view of it, you're going to see it's still moving towards the left. Now, as it moves from C to D, it's going to appear as if it's moving towards the right, which it is.

And from D to A. it's moving this way. So if you can view it from the side, the circular motion would seem similar to the oscillation of a spring as it moves back and forth.

So knowing that, we can say that the speed of the object around a circle is basically the distance that it travels divided by the time. And the distance around a circle is the circumference, which is 2 pi r. The time it takes to make one revolution or one cycle is the period, which is represented by capital T.

Now, notice that the radius of the circle is basically the amplitude. So, we can replace r with a. Now, we no longer need this picture.

So now at this point, let's multiply both sides by t. So vt is equal to 2 pi a, and now let's divide both sides by v. So the period is 2 pi times the amplitude divided by the velocity. And if you recall, the maximum velocity we said was equal to... the square root of k divided by m times a. So to solve for a, we can multiply both sides by the square root of m divided by k so that m will cancel on the right and k will cancel on the right as well.

So v times the square root of m over k is equal to a. So now let's replace this a with this expression. So the period T is 2 pi divided by V times V max times the square root of m over k.

So we can cancel V. So therefore, the period... of a spring is 2 pi times the square root of m divided by k. So now let's talk about this equation.

If we increase the mass of the spring, what's going to happen to the period? Well the time it takes to complete one cycle will it increase or decrease? Because the mass is an enumerate of the fraction, the period will increase as well. So let's say if you double the mass of the spring, what effect will it have on the period? So for these questions, plug in the value that is being changed, everything else plug in the 1. So if we double the mass, the period will increase by the square root of 2. Now what if we quadruple the mass?

The period will increase by the square root of 4, which is 2. So if you quadruple the mass, the period will double. If you increase the mass by a factor of 9, the period will triple in value. What if you reduce the mass by 1 half? The square root of 1 half, which is 1 over root 2, that's root 2 over 2. So the period will increase.

Well, actually, it's going to... decreased by root 2 over 2. That's about 0.707. So it's going to be 70% of its value. Or you could say it's reduced by a factor of square root 2, because you're dividing it by root 2. So if you decrease the mass to half of its value, the period is going to be reduced by a factor of square root 2. It's going to be 70% of its original value.

or 70.7. Now what about the spring constant? Let's say if we increase the value of k, what happens to the period?

The period will decrease since k is on the bottom. So if you double the value of k, the period will decrease by the square root of 2. It's going to be 70% or 70.7% of its original value. If you quadruple the value of k, the period is going to be 1 half of its original value.

Now what if you reduce the spring constant by... 1 4th. What effect will that have on the period? 1 divided by 1 4th is 4. So if you decrease k to 1 4th its value, the period will double. Now let's think about what this means.

So we could see from the equation why the period will increase if we increase the mass, but let's understand it. conceptually as you increase the mass of the block is going to have more inertia so it's going to be harder to move as a result it's going to have less acceleration f is equal to ma if the force is constant whenever you increase the mass the acceleration will decrease so if it has a lower acceleration it's going to move more slowly And therefore, an object that moves slowly takes a longer time to get to its destination. Therefore, as you increase the mass, the period increases.

Since the object is heavier, it's going to take a longer time to move back and forth. And so that's why the period increases. Now what about the second situation, where if we increase the value of k, the period decreases?

A large k value means that... The spring requires a greater force to compress or stretch it, so the spring is stiff. Because the spring is stiff, it doesn't really move very far. It doesn't really, it's not easily stretched or compressed, so it can vibrate faster.

As a result, because it can vibrate faster, its period will be reduced. So keep in mind k is f over x. So a large k constant means that a greater force is required to stretch it with the same distance.

And if the force is greater, the acceleration will be greater as well. And a higher acceleration means a fast moving object. It can quickly gain speed.

And if the object is moving faster, then it's going to take a longer time. I mean a shorter time actually, to oscillate back and forth. And so that's why as you increase the value of k, it oscillates faster and the period is reduced.

