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# Period dependence for mass on spring

David explains what affects the period of a mass on a spring (i.e. mass and spring constant). He also explains what does not affect the period of a mass on a spring (i.e. amplitude and gravitational acceleration). Created by David SantoPietro.

## Want to join the conversation?

- Why isn't dependent on gravity? There must be some tension and hence gravitational acceleration that may affect the period.(22 votes)
- @Eesh Gupta,

It is beyond the scope of a high school physics course, but if you solve the differential equation for a vertical oscillator with the force of gravity included then the solution you obtain includes two additional terms: 1. An additional cosine term with amplitude proportional to the acceleration of gravity but with the same period as the horizontal oscillator (dependent only on the mass and the spring constant), and 2. a constant term equal to the negative amplitude of the additional cosine term; such that the effect of gravity is cancelled out at time = 0. The short answer to your question is that gravity does not affect the period of the oscillator.(22 votes)

- How is the period not dependent on the friction constant?(9 votes)
- It is, but as you may have noted in previous videos in the series, friction has not been considered at all as part of the equations, and such.(6 votes)

- If I double the mass, will it double the period? If I decrease the spring constant by half, would the period double?(5 votes)
- Lets look at the equation:

T = 2π * √(m/k)

If we double the mass, we have to remember that it is under the radical. So this will increase the period by a factor of √2. If we cut the spring constant by half, this still increases whatever is inside the radical by a factor of two. So this also increases the period by √2. Hope this helps!(7 votes)

- I think David has a mistake: T = 2π√(
**k/m**), not T = 2π√(**m/k**).

Am I right?(4 votes)- The equation that is written in the video at about3:45T = 2 π * √(m/k) is correct. As is mentioned in the video increasing the mass increases the period (time it takes for an oscillation) and increasing the spring constant will decrease the period.(6 votes)

- If there is no spring and only a normal string it does not depend on mass, why?(3 votes)
- Because heavy things fall with the same acceleration as light things.(5 votes)

- From the equation T=2π√(m/k) , can we say that T² is directly proportional to m and inversely proportional to k?(3 votes)
- Yes, this is an accurate analysis of proportionality.(1 vote)

- At7:06, David says that the time period isn't dependent on gravity. Then what about the formula for time period which says T= 2*pi* (l/g)^1/2?(3 votes)
- That formula is not valid. The equation should be T = 2 * pi * sqrt(m / k). It is likely that you are thinking of the time period of a gravitational pendulum, which is equal to 2 * pi * sqrt(L / g). This problem is analogous but not equivalent to the spring problem discussed in the video.(1 vote)

- Hey I tried kinematic equations but it had mass.

Is there any way to calculate the period doesn't depend on the amplitude? I'm still trying to prove/derive it...(3 votes)- The period does not depend on the Amplitude. The period depends on k and the mass. The more amplitude the more distance to cover but the faster it will cover the distance. The distance and speed will cancel each other out, so the period will remain the same.(3 votes)

- Can someone direct me to the derivation videos mentioned at4:00? I checked this video: https://www.khanacademy.org/science/physics/mechanical-waves-and-sound/simple-harmonic-motion-with-calculus/v/harmonic-motion-part-2-calculus, but am unclear on how to get from x(t)=Acos(sqrt(k/m)t) to the equation depicted in this video?(4 votes)
- Just curious,

F=ma,

For Simple Harmonic Oscillators,

F=-kx,

Now since x is greater, F is greater

F=m(v-u/t)

Since F is now greater shouldn't t be smaller,

I mean shouldn't they be related?(2 votes)- F = ma is Newton's second law. It relates a
*constant force*with a*constant acceleration*. Hooke's law (F = -kx) shows the recoil force of the spring as a function of the displacement of the spring, this means that the recoil force of the spring is*constantly changing*as the spring is in motion.(4 votes)

