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## Physics library

### Unit 8: Lesson 2

Simple harmonic motion (with calculus)# Harmonic motion part 2 (calculus)

We test whether Acos(wt) can describe the motion of the mass on a spring by substituting into the differential equation F=-kx. Created by Sal Khan.

## Want to join the conversation?

- in some books its given x= A sin wt and in some books it is x= Acos(wt). which one is right or are both of them correct??(11 votes)
- The thing is, the only difference between the two is where you start. The function A sin wt is just the function A cos wt displaced by 90 degrees (graph it on a calculator, you'll see). So, both are right. It just depends on how you decide to graph it.

If you start the oscillation by compressing a spring some distance and then releasing it, then x = A cos wt, because at time t=0, x=A (A being whatever distance you compressed the spring). But if you start the oscillation by suddenly applying a force to the spring at rest and then letting it oscillate, at t=0, x must equal 0. So in that case, x = A sin wt (sin0 = 0)(38 votes)

- what is the purpose of omega in the equation?

x(t) = A cos(wt)(9 votes)- The omega is a constant in the equation that stretches the cosine wave left and right (along the x axis), just as the A at the front of equation scales the cosine wave up and down. The bigger the omega, the more squashed the cosine wave showing the spring's position (and thus quicker the spring's movement).(5 votes)

- If my knowledge of Calculus is correct, derivatives lower the power of a function and integrals raise the power of a function. In this lesson, a (acceleration) is depicted as a second derivative, which is true, but the actual work appears to be that of an integral equation. In short, my question is this:

Why would the acceleration equation have a higher power than the distance equation?(2 votes)- In addition, the power of a function is given by the exponent on the variable (in this case, x). The w, on the other hand, is a constant, so the power of w isn't related to the power of the function.(5 votes)

- I can understand the fact that to make position-time equation dimensionally correct you introduce a term w(omega) multiplied with t in 'sin(wt)'. But I can't justify the appearance of a rotational quantity 'w' in this equation which is actually the representation of linear motion of spring-mass system performing SHM.

Thanking You.(3 votes)- the omega (curvy w) comes from the explanation of SHM using a reference circle. Omega is the angular velocity of the displacement phasor as it travels in a circular motion. This velocity is constant, unlike the linear motion of SHM. wt (omega x time) therefore equals angualr displacement, represented by the letter theta. For a little clearer understanding, the diagrams on this page may help. http://www.tutorvista.com/content/physics/physics-iii/oscillations/circle-reference.php(2 votes)

- Why w=square root of k/m not plus or minus square root of k/m ?. Isn't angular velocity a vector(3 votes)
- angular velocity is a vector but w is a magnitude of it(3 votes)

- why x(t)= A*cos(omega*t)? and wat is 'T'?(2 votes)
- x(t) = A*cos(omega*t) represents the function of the SHM. and T is the time period of the SHM, that is the time taken by the system to complete on cycle.. from A to O and to -A and again to O and again to A.(4 votes)

- If I keep a block attached to a horizontal spring on the floor of an elevator going up with an acceleration ‘a’, and then displace the block slightly by stretching the spring-block system horizontally, will the time period of oscillation change as compared to the (T=2π√m/k)?

I don't think it should, because the time period depends on the horizontal force, and the elevator changes it in the vertical direction, but my friend asserts that it should change, though she can't explain why. Which of us, if either, is correct?(2 votes)- Here is a simulation of a mass on a vertical spring

https://phet.colorado.edu/sims/mass-spring-lab/mass-spring-lab_en.html

Go to the simulation and try it out. You can measure the period of oscillation by clicking on the box for "stopwatch". To measure the period, measure the time for 10 or 20 bounces and divide by 10 or 20.

Now look over on the right side and you will see that you can change gravity from that of Earth to that of Jupiter. Go ahead and do that and see if gravity makes a difference to the period.

Now how does this apply to acceleration? When you are in an elevator accelerating up at a rate of a, that is exactly the same as if gravity increased from g to g+a. You can tell this by figuring out what your weight would be in the elevator - if you work it out you will see it will be m(g+a).

So if the period is the same on jupiter, that means it will also be the same in an accelerating elevator, and if it is different on jupiter, that means it will also be different in the elevator.

Once you know the answer by experiment, see if you can figure out why the answer is what it is.(4 votes)

- How would we solve a problem that is an oscillation but doesn't start where cos(wt) and sin(wt) start?(3 votes)
- That's a good question. The function would then be Asin(wt+k) where k is some constant.

If the graph is just shifted (horizontally) slightly, you add or subtract a constant that is equal to the amount by which it has been shifted on the x-axis.

When to add and when to subtract?