Now going back to this equation, there's one more equation that you need, and that is frequency. Frequency is 1 divided by period. So the frequency is going to be 1 divided by 2 pi. times the square root of k divided by m. So this time, if you increase the k value, if you increase the spring constant, a greater force is required to stretch or compress the spring, which means that there's going to be a greater acceleration, which means that it's oscillating faster.

And because it's oscillating faster, the frequency will increase. Now if you increase the mass, the object requires a greater acceleration to move it back and forth. But if the force is constant, as you increase the mass, the acceleration will be reduced. And so it's going to be moving slower, so to speak.

So therefore, the frequency will be reduced. Frequency and speed, they're related. An object that can oscillate back and forth at a faster rate will have a higher frequency.

And an object that oscillates at a slower rate will have a low frequency. So as you increase the value of k, the speed of the oscillations will increase, and so the frequency will increase. If you increase the mass, heavy objects tend to move slower, and so the frequency will decrease.

A spring attached to a 0.25 kilogram block is stretched horizontally across a frictionless floor 0.25 meters by an applied force of 200 N. Calculate the frequency and period of the oscillations of this mass spring system. So let's begin by drawing a picture.

So here we have a spring attached to a mass. and here is the floor So as we stretch it with an applied force, once we release it, the spring will oscillate back and forth. So how can we find the frequency of the oscillations? Well, we can use this equation, 1 divided by 2 pi times the square root of k divided by m. Now, in this problem, we don't have the spring constant, k, so we need to find it. But we do have the applied force and the distance that it's stretched by.

So, k is equal to the force divided by x. So, that's 200 newtons divided by 0.25 meters. So, let's go ahead and divide those two numbers.

So, k is 800 newtons per meter. Now that we have the spring constant, we can calculate the frequency. So it's 1 divided by 2 pi times the square root of k, which is 800, divided by the mass, which is 0.25 kilograms. 800 divided by 0.25 is 3200, and the square root of 3200 is about 56.57.

And let's divide that by 2 pi. And you should get a frequency of 9 Hz. Now that we have the frequency, we can easily calculate the period. The period is simply 1 over f, which is basically 1 divided by 9. 1 divided by 9 is about 0.1 repeated. So the period is about 0.11 seconds.

So it takes 0.11 seconds to make a full cycle. That is, it takes 0.11 seconds for the block to travel towards the left and back to its original position. And in one second, the spring will oscillate nine times.

That's the frequency, the number of cycles that occurs in one second. So in one second, it's going to go backwards and forwards nine times in that single second. Here's another problem you can try. When a 70 kilogram person gets on a 1200 kilogram car, the spring is compressed by 2 centimeters, what will be the frequency of vibration when the car hits a bump.

So first, let's find K. So F is equal to Kx. The force that compresses the spring by 2 centimeters is the weight of the person, because once he gets on the car, X changes by 2 centimeters, so we only need to use the weight of that person, and not the weight of the car and the person, to find K.

So K is equal to mg divided by X. So it's the mass of the person times 9.8 divided by 2 centimeters converted to meters. To convert centimeters to meters, divide by 100. So this is 0.02 meters.

So it's 70 times 9.8 divided by that number and the spring constant is 34,300 newtons per meter. Now that we have the spring constant, we can calculate the frequency of vibration. Now when the car hits a bump, you have to consider the mass of the person and the car because the spring... is attached to both, or it's affected by the mass of the person and the car. It has to support the weight of both.

So, using the equation F is equal to 1 over 2 pi times the square root of k divided by m. This is 1 over 2 pi times root 34,300 divided by the total mass. That's 70 plus 1,200, or 1,270. 34,300 divided by 1270 is about 27. And the square root of 27 is about 5.2.

And if you divide that by 2 pi, you should get a frequency of 0.83 hertz. So that's it for this particular problem. So here's another problem. Here we have an insect caught in a spider web. We have the mass of the insect and the frequency and we need to find the spring constant k.

So let's start with the equation f is equal to 1 over 2 pi times the square root of k divided by m. Now let's solve for k. So first let's multiply both sides by 2 pi. So 2 pi f is equal to the square root of k divided by m. Now at this point, we're going to square both sides, so we can get rid of the square root on the right.

