Transcript for:
Periodic Motion and SHM Overview

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 but let's talk about the mass spring system so 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 is 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 equivalent position x is zero so the restoring force will always be zero 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 a hundred newtons of force to either stretch or compress the spring by one meter from its equilibrium so if you wish to stretch it by two 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 a thousand so for this spring it requires a thousand newtons of force just to stretch it by one 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 f r 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 equal position so if the equivalent 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 uh move it towards the right back to its equal position and remember at the equilibrium position the restoring force is zero since x is equal to zero and f r is negative kx so negative k times zero is zero 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 newtons per meter 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 we need 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 percent 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 stress i mean 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 two times the amount the force will increase by a factor of two 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 newtons 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 the 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 three if you triple the distance by a factor of three the force is going to increase by a factor of 3. so the force that's required is 600 newtons 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 newtons 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 three so therefore f2 divided by 200 is equal to three 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 1 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 anti-derivative of k times x dx to find the anti-derivative of kx add 1 to the exponent and divide by that number the anti-derivative of x to the first power is x squared divided by two so the work required to stretch a spring is going to be basically one 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 one 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 x is 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 two meters here because 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 to spring is the apply 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 string 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 it's is at its maximum value at this point as well so make sure you understand is 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 zero friction will be zero 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 decrease 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 zero at the equilibrium position and let's say at this point x is equal to one 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 one 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 point five okay that does not look like a five it's going to be negative a hundred times negative point five 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 movement 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 zero when it's fully stretched or compressed now at the equilibrium position the velocity is at a max so therefore ke is at its maximum value at the equilibrium position as you mentioned before the acceleration is zero at the equilibrium position and the restoring force is equal to zero since x is zero and f equals k x so k times zero is zero 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 zero at the equivalent position the potential energy is zero because one-half k times zero squared is zero 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 mind when the amplitude is one x can vary anywhere between negative one to one the mechanical energy depends on the amplitude that is the maximum value of x or the maximum displacement so the mechanical energy is one half k a squared the kinetic energy is one half mv squared and the potential energy is one 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 zero and whenever x is equal to zero that is at the equilibrium position potential energy is equal to zero so let me organize this information so let's say this is the equilibrium position where x is zero at that point the potential energy will be equal to zero 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 zero and when it's fully compressed pe is still equal to the mechanical energy and ke is equal to zero 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 zero so when x is equal to zero the potential energy is zero so this disappears so therefore one half k a squared is equal to one-half mv squared we can multiply both sides by two 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 m a 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 0.4.7 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 two two times a half is one this will cancel all of the fractions so we have k a squared is equal to mv squared plus kx squared now let's subtract kx squared by both sides squared minus kx squared is equal to m v squared and now let's factor out the gcf the greatest common factor which is king 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 one 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 now as the spring 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 equilibrium it travels the distance of x and then when it's fully compressed it travels another distance of x and then when it goes back to its equilibrium 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 three seconds to complete a single cycle therefore the period is three 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 divided by the total time you can get the frequency so for example let's say if it takes let me choose a nice number um 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 one 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 m e is equal to one half k a 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 one so one 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 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 one the velocity will increase by a factor of two if you triple the displacement the velocity will increase by a factor of three if you cut the displacement in a half the velocity will decrease by a factor of two or it's going to be one 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 two 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 friction 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 a horizontal surface and 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 eight point two one six 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 a 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 three times two is six so 300 times 0.2 must be 60. so the restoring force is negative 60 newtons it's negatives 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 point 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 a 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 point 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 point 20 from this position so it's going to go up point 20 and down 0.20 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 0.2 that's about 0.99 meters per second so now what about part e what is the kinetic energy potential energy and the mechanical energy when it's 0.10 meters from its equilibrium position so let's find the potential energy first the potential energy is one half kx squared which is basically one 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 one-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 zero at the equilibrium position but we're not at the equilibrium position x is point 10. so we got to find the velocity first at point n 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 4. 1 minus 1 fourth is three fourths and the square root of three fourths is about point eight six six times point ninety nine you should get point eight five seven four 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 e is equal to one-half times the mass which is two kilograms times the speed point eight five seven four squared so half of two is one and one times point eight 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 one-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 nine eight joules another way in which we could find it is by adding the potential energy and the kinetic energy so if we have 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 units 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 a 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 newtons 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 gonna 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 14 29 rounded 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 one half ka squared if you use the potential energy equation it's one half kx squared but in this case x is equal to a because it's stretch 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 one half ka squared is equal to one half mv squared so if we multiply both sides by two we can get rid of the one-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 0.25 we can 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 one 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 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 zero 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 one-half base times height or one-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 one 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 gonna see it's still moving towards the left now as it moves from c to d it's going to appear as if it 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 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 the circle is basically the distance that it travels divided by the time and the distance around the circle is the circumference which is two 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 v t 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 in the numerator 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 a value that is being changed everything else plug into one so if we double the mass the period will increase by the square root of two now what if we quadruple the mass the period will increase by the square root of four which is two so if you quadruple the mass the period will double if you increase the mass by a factor of nine the period will triple in value what if you reduce the mass by one half the square root of one half which is one over root two that's root two over two so the period will increase well actually it's gonna decrease by root two over two 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 two it's going to be seventy percent or seventy point seven percent of its original value if you quadruple the value of k the period is going to be one half of its original value now what if you reduce the spring constant by 1 4 what effect will i have on the period 1 divided by 1 4 is 4. so if you decrease k to 1 4 of 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 it's 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 m a 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 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 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 newtons 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 nine hertz now that we have the frequency we can easily calculate the period the period is simply one over f which is basically one divided by nine one divided by nine is about point one repeating so the period is about point eleven seconds so it takes point eleven seconds to make a full cycle that is it takes point 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 springs compress by two 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 two centimeters is the weight of the person because once he gets on the car x changes by two centimeters so we only need to use the weight of that person and not the weight of the car and a 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 34 000 300 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 uh seventy plus twelve hundred or twelve seventy thirty four thousand three hundred divided by twelve seventy 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 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 a thousand or instead of writing 0.25 g you can write 0.25 times 10 to the minus 3 kilograms one gram is about .