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### Course: Class 11 Physics (India)>Unit 18

Lesson 5: Simple harmonic motion (with calculus)

# Harmonic motion part 3 (no calculus)

Figuring out the period, frequency, and amplitude of the harmonic motion of a mass attached to a spring. Created by Sal Khan.

## Want to join the conversation?

• Sal talks about the simple harmonic motion of a spring and how its formulas are derived. Can anyone please do that for a pendulum? thanks.
(4 votes)
• Set the parallel component of the force of gravity as the source of the torque on the pendulum.
τ = r x F = r*mg*sin(Θ) = Iα = mr²α = mr²*d²(Θ)/dt²
where m is the mass of the pendulum and r is the length of the string on the pendulum.
Use a small angle approximation to let sin(Θ) ~= Θ to make the differential equation linear and solvable.
gΘ = r*d²(Θ)/dt²
This equation is now in the same form as the mass spring equation of motion
kx = m*d²(x)/dt²
So solving the 2nd order differential equation you get
Θ(t) = Θi*cos(√(g/r)*t)
where Θi is the initial angle of the pendulum at t=0.
(15 votes)
• is the angular velocity a constant value.
if it is a constant value then why is angular acceleration present. because acceleration is just change in velocity and if velocity is constant then there should be no acceleration.
pls help
(7 votes)
• You are confusing angular motion with linear. Acceleration, a, is the time rate of change of velocity in in some straight line direction. The spring for example accelerates the mass along a line. Angular frequency, omega, is the number of radians per second (thus the angular) which is just 2*pi*f. The frequency, f, is the number of full cycles per second. Angular frequency is used because it works best with trig functions.

Angular acceleration is not part of SHM.
(5 votes)
• does harmonic motion means oscilatting motion?
(4 votes)
• What if gravity's effect is taken into account? Will the motion of the spring still be considered simple harmonic?
(4 votes)
• Yes, it will, because gravity has the same effect at the top as it does at the bottom, so it sort of cancels out.
(2 votes)
• Sal and David (see previous section video on equation for...) give different equations for the SHM. They are similiar, but what are the differences? Is there places where I should only use one or is one more general?
David's Equation: x(t)=A sin (2pi/T t)
Sal's Equation: x(t)=A cos sqr root(k/m) *t (see above)
(3 votes)
• Sal's is for a mass/spring system.
David's is for any SHM system. By comparison you can tell what the period of Sal's mass spring system must be.
Also in sal's system he is setting the time = 0 when his mass is at a peak. David's equation assumes t = 0 when the system is at equilibrium. That's why one is sin and the other cos.
(3 votes)
• I learned that the equation for a harmonic series of waves in a pipe was f=n(v/2L) where f is the frequency, n is the harmonic number, v is the velocity, and L is the length. Where did the equation t=2pi times square root of m divided by k come from?
(3 votes)
• According to the equation T=2pi(l/g)^1/2 (time period of a simple pendulum, time period is inversely proportional to the square of gravity and independant of mass, But according to the equation T=2pi(m/k)^1/2, time period is independent of gravity ans directly proportional to the square of mass. What are we supposed to consider?
(3 votes)
• Are you comparing a pendulum to a mass/spring system? They're similar, but not the same.
(2 votes)
• I have two questions actually:
1. What exactly is omega in this particular equation? I am used to it being angular velocity but because it does not seem like there is angular velocity in this I am a bit confused.
2. Is the t in the equation x(t) = Acos(SQRT(k/m)*t) the period? If no or not necessarily then could you explain how to get t without it being given to you?
(1 vote)
• Omega is still angular velocity. Harmonic motion corresponds to circular motion. Each full cycle is once around a circle.