So 2 pi f squared is equal to k divided by m. Now the last thing we need to do is multiply both sides by m. So now we have the equation that we need.

So the spring constant K is equal to the mass times 2 pi f squared. And let's assume that the mass of the spider web is negligible. So we're only concerned with the mass of the insect. Now, we need to plug in the mass in kilograms.

To convert grams to kilograms, you can divide by 1,000, or instead of writing 0.25g, you can write 0.25 times 10 to the minus 3 kilograms. 1 gram is about 0.001 kilograms, or 1 times 10 to the minus 3 kilograms. And the frequency is 20 hertz. 2 pi times 20 is about 40 pi or 125.66. If you square it, you should get 15,791.

And then if you multiply it by 0.25 times 10 to the minus 3, this should give you a spring constant of 3.95 newtons per meter. Now that we have the spring constant K, how can we answer the second part of the problem? So, if the mass is now 0.10 grams, what is the frequency? So, using the equation 1 over 2 pi times the square root of K over M, it's going to be different. So, K is about 3.9478.

3.95 is the right answer. and the mass is going to be 0.1 times 10 to the minus 3 kilograms. So 3.9478 divided by 0.1 times 10 to the minus 3, that's 39,478.

If you take the square root of that, you should get 198.69, and then divide that by 2 pi, and this will be equal to 31.6 hertz. Now, you can get the same answer. Even if you don't have the value of K, and let me show you so let's say if we had a ratio F2 divided by F1 F2 is 1 over 2 pi square root K divided by M2 and we're assuming that K is the same since we're dealing with the same spider web and this is going to be same thing but m1.

So we can cancel the 2 pi and we can cancel k. So we get the equation f2 divided by f1 is equal to the square root of 1 over m2 divided by the square root of 1 over m1. So how can we simplify the expression that we have on the right?

What would you do to simplify it? Let's multiply the top and the bottom by the square root of m1. If we do that, m1 will cancel on the bottom. And so we're going to have f2 divided by f1. And if you multiply 1 over m2 times m1, this is just going to be the square root of m1 over m2.

So this is the equation that we can use to get the answer to the second part of the problem if we don't have the spring constant K. So notice that a frequency of 20 Hz corresponds to a mass of 0.25 g. Because we have a ratio of the masses, we don't need to convert it to kilograms. So m1 is going to be 0.25 grams.

Now we're looking for the frequency f2 when the mass is 0.10. 0.25 divided by 0.10 is 2.5. And the square root of 2.5 is about 1.581.

So f2 divided by 20 is equal to... 1.581. To solve for F2, multiply both sides by 20. So 1.581 times 20 is 31.62 Hz, which is about the answer that we got a few minutes ago. A block of mass attached to a spring with a constant of 200 Nm vibrates at 15 Hz. What is the frequency of vibration if the same block is attached to a spring of 500 Nm?

So the mass is constant here. The only thing that changes is the spring constant and f. So we know that f is equal to 1 over 2 pi times the square root of k divided by m. So we can write a ratio, f2 divided by f1 is 1 over 2 pi square root k2 over m. So we're going to write the subscript for k since k changes, but m is constant, so we don't need to write the subscript for m.

So we can cancel the 2 pi and the mass. So we're going to get this equation, f2 divided by f1 is equal to the square root. of k2 divided by k1. So as you increase the spring constant, the force will increase. So, I mean not the force, but the frequency will increase.

So the frequency should be higher than 15 hertz. So f1 is 15, and that corresponds to a k constant of 200. k2 is 500, and we're looking for f2. So if we divide 500 by 200, that's 2.5 and the square root of 2.5 is 1.581 so f2 divided by 15 is that number so to find f2 we need to multiply 1.581 by 15 and you should get a frequency of 23.7 Hertz The block undergoes SHM with amplitude 0.4, that simple harmonic motion.

If you see SHO, that's a simple harmonic oscillator. What is the total distance that it travels in 8 periods? Now, keep this in mind.