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 over 125.6 if you square it you should get 15791 and then if you multiply 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 routed answer and the mass is going to be 0.1 times 10 to the minus 3 kilograms so three point nine four seven eight divided by point one times ten to the minus three that's thirty nine thousand four hundred seventy eight if you take the square root of that you should get 198.69 and then divide that by two 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 one over two 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 the 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 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 hertz corresponds to a mass of 0.25 grams 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 hertz 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 newtons per meter vibrates at 15 hertz what is the frequency of vibration if the same block is attached to a spring of 500 newtons per meter 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 one over two pi square root k2 over m so we 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 a block undergoes shm with amplitude 0.4 that's simple harmonic motion if you see show that's a simple harmonic oscillator what is the total distance that it travels in eight periods now keep this in mind the distance that it travels in one period or one cycle is four times the amplitude let's say this is the equilibrium position this is the amplitude the amount that it's fully stretched by so fortu 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 equivalent position and back towards where it started so as you can see the distance that it travels in one cycle or one period is four times a 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 eight 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 a 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 bounce 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 so 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 uh sokotoa 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 cop heart 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 v t 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 angular 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 two 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 d 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 sign now you have to keep the angle the same the 2 pi ft and then you have to differentiate the nested function or the inside part of sine which is 2 pi ft the derivative of 2 pi ft is simply 2 pi f t is the variable that you're differentiating with respect to the derivative of five x is five the derivative of nine x is nine the derivative of eight t is eight so the derivative of two pi f t is two 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 and 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 f t 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 v max times sine 2 pi f t 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 v max some textbooks will use the symbol v o 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 v max is a constant we don't have to change that the derivative of sine is cosine and the derivative of two pi f t is two pi f now let's replace v max with what we wrote before we said it's equal to negative square root k over m times a and the frequency as you 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 two pi f t 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 two pi f t 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 two pi f t 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 9.2 t axis 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 two pi f t 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 kilogram 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 could find a period which is one divided by the frequency or one divided by one point four six four which is about point six eight three seconds now how can we find a spring constant what equation would you use to find it you can use this equation the spring constant is the mass times two 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 two 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 63.5 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 one 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 zero when it's fully stretched or compressed x is equal to a which is about 0.6 when x is zero we know that the velocity is at its maximum which means kinetic energy it's at its maximum so at x equals zero the kinetic energy is equal to the total energy which is 11.42 and when x is zero the potential energy is therefore equal to zero so that's the answer for the first part so when x is zero ke equals me that's 11.42 joules and pe is equal to zero now what about the other extreme when x is 0.6 when it's fully compressed or stretched the kinetic energy will be zero 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 0.2 i mean somewhere between 0 and 0.6 which is at point 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 one half kx squared so that's one 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 0.2 so it's v max times the square root of 1 minus x squared divided by a squared now we've got to find v max first so v max 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 0.75 times the amplitude which is 0.6 63.46 divided by 0.75 that's like 84.61 if you take the square root of that you should get 9.2 times 0.6 so the maximum velocity is 5.519 so now we could find a velocity at point 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 repeating 1 minus that is 0.8 repeating 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 one-half mv squared so one-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 used in both techniques a 0.55 kilogram block vibrates according to the equation x is equal to 1.5 cosine 12.4 t 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 two 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 v o which is v max times sine two pi f t 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 got to find the frequency 12.4 t is equal to 2 pi fc 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 of 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 multiplied 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 .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 12.4 t now let's find the acceleration function so we got to 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 it's equal to the negative maximum acceleration times cosine pi ft so that's going to be negative 230.64 cosine 12.4 t 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 so using the equation that we have v of t is equal to negative 18.6 sine 12.4 t so we're looking for v of 0.5 so it'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 we multiply that by negative 18.6 the velocity at this point it's 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 12.4 t so this is going to be cosine 6.2 just like sine 6.2 cosine 6.2 is about 0.9965 multiplied 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 described in this motion if the mass passes through the equilibrium point with positive velocity at t equals zero 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 zero 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 zero 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 sign so the position function is going to equal positive a sine 2 pi f t 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 neg the blue sine wave would represent this function instead of being positive sign it's going to be negative a sine two pi f t 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 sign if it's going down from the equilibrium point use negative sign so let's get back to the the stuff that we're trying to figure out which is positive a sine two pi ft so we have the amplitude it's 45 centimeters which is 0.45 meters now what we need to figure out is two 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 uh 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 it 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 two pi f t so the equation that describes this motion is negative point four five 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 you could start from the top or you could start from the bottom any time it starts from the center you have the positive sign function if it's going up if it starts from the center and goes down it's negative sign 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 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 a 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 damping 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 damping that you need to be familiar with 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 0. 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 damping where it reaches zero faster and it doesn't make multiple cycles critical damping is useful for the shock absorbers in the 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 damping in a situation where you have an under damp 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 equilibrium position you probably will not get the greatest effects and increase in 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 and the applied force will be most effective in increasing the amplitude of the oscillator you