the t is the time at which you are trying to find the position, x(t)
You could be given a t and then asked to find x. Or you might know x and want to figure out the time at which the system was at point x.
(4 votes)
• how is it root of k/m ? wasn't it omega ?
(2 votes)
• w = sqrt ( k / m ) or sq. w = ( k / m ) .
(2 votes)
• at my beloved Mr.Khan said if there is g the situation will be little bit different but from equation T = 2pi sqrt(m/k) , must not g have any effect on this case?
(2 votes)
• Good observation :-). He said that because the figure that he made wouldn't be as simple as it is if we were dealing with gravity. There would be the weight of the object acting downwards (which would affect its position) and a whole lot other things. But eventually the formula would turn out to be the same.(T=2pi sqrt(m/k)). So I think he said that so that we don't get confused by the diagram. That's it. :-)
(2 votes)

## Video transcript

Welcome back. And if you were covering your eyes because you didn't want to see calculus, I think you can open your eyes again. There shouldn't be any significant displays of calculus in this video. But just to review what we went over, we just said, OK if we have a spring-- and I drew it vertically this time-- but pretend like there's no gravity, or maybe pretend like we're viewing-- we're looking at the top of a table, because we don't want to look at the effect of a spring and gravity. We just want to look at a spring by itself. So this could be in deep space, or something else. But we're not thinking about gravity. But I drew it vertically just so that we can get more intuition for this curve. Well, we started off saying is if I have a spring and 0-- x equals 0 is kind of the natural resting point of the spring, if I just let this mass-- if I didn't pull on the spring at all. But I have a mass attached to the spring, and if I were to stretch the spring to point A, we said, well what happens? Well, it starts with very little velocity, but there's a restorative force, that's going to be pulling it back towards this position. So that force will accelerate the mass, accelerate the mass, accelerate the mass, until it gets right here. And then it'll have a lot of velocity here, but then it'll start decelerating. And then it'll decelerate, decelerate, decelerate. Its velocity will stop, and it'll come back up. And if we drew this as a function of time, this is what happens. It starts moving very slowly, accelerates. At this point, at x equals 0, it has its maximum speed. So the rate of change of velocity-- or the rate of change of position is fastest. And we can see the slope is very fast right here. And then, we start slowing down again, slowing down, until we get back to the spot of A. And then we keep going up and down, up and down, like that. And we showed that actually, the equation for the mass's position as a function of time is x of t-- and we used a little bit of differential equations to prove it. But this equation-- not that I recommend that you memorize anything-- but this is a pretty useful equation to memorize. Because you can use it to pretty much figure out anything-- about the position, or of the mass at any given time, or the frequency of this oscillatory motion, or anything else. Even the velocity, if you know a little bit of calculus, you can figure out the velocity at anytime, of the object. And that's pretty neat. So what can we do now? Well, let's try to figure out the period of this oscillating system. And just so you know-- I know I put the label harmonic motion on all of these-- this is simple harmonic motion. Simple harmonic motion is something that can be described by a trigonometric function like this. And it just oscillates back and forth, back and forth. And so, what we're doing is harmonic motion. And now, let's figure out what this period is. Remember we said that after T seconds, it gets back to its original position, and then after another T seconds, it gets back to its original position. Let's figure out with this T is. And that's essentially its period, right? What's the period of a function? It's how long it takes to get back to your starting point. Or how long it takes for the whole cycle to happen once. So what is this T? So let me ask you a question. What are all the points-- that if this is a cosine function, right? What are all of the points at which cosine is equal to 1? Or this function would be equal to A, right? Because whenever cosine is equal to 1, this whole function is equal to A. And it's these points. Well cosine is equal to 1 when-- so, theta-- let's say, when is cosine of theta equal to 1? So, at what angles is this true? Well it's true at theta is equal to 0, right? Cosine of 0 is 1. Cosine of 2 pi is also 1, right? We could just keep going around that unit circle. You should watch the unit circle video if this makes no sense to you. Or the graphing trig functions. It's also true at 4 pi. Really, any multiple of 2 pi, this is true. Right? Cosine of that angle is