The distance that it travels in 1 period, or 1 cycle, is 4 times the amplitude. Let's say this is the equilibrium position. This is the amplitude, the amount that it's fully stretched by.

So for this block to travel one complete cycle, it has to travel A towards the equilibrium position and another distance of A, and then back towards the equilibrium position and back towards where it started. So as you can see, The distance that it travels in one cycle or one period is 4 times 8. So 4 times 0.4 is 1.6 meters. So in one period, it's going to travel a distance of 1.6 meters.

In 8 periods, it's going to be 1.6 times 8, which is about 12.8 meters. So that's the total distance that... the block will travel in eight periods. Now there's some other functions that can describe the motion of a mass spring system.

So let's say if we have a vertical spring and currently it's compressed. Let's say this is the equilibrium position so it's capable of bouncing Up and down so therefore This is going to be equal to the amplitude and this will be negative A. Now, as it bounces up and down, if you plot this on a graph, let's say time is the x-axis and the displacement is on the y-axis. So this is going to be the maximum displacement, and here's negative a. So according to this graph, it's starting at its compressed position.

So that's at the top. If you graph it over time, it's going to form a sinusoidal function. It's basically a sine wave.

Now, because it started at the top, this is the cosine function. If it started at the middle, it would be sine. Now if we draw the circle that we had earlier in this video, this is X and the radius of the circle is A.

On the inside, this is the angle theta, and this is the right triangle. According to Circulator, if you've taken trigonometry at this point, you know that cosine of the angle theta is equal to the adjacent side divided by the hypotenuse. That's the Ka part in SOHCAHTOA. The adjacent side is the side that's very close to the angle. So that's x.

The hypotenuse is across the box. That's a. So cosine theta is x divided by a.

So if you multiply both sides by a, you can get the value of x. So therefore x is equal to a cosine theta. Now we know that linear displacement is equal to velocity multiplied by time, d equals vt. Angular displacement is equal to omega, which is angular velocity multiplied by time. Now, perhaps you have seen this equation as you've studied rotational kinematics.

If you haven't, you can look up my video that I've created on YouTube, you can check that out, and you'll see that equation. Omega which is the angle of velocity, it's equal to 2 pi times the frequency. So therefore, we could say x is equal to a cosine omega times t, since theta equals omega times t, and then we can replace omega with 2 pi f. So x, the position function with respect to time, is equal to the amplitude times cosine. 2 pi f times t.

So make sure that you write this equation down. This is an equation that you need to know. Now if you've taken calculus, you know that the derivative of the position function x gives you the instantaneous velocity.

So let's find the derivative. The derivative of cosine is equal to a negative sine. Now you have to keep the angle.

the same, the 2 pi f t. And then you have to differentiate the nested function, or the inside part of sine, which is 2 pi f t. The derivative of 2 pi f t is simply 2 pi f. t is the variable that you're differentiating with respect to. The derivative of 5x is 5. The derivative of 9x is 9. The derivative of 8t is 8. So the derivative of 2 pi f t is 2 pi f.

By the way, for those of you who haven't taken calculus and who want to understand how to find the derivative of a function to understand a step, you can find my video on derivatives in YouTube. You just gotta search for it. You may have to type in like power rule, product rule, quotient rule, but you only need the power rule for this function.

And also, you need to know how to find the derivative of trigonometric functions as well, which you can also find out on YouTube. So now let's continue. So we need to replace a few things at this point. Let's replace f.

We know that the frequency is 1 divided by 2 pi times the square root of k divided by m. In addition, earlier in this video we mentioned that the amplitude is equal to the maximum velocity. times the square root of m divided by k.

This came from the equation v max is equal to square root k over m times a. So let's replace a with this expression, and let's replace f with that expression. So the velocity as a function of time is going to be a, which is v max, times the square root of m divided by k times negative sine 2 pi ft times 2 pi times f, which is 1 divided by 2 pi, square root k divided by m.

So we can cancel 2 pi, we can cancel k, and we can cancel m. So now we have the velocity function with respect to time. So here it is.

