in this video we're going to talk about how to solve some basic simple harmonic motion problems in physics so let's start with number one a horizontal spring with a mass of 0.75 kilograms attached to it is undergoing simple harmonic motion calculate the period frequency and angular frequency of this oscillator so if you want to try this problem feel free to pause the video and work on it the formula that we need to calculate the period is this equation it's equal to 2 pi times the square root of the mass divided by the spring constant which we have all of that in this problem so the mass is 0.75 kilograms and the spring constant is 300 newtons per meter so let's go ahead and type that into the calculator so the period is .314 two seconds so that's how you could find the period of a simple harmonic oscillator all you need is the mass and the spring constant now part b what is the frequency frequency is 1 divided by the period so it's 1 divided by 0.3142 seconds so the frequency in this problem is 3.183 hertz now let's move on to part c let's calculate the angular frequency the angular frequency represented by the symbol omega it's 2 pi times the frequency so that's going to be 2 pi times 3.183 hertz and so that's going to be about 20 radians per second and that's it for this problem number two a force of 500 newtons is used to stretch a spring with a 0.5 kilogram mass attached to it by 0.35 meters what is the value of the spring constant and calculate the frequency of the oscillator so let's say this is a wall and we have a spring attached to it and there's a mass now we're going to stretch the spring using a force and so this is a 0.5 kilogram mass and at this point we're applying a force of 500 newtons to stretch it by point 35 meters so how can we calculate the spring constant well we know that the force is equal to kx based on hooke's law so the spring constant k is the ratio between the applied force and the amount that the length of this spring changes so it's going to be 500 newtons divided by 35 meters and so that's going to be 14 28.6 newtons per meter so that's how you can calculate the spring constant of a spring it's simply the force divided by the change in life now let's calculate the frequency of the oscillator the frequency of this spring is going to be one over two pi times the square root of k over m so it's 1 over 2 pi times the square root of the spring constant which is 1428.6 divided by the mass of 0.5 so the frequency is 8.51 hertz and so that's the answer to part b of that problem number three a spring with a constant of 100 newtons per meter vibrates at 25 hertz what is the frequency of vibration of a spring with a constant of 400 newtons per meter so what happens to the frequency if we increase the spring constant as the spring constant increases the frequency increases and we know that the frequency of a spring is one over two pi times the square root of k over m so notice that the frequency is proportional to the square root of k and in this example the spring constant increases from 100 to 400 so it increases by factor four and the square root of four is two so therefore the frequency should increase by a factor of two so the answer is 50 hertz now if you want a formula for this type of problem here's how you could derive it since we're dealing with two frequencies let's write a ratio between the two frequencies f2 and f1 so f2 is going to be 1 over 2 pi times the square root of k2 divided by m f one is going to be one over two pi times the square root of k one over m now i didn't write a subscript for m because the mass doesn't change in this problem so we could cancel it and at the same time we could cancel one over two pi so thus we have this expression f2 divided by f1 is equal to the square root of k2 divided by the square root of k1 or we could just simply write it like this f2 over f1 is simply the square root of k2 over k1 all within the single fraction so now let's calculate f2 so f1 in this problem that's 25 hertz and that corresponds to a a spring constant of 100 so k1 is 100 in this problem we're trying to find a new frequency at this new k value so k2 is going to be 400 so 400 divided by 100 is 4 and the square root of 4 is 2. so f2 over 25 is equal to 2. now let's cross multiply so f2 times 1 is simply f2 and then we have 25 times 2 which is 50. and so that's the new frequency so if you have a problem that relates frequency to the spring constant you could use that formula now let's work on this problem a 0.75 kilogram mass vibrates according to the equation x is equal to 0.65 cosine 7.35 t determine the amplitude frequency period and the spring constant so x represents the position of the oscillator so let's say the oscillator is at equilibrium it's right here so right now the position is x equals zero it can oscillate this way and that way so here x could be one and here x could be negative one it could be two it could be three it can vary the amplitude is the maximum displacement for the spring so let's say if the most that the spring will stretch to is up to this line and let's say that's 1.5 and so the most that you can stretch to in the other direction will be here so in this case the amplitude the maximum x value is 1.5 now x could be anywhere between negative 1.5 and positive 1.5 so x represents the current displacement a the amplitude is the maximum displacement so make sure you understand the difference between the two now you need to know the generic form of that formula the current displacement is equal to the maximum displacement times cosine omega t now sometimes you may have a phase angle but we don't have it for this problem so all we have is this formula so the amplitude is whatever number you see in front of cosine so the maximum displacement is 0.65 which means that x can be anywhere between negative 0.65 and positive 0.65 so if you graph this cosine equation the amplitude is 0.65 and negative 0.65 on the graph and cosine starts at the top so the cosine wave is going to look something like this and it's going to keep oscillating between these two points now what does this number represent 7.35 notice that the variable in front of t is omega so the 7.35 represents the angular frequency of this oscillator so it's 7.35 radians per second and if we know the angular frequency we can now find the frequency the angular frequency is equal to 2 pi f and so 7.35 radians per second is equal to 2 pi times the frequency so the regular frequency is just 7.35 divided by 2 pi and so you should get 1.17 hertz now how can we calculate the period if you know the frequency then you know the period the period is simply 1 over the frequency so in this example it's 1.17 hertz it's 1 over 1.17 hertz so 1 over 1.17 that's 0.855 seconds so we have the amplitude we have the frequency and we have the period so the last thing we need to calculate is the spring constant so what formula can help us to do that well we know that frequency is 1 over 2 pi times the square root of k over m so the first thing i'm going to do is multiply both sides by 2 pi so i can get rid of this so on the left i have 2 pi f and that's equal to the square root of k over m now i'm going to take the square of both sides to get rid of the radical on the right so i have 2 pi f squared and when you combine a square and a square root those two will cancel so on the right all i have is k over m so i'm going to multiply both sides by m so let's get rid of that so the spring constant is equal to the mass times 2 pi times the frequency squared so if you have the mass and the frequency you could use that equation to calculate the spring constant now keep in mind that omega is 2 pi f so we can replace 2 pi f with omega so the spring constant is the mass times the square of the angular frequency which we know it's 7.35 so we have a mass of 0.75 kilograms multiplied by 7.35 radians per second squared and so the spring constant is 40.5 newtons per perimeter and so that's the final answer for this problem so hopefully this video gave you a decent understanding of all the formulas that you need for solving basic simple harmonic motion problems thanks for watching you