- Add when it is shifted to the left

- Subtract when it is shifted to the right

Hope this answers your doubt! :)(1 vote)

- why acceleration= second derivative of the function x(t)?(1 vote)
- Why do we have to take a double derivative?(2 votes)
- Because in the problem we are given the position of the particle. And it's known that if you differentiate the position of a particle (with respect to time), you get the velocity of the particle. You differentiate the result (the velocity of the particle) which then gives you the acceleration of the particle, which is why we have to take the double derivatie.(2 votes)

## Video transcript

So where I left off in the last
video, I'd just rewritten the spring equation. And I just wrote force is
mass times acceleration. And I was in the process of
saying, well if x is a function of t, what's
acceleration? Well, velocity is this
derivative of x with respect to time, right? Your change in position
over change of time. And acceleration is the
derivative of velocity, or the second derivative of position. So you take the derivative
twice of x of t, right? So let's rewrite this equation
in those terms. Let me erase all this--I actually want to
keep all of this, just so we remember what we're talking
about this whole time. Let me see if I can
erase it cleanly. That's pretty good. Let me erase all of this. All of this. I'll even erase this. That's pretty good, all right. Now back to work. So, we know that-- or hopefully
we know-- that acceleration is the second
derivative of x as a function of t. So we can rewrite this as
mass times the second derivative of x. So I'll write that as-- well,
I think the easiest notation would just be x prime prime. That's just the second
derivative of x as a function of t. I'll write the function
notation, just so you remember this is a function of time. Is equal to minus
k times x of t. And what you see here, what
I've just written, this is actually a differential
equation. And so what is a differential
equation? Well, it's an equation where,
in one expression, or in one equation, on both sides of
this, you not only have a function, but you have
derivatives of that function. And the solution to a
differential equation isn't just a number, right? A solution to equations that
we've done in the past are numbers, essentially, or a set
of numbers, or maybe a line. But the solution to differential
equations is actually going to be a function,
or a class of functions, or a set
of functions. So it'll take a little time to
get your head around it, but this is as good an example as
ever to be exposed to it. And we're not going to solve
this differential equation analytically. We're going to use our intuition
behind what we did earlier in the previous video. We're going to use that to guess
at what a solution to this differential equation is. And then, if it works out, then
we'll have a little bit more intuition. And then we'll actually know
what the position is, at any given time, of this mass
attached to the spring. So this is exciting. This is a differential
equation. When we drew the position-- our
intuition for the position over time-- our intuition tells
us that it's a cosine function, with amplitude A. So we said it's A cosine omega
t, where this is the angular velocity of-- well, I don't want
to go into that just yet, we'll get a little bit more
intuition in a second. And now, what we can do is,
let's test this expression-- this function-- to see if it
satisfies this equation. Right? If we say that x of t is equal
to A cosine of wt, what is the derivative of this?
x prime of t. And you could review
the derivative videos to remember this. Well, it's the derivative of the
inside, so it'll be that omega, times the
outside scalar. A omega. And then the derivative-- I'm
just doing the chain rule-- the derivative of cosine of t is
minus sine of whatever's in the inside. I'll put the minus outside. So it's minus sine of wt. And then, if we want the second
derivative-- so that's x prime prime of t. Let me do this in a different
color, just so it doesn't get monotonous. That's the derivative
of this, right? So what's the derivative
of-- these are just scalar values, right? These are just constants. So the derivative of the
inside is an omega. I multiply the omega times
the scalar constant. I get minus A omega squared. And then the derivative of
sine is just cosine. But the minus is still there,
because I had the minus to begin with. Minus cosine of omega t. Now let's see if this is true. So if this is true, I should
be able to say that m times the second derivative of x of
t, which is in this case is this, times minus Aw
squared cosine wt. That should be equal to minus k
times my original function-- times x of t. And x of t is a cosine wt. I'm running out of space. Hopefully you understand
what I'm saying. I just substituted x prime
prime, the second derivative, into this, and I just
substituted x of t, which I guess that's that, in here. And now I got this. And let me see if
I can rewrite. Maybe I can get rid of
the spring up here. I'm trying to look for space. I don't want to get rid of this,
because I think this gives us some intuition
of what we're doing. One of those days that I wish
I had a larger blackboard. Erase the spring. Hopefully you can remember
that image in your mind. And actually, I can
erase that. I can erase that. I can erase all of this, just so
I have some space, without getting rid of that nice curve I
took the time to draw in the last video. Almost there. OK. Back to work. Make sure my pen feels
right, OK. So all I did is I took-- we
said that by the spring constant, if you rewrite force
as mass times acceleration, you get this. Which is essentially a
differential equation, I just rewrote acceleration as
the second derivative. Then I took a guess, that this
is x of t, just based on our intuition of the drawing. I took a guess. And then I took the second
derivative of it. Right? This is the first derivative,
this is the second derivative. And then I substituted the
second derivative here, and I substituted the function here. And this is what I got. And so let me see if I can
simplify that a little bit. So if I rewrite there, I get
minus mAw squared cosine of wt is equal to minus
kA cosine of wt. Well it looks good so far. Let's see, we can get rid of the
minus signs on both sides. Get rid of the A's
on both sides. Right? We can divide both sides by A. Let me do this in black, just
so it really erases it. So if we get rid of A on both
sides, we're left with that. And then-- so let's see, we
have mw squared cosine of omega t is equal to k
cosine of omega t. So this equation holds
true if what is true? This equation holds true if mw
squared-- or omega squared, I think that's omega. I always forget my--
is equal to k. Or another way of saying
it, if omega squared is equal to k over m. Or, omega is equal to the
square root of k over m. So there we have it. We have figured out what
x of t has to be. We said that this differential
equation is true, if this is x of t, and omega is
equal to this. So now we've figured out the
actual function that describes the position of that spring
as a function of time. x of t is going to be equal to--
we were right about the A, and that's just intuition,
right, because the amplitude of this cosine function is A--
A cosine-- and instead of writing w, we can now write the
square root of k over m. The square root of k over m t. That to me is amazing. We have now, using not too
sophisticated calculus, solved a differential equation. And now can-- if you tell me at
5.8 seconds, where is x, I can tell you. And I just realized that I am
now running out of time, so I will see you in the
next video.