The instantaneous velocity is equal to the maximum velocity. Let's put the negative in front. So negative Vmax times sine 2 pi ft.

And that's it. So make sure that you add this to the list of equations that you need to know. Now instead of writing Vmax, some textbooks will use the symbol VO, but it's the same thing, it's the maximum velocity.

Now let's find the acceleration. The acceleration is the derivative of the velocity function. So Vmax is a constant, we don't have to change that.

The derivative of sine is cosine, and the derivative of 2 pi ft is 2 pi f. Now let's replace Vmax with what we wrote before. We said it's equal to negative square root k over m times a. And the frequency, as we mentioned before, is 1 over 2 pi square root k over m. So we can cancel 2 pi.

k times k is k squared. And m times m is m squared. This time they don't cancel.

So we have k squared over m squared, a cosine 2 pi ft. The square root of k squared is simply k, and the square root of m squared is simply m. So the acceleration as a function of time is negative k over m times a. cosine 2 pi ft.

Now if you recall, f is equal to kx, and the force is basically mass times acceleration. And let's replace x with the amplitude a. So the maximum acceleration is k times the amplitude divided by m, which is exactly what we have here.

So therefore, we could say that the acceleration function with respect to time, is equal to the maximum acceleration, you can say a-naught, or simply just a-max, times cosine 2 pi ft. So t is the variable, f is a constant, and the maximum acceleration is a constant. So here's the third equation that you need.

A 0.75 kilogram mass vibrates according to the equation. x is equal to 0.6 cosine 9.2t. x is in meters, t is in seconds.

So, what is the amplitude? Now, we have the position function with respect to time, and we know the equation is a times cosine 2 pi ft. So, the amplitude is simply the number in front of cosine.

Therefore, the amplitude is 0.6. That's the maximum displacement of the 0.75 kg mass. Now what about part B, the frequency? Notice that the 9.2, the number in front of T, is equal to 2 pi f. So let's set 2 pi f equal to 9.2 and let's solve for f.

To solve for f, divide both sides by 2 pi. So f is 9.2 divided by 2 pi. So the frequency is about 1.464 hertz.

Now that we have the frequency, we can find the period, which is 1 divided by the frequency, or 1 divided by 1.464, which is about 0.683 seconds. Now, how can we find the spring constant? What equation would you use to find it? You can use this equation. The spring constant is the mass. times 2 pi f squared.

We've established this equation earlier in the video. So k is going to be the mass, which is 0.75 times 2 pi times the frequency 1.464 squared. 2 pi times 1.464 is 9.198, and if you square that, you should get about 84.61 times 0.75.

So k is equal to 63.46 newtons per meter. So now, the next thing we need to do is find the total energy, which is really the mechanical energy of the system. And that's 1 half times k times the amplitude squared. So, one half times 63.46 times 0.6 squared.

Half of 63.46 is about 31.73. And if you multiply that by 0.6 squared, you should get 11.42 joules. So, that is the total energy of the system. Now what is the potential and the kinetic energy at 0.2 and 0.6?

So let's say if this is the spring, and let's say this is its compressed position, let's say this is the equilibrium position. At equilibrium, we know that x is 0. When it's fully stretched or compressed, x is equal to 8, which is about 0.6. When x is 0, we know that the velocity is at its maximum, which means kinetic energy is at its maximum. So at x equals 0, the kinetic energy is equal to the total energy, which is 11.42. And when x is 0, the potential energy is therefore equal to 0. So that's the answer for the first part.

So when x is 0, Ke equals Me, that's 11.42 Joules, and Pe is equal to 0. Now what about the other extreme, when x is 0.6? When it's fully compressed or stretched, the kinetic energy will be 0 at this point. But the potential energy will not be zero.

It's going to be 11.42 at a displacement of 0.6. So whenever it's fully stretched or compressed, it's not moving. The velocity is zero.

And all of the energy is stored in the form of potential energy. But at the equilibrium position, it's moving at its greatest speed. There's no restoring force at that point. So all of the energy is in the form of kinetic energy and there's no potential energy. Now what about in the middle, somewhere between 0.6, which is at 0.2?