equal to 1. So the same thing is true. This function, x of t, is equal to A at what points? x of t is equal to A whenever this expression-- within the cosines-- whenever this expression is equal to 0, 2 pi, 4 pi, et cetera. And this first time that it cycles, right, from 0 to 2 pi-- from 0 to T, that'll be at 2 pi, right? So this whole expression will equal A, when k-- and that's these points, right? That's when this function is equal to A. It'll happen again over here someplace. When this little internal expression is equal to 2 pi, or really any multiple of 2 pi. So we could say, so x of t is equal to A when the square root of k over m times t, is equal to 2 pi. Or another way of thinking about it, is let's multiply both sides of this equation times the inverse of the square root of k over m. And you get, t is equal to 2 pi times the square root-- and it's going to be the inverse of this, right? Of m over k. And there we have the period of this function. This is going to be equal to 2 pi times the square root of m over k. So if someone tells you, well I have a spring that I'm going to pull from some-- I'm going to stretch it, or compress it a little bit, then I let go-- what is the period? How long does it take for the spring to go back to its original position? It'll keep doing that, as we have no friction, or no gravity, or any air resistance, or anything like that. Air resistance really is just a form of friction. You could immediately-- if you memorize this formula, although you should know where it comes from-- you could immediately say, well I know how long the period is. It's 2 pi times m over k. That's how long it's going to take the spring to get back-- to complete one cycle. And then what about the frequency? If you wanted to know cycles per second, well that's just the inverse of the period, right? So if I wanted to know the frequency, that equals 1 over the period, right? Period is given in seconds per cycle. So frequency is cycles per second, and this is seconds per cycle. So frequency is just going to be 1 over this. Which is 1 over 2 pi times the square root of k over m. That's the frequency. But I have always had trouble memorizing this, and this. You always [UNINTELLIGIBLE] k over m, and m over k, and all of that. All you have to really memorize is this. And even that, you might even have an intuition as to why it's true. You can even go to the differential equations if you want to reprove it to yourself. Because if you have this, you really can answer any question about the position of the mass, at any time. The velocity of the mass, at any time, just by taking the derivative. Or the period, or the frequency of the function. As long as you know how to take the period and frequency of trig functions. You can watch my videos, and watch my trig videos, to get a refresher on that. One thing that's pretty interesting about this, is notice that the frequency and the period, right? This is the period of the function, that's how long it takes do one cycle. This is how many cycles it does in one second-- both of them are independent of A. So it doesn't matter, I could stretch it only a little bit, like there, and it'll take the same amount of time to go back, and come back like that, as it would if I stretch it a lot. It would just do that. If I stretched it just a little bit, the function would look like this. Make sure I do this right. I'm not doing that right. Edit, undo. If I just do it a little bit, the amplitude is going to be less, but the function is going to essentially do the same thing. It's just going to do that. So it's going to take the same amount of time to complete the cycle, it'll just have a lower amplitude. So that's interesting to me, that how much I stretch it, it's not going to make it take longer or less time to complete one cycle. That's interesting. And so if I just told you, that I actually start having objects compressed, right? So in that case, let's say my A is minus 3. I have a spring constant of-- let's say k is, I don't know, 10. And I have a mass of 2 kilograms. Then I could immediately tell you what the equation of the position as a function of time at any point is. It's going to be x of t will equal-- I'm running out of space-- so x of t would equal-- this is just basic subsitution-- minus 3 cosine of 10 divided by 2, right? k over m, is 5. So square root of 5t. I know that's hard to read, but you get the point. I just substituted that. But the important thing to know is this-- this is, I think, the most important thing-- and then if given a trig function, you have trouble remembering how to figure out the period or frequency-- although I always just think about, when does this expression equal 1? And then you can figure out-- when does it equal 1, or when does it equal 0-- and you can figure out its period. If you don't have it, you can memorize this formula for period, and this formula for frequency, but I think that might be a waste of your brain space. Anyway, I'll see you in the next video.