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## Physics library

### Unit 8: Lesson 2

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?

- @7:34, Sal says that the expressions of time period and frequency are independent of the displacement but it is clear that the two expressions are dependent on 'k' which is the spring constant. My question is that isn't 'k' dependent on the displacement 'A' if we go back to the equation F= -kx, if so then the expressions for time period and frequency are indirectly dependent on the displacement. I hope this makes sense and is my reasoning right?(2 votes)
- The spring constant 'k' is dependent on the material of which the spring is made and not on the displacement.(21 votes)

- 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.(2 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.(12 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(5 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.(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)

- does harmonic motion means oscilatting motion?(3 votes)
- From what I've learned,
*simple*harmonic motion has these characteristics:

1. Force is directly proportional to the displacement (`F = -kx`

in a spring, for example)

2. This causes the motion to be oscillatory.

3. It can thus be described by the formula`x(t) = A * cos(wt - φ)`

(3 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)

- at0:22my 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)

- In the example stated above, shouldn't "A"=3 while substituting in the equation, as amplitude is the 'magnitude' of maximum displacement from the mean position, even if it is '-3' in the graph?(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.