In the middle, it's going to have kinetic and potential energy. So we need to use the equations to get that answer. Now there's two ways you can do this. You can do it the fast way or the long way. But it's good to understand both, so you can fully get the sense of this chapter.

So let's do it the fast way first. Let's find the potential energy when it's 0.2 meters from the equilibrium position. So we can use the equation 1 half kx squared. So that's 1 half times the value of k, which... I forgot what that is since I erased the board.

But let me recalculate it. So k was 63.46. So let me just write that here, just in case I need that again. And x is 0.2.

So therefore... The potential energy at this point is about 1.27 joules. Now, if you know the mechanical energy and the potential energy, you can find the kinetic. The kinetic energy is simply the difference between a mechanical energy and a potential energy.

So 11.42 minus 1.27 is about 10.15. So that is the kinetic energy. That's how you can find it the fast way.

Now let's use the long way to get the kinetic energy. So before we can find the kinetic energy, we need to use this equation. We've got to find the velocity at point 2. So it's Vmax times the square root of 1 minus x squared divided by a squared.

Now we've got to find Vmax first. So Vmax is equal to the square root of k divided by m times a. So that's going to be the square root of...

63.46 divided by the mass which is.75 times the amplitude which is.6. 63.46 divided by.75, that's like 84.61. If you take the square root of that, you should get 9.2 times.6. So the maximum velocity is 5.519. So now we could find the velocity at 0.2.

So that's going to be 5.519 square root 1 minus, let's plug in the x value of 0.2. So 0.2 squared divided by 0.6 squared. 0.2 squared divided by 0.6 squared is basically 1 over 9, which is 0.1 repeated. 1 minus that is 0.8 repeated.

Take the square root, and then that's 0.9428 multiplied by 5.519. So, V... at this position is about 5.203.

It's slightly less than the maximum velocity. So now we can find the kinetic energy. It's 1 half mv squared, so 1 half times the mass, times the velocity squared.

5.203 squared is about 27.07 times 0.75 times 0.5. And you should get... 10.15 joules.

So you get the same answer. But as you can see, taking the difference between a mechanical energy and a potential energy was a lot easier. So make sure you know how to find the kinetic energy using both techniques.

A 0.55 kilogram block vibrates according to the equation x is equal to 1.5 cosine 12.14. Determine the equations of the velocity and acceleration as a function of time. So let's write the generic equations.

So we know the position function is a cosine 2 pi f t. Now, by the way, you can describe the position function in terms of sine as well. It all depends on where the block starts at t equals 0. For this function... If you're using positive cosine, the block has to start from the top if it's a vertical spring. If it's horizontal, it has to start towards the right of the equilibrium position.

If it's starting at the equilibrium position, you need to use the sine function. Now, the velocity function, we said... It was negative VO, which is Vmax, times sine 2 pi FT.

So notice for the velocity function, we don't need to change the inside part of the sine and cosine function. So 2 pi f t is 12.14. So that's going to be the same for the velocity function.

All we need to find is the maximum velocity. So let's go ahead and do that. Now v max is equal to the square root of k divided by m times a. We know that the amplitude is the number in front of cosine, which is 1.5.

We know the mass, it's 0.55. The only thing that we're missing is the spring constant K, which we can use this equation, 2 pi f squared times the mass. So we've got to find the frequency. 12.4t is equal to 2 pi ft, which means that 2 pi f is 12.4.

So therefore, the frequency... is 12.4 divided by 2 pi. Well, actually, we really don't need to find a frequency. What we could do is simply replace 2 pi f with 12.4, since we have 2 pi f in the equation. So 2 pi f is going to be 12.4, and then we're going to square it and multiply it by 0.55.

So the spring constant... is 84.568. So with that information, we can now find Vmax.

So that's going to be the square root of K, which is 84.568, divided by M, which is 0.55, times the amplitude of 1.5. So 84.568 divided by 0.55, that's 153.76. And the square root of that is 12.4 times 1.5 therefore the maximum velocity is 18.6 meters per second so the equation of the velocity as a function of time it's negative 18.6 sine and this part is just 12.4 so sine 12.4t Now let's find the acceleration function. So we gotta find the maximum acceleration first. And the maximum acceleration is equal to k times a divided by m.

So k is 84.568. The amplitude is 1.5 and the mass is 0.55. Therefore, the maximum acceleration is 230.64.

The equation for the acceleration as a function of time is equal to the negative Maximum acceleration times cosine 2 pi ft. So that's going to be negative 230.64 cosine 12.4t. So if you have the position function, it's not that difficult to write the velocity function and the acceleration function.

You need to know the generic equation. This part is going to stay the same. And you need to know the equations to find the maximum velocity and the maximum acceleration.

So now let's work on the next part. Let's find the velocity 0.5 meters from its equilibrium position. Using the equation that we have, V of t is equal to negative 18.6 sine 12.4t.

So we're looking for V of 0.5. So that's negative 18.6 sine 12.4 times 0.5. That's going to be 6.2. So make sure the calculator is in radian mode for this to work.

sine of 6.2 in radian mode is about negative 0.083 and if you multiply that by negative 18.6 The velocity at this point is 1.545 meters per second. So now let's find the acceleration. So let's write the function first.

It's negative 230.64 cosine 12.4t. So this is going to be cosine 6.2, just like... sine 6.2.

Cosine 6.2 is about 0.9965. Multiply by negative 230.64. The acceleration at this point is negative 229.84 meters per second squared. Now that we have the acceleration, we can calculate the restoring force. So the restoring force is simply going to be the mass times the acceleration.

So the mass is 0.55 and the acceleration negative 229.84. So the restoring force at this point is about negative 126.4 newtons. So that is it for this problem.

A spring of force constant 300 newtons per meter vibrates with an amplitude of 45 centimeters when a 0.35 kilogram mass hangs from it. What is the equation describing this motion? If the mass passes through the equilibrium point with positive velocity at t equals 0. So we have a vertical spring, we have a mass hanging from it. Now, there's four points at which...

There's four ways that the system can start. It can start above the equilibrium position, in which case it has no choice but to go downward. So at t equals 0, for this particular shape, or this particular situation, it has negative velocity because it's going in the negative y direction. So therefore, to graph it, it is starting at the top.

with an amplitude of positive a. So this would be the cosine graph, positive cosine to be exact. So x would be equal to positive a cosine 2 pi ft.

But we don't have that situation. This is if it's starting above the equilibrium point. and if it's going down with negative velocity at t equals 0. The situation that we have for part A is that it's starting at the equilibrium position, and it's moving with positive velocity, that means it's going this way, in the positive y direction. So to graph it, because it's at the equilibrium point, it starts at the center. And it's going in the positive y direction, so it's going to go up.

This graph is positive sine, so the position function is going to equal positive a sine 2 pi ft. So that's what we have for part a. But let's say if part A was different. Let's say if everything was the same, the mass passes through the equilibrium point, but with negative velocity, meaning that it's going in a downward direction. I'm going to graph it in blue.

So, in that case, it starts at the equilibrium position that is the origin, but it's going down, and then up. So, the blue sine wave would represent this function. Instead of being positive sine, it's going to be negative.

A sine 2 pi FT, if it was going in the negative y direction. So anytime it starts at the equilibrium position, use the sine function. If it goes up, use positive sine. If it's going down from the equilibrium point, use negative sine.

So let's get back to the stuff that we're trying to figure out, which is positive A sine 2 pi FT. So we have the amplitude. It's 45 centimeters, which is 0.45 meters.

Now what we need to figure out is 2 pi f. We know that k is equal to 2 pi f squared times m. So k divided by m is 2 pi f squared. So 2 pi f is the square root of k over m.

K is 300, the mass is 0.35. 300 divided by 0.35 is 857.1. If you take the square root of that, you should get 29.28, or 29.3. Therefore, the position function is going to be A, which is 0.45, times sine.

2 pi f, which is 29.28 times t, and this is positive 0.45. Now, what about the second part of the question? What if the mass starts at the lowest point below the equilibrium point?

So let's say this is the equilibrium point. If it starts below it, It has no choice but to go up. It can't go down if it's at the lowest point.

So it has to be moving up with positive velocity. So if we graph in, we're starting at negative a. So if you don't start at the center, it's going to be the cosine function.

Sine always starts at the center. Since we're starting at the bottom, this is negative cosine. So x is negative a cosine 2 pi ft.

So the equation that describes this motion is negative 0.45 cosine 29.28 times t. So that's the answer for part b. So let's review.

There's four functions that you need to know. So it can start at the equilibrium position and go up with positive velocity. Or. It can go down with negative velocity, starting from the equilibrium position. Or it can start from the top, or it can start from the bottom.

Anytime it starts from the center, you have the positive sine function if it's going up. If it starts from the center and goes down, it's negative sine. If it starts at the top, it's positive cosine. If it starts at the bottom, negative cosine.

And remember, the origin represents the equilibrium position. Now let's talk about damp harmonic motion. This form of motion occurs when friction is present.

But let's draw two graphs, when friction is not present and when it is present. So let's say this is positive A and negative A. And let's say the graph starts at the top.

The mass is pulled to the right and then released from rest. So if there's no friction, the amplitude will be constant. The energy in the system remains the same. Notice that the amplitude is not decreasing. That is the maximum amplitude, or the maximum displacement.

Now what if friction is present? How will that change? the appearance of the graph.

So if friction is present, it's going to lose energy. And so the amplitude is going to decrease over time until eventually it reaches zero. In this case, dampen has occurred. So dampen is caused by Friction. It could be internal friction or friction due to air molecules, but it reduces the energy of the system, and this is seen by a decrease in amplitude over time.

Now, there's three different types of dampen that you need to be familiar with. There's over dampen, under dampen, and critical dampen. When the system is under damp, The object will make a few swings before it comes to rest. So it's going to make a few cycles, and eventually it's going to become zero.

Now, if it's over damp, it won't get the chance to make multiple cycles. Within its first cycle, it's going to lose all of its energy and go to zero. But it's going to take a long time. for it to do that.

And then there's critical dampen, where it reaches zero faster, and it doesn't make multiple cycles. Critical dampen is useful for the shock absorbers in a car. You don't want your car bouncing up and down every time it hits a bump. You want it to absorb that energy fast, and so you want critical dampen. In a situation where you have an underdamp system, let's say like a door that's old, it can bounce back and forth before it closes.

And you don't want that. So whenever you want to reduce bouncing in a device, you want critical damping. You want to remove the energy as quickly as possible from the system.

Now there's one other topic that we need to discuss and that is resonant frequency. When you apply a force to an oscillator, you can cause forced vibrations to occur. And as you force the oscillator to move, you can increase the amplitude.

However, if you apply the force at the natural frequency of the object, it's going to be most effective when increasing the amplitude. So if the applied frequency matches the natural frequency or if it's very close to it, the amplitude can increase greatly. When the applied frequency matches the natural frequency, it's known as the resonant frequency.

A good example to illustrate this, is a swing. Imagine if you have a child on a swing and to increase the amplitude of the swing you need to push the child at the right position and that is while the child is swinging back towards you. That's when you apply the push and at that point the child can go higher and higher.

However, let's say if you try to push the child at the middle at the equal position. you probably will not get the greatest effect in increasing the amplitude. You want to apply the force at a rate that's equal to the natural frequency.

If you do that, then the applied force is at the resonant frequency and that's when you can achieve maximum amplitude. That's when you can increase the amplitude of the oscillator most effectively. So just make sure that... You keep this fact in mind. Whenever the applied force matches the natural frequency of the system, that is the resonant frequency, the applied force will be most effective in increasing the amplitude of the oscillator.

Transcript for:Understanding Periodic Motion and SHM

Transcript for:
Understanding Periodic Motion and SHM