## Video transcript

- [Instructor] So, we saw that for a mass oscillating on a spring, there's a certain amplitude and that's the maximum
displacement from equilibrium. But there's also a certain period, and that's the time it takes
for this process to reset. In other words, the time it takes for this mass to go
through an entire cycle. But what do these things depend on? We know the definitions of them, but what do they depend on? Well, for the amplitude,
it's kind of obvious, the person pulling the mass back. Whoever or whatever is
displacing this mass is the thing determining the amplitude. So if you pull the mass back far, you've given this oscillator
a large amplitude, and if you only pull it back a little bit, you've given it a small amplitude. But it's a little less obvious
in terms of the period. What does the period depend on? Who or what determines the period? Maybe it depends on the amplitude, so let's just check. If I asked you, if I asked you, if I pulled this back farther, if I increase the amplitude farther, will that change the
period of this motion? So, let's think about it. Some of you might say, yes, it should increase the period because look, now it has
farther to travel, right? Instead of just traveling
through this amount, whoa that looked horrible, instead of just traveling
through this amount back and forth, it's gotta travel through
this amount back and forth. Since it has farther to travel, the period should increase. But some of you might
also say, wait a minute. If we pull this mass farther, we know Hooke's law says that
the force is proportional, the force from the spring, proportional to the amount
that the spring is stretched. So, if I pulled this mass back farther, there's gonna be a larger force that's gonna cause this mass
to have a larger velocity when it gets to you, a larger speed when it gets
to the equilibrium position, so it's gonna be moving
faster than it would have. So, since it moves faster, maybe it takes less time for
this to go through a cycle. But it turns out those two
effects offset exactly. In other words, the fact that this mass has farther to travel and the fact that it will
now be traveling faster offset perfectly and it doesn't
affect the period at all. This is kinda crazy but something you need to remember. The amplitude, changes in the amplitude do not affect the period at all. So pull this mass back a little bit, just a little bit of an amplitude, it'll oscillate with a certain period, let's say, three seconds, just to make it not abstract. And let's say we pull
it back much farther. It should oscillate
still with three seconds. So it has farther to travel, but it's gonna be traveling faster and the amplitude does not affect the period for a mass
oscillating on a spring. This is kinda crazy, but it's true and it's
important to remember. This amplitude does not affect the period. In other words, if you were
to look at this on a graph, let's say you graphed this,
put this thing on a graph, if we increase the amplitude, what would happen to this graph? Well, it would just
stretch this way, right? We'd have a bigger amplitude, but you can do that and
there would not necessarily be any stretch this way. If you leave everything else the same and all you do is change the amplitude, the period would remain the same. The period this way would not change. So, changes in amplitude
do not affect the period. So, what does affect the period? I'd be like, alright, so the
amplitude doesn't affect it, what does affect the period? Well, let me just give
you the formula for it. So the formula for the
period of a mass on a spring is the period here is gonna be equal to, this is for the period
of a mass on a spring, turns out it's equal to two pi times the square root of the mass that's connected to the spring divided by the spring constant. That is the same spring constant that you have in Hooke's law, so it's that spring constant there. It's also the one you
see in the energy formula for a spring, same spring
constant all the way. This is the formula for the
period of a mass on a spring. Now, I'm not gonna derive this because the derivations
typically involve calculus. If you know some calculus and you want to see how this is derived, check out the videos we've
got on simple harmonic motion with calculus, using calculus, and you can see how this
equation comes about. It's pretty cool. But for now, I'm just gonna quote it, and we're gonna sort of just
take a tour of this equation. So, the two pi, that's
just a constant out front, and then you've got mass here and that should make sense. Why? Why does increasing the
mass increase the period? Look it, that's what this says. If we increase the mass, we
would increase the period because we'd have a larger
numerator over here. That makes sense 'cause a larger mass means that this thing
has more inertia, right. Increase the mass, this
mass is gonna be more sluggish to movement, more
difficult to whip around. If it's a small mass, you can
whip it around really easily. If it's a large mass,
very mass if it's gonna be difficult to change its
direction over and over, so it's gonna be harder
to move because of that and it's gonna take longer to
go through an entire cycle. This spring is gonna
find it more difficult to pull this mass and then slow it down and then speed it back up
because it's more massive, it's got more inertia. That's why it increases the period. That's why it takes longer. So increasing the period
means it takes longer for this thing to go through a cycle, and that makes sense in terms of the mass. How about this k value? That should make sense too. If we increase the k value, look it, increasing the k would
give us more spring force for the same amount of stretch. So, if we increase the k value, this force from the
spring is gonna be bigger, so it can pull harder and
push harder on this mass. And so, if you exert a
larger force on a mass, you can move it around more quickly, and so, larger force means
you can make this mass go through a cycle more quickly and that's why increasing this
k gives you a smaller period because if you can whip this
mass around more quickly, it takes less time for
it to go through a cycle and the period's gonna be less. That confuses people sometimes, taking more time means it's
gonna have a larger period. Sometimes, people think if this mass gets moved around faster, you should have a bigger period, but that's the opposite. If you move this mass around faster, it's gonna take less time to move around, and the period is gonna decrease if you increase that k value. So this is what the period of
a mass on a spring depends on. Note, it does not depend on amplitude. So this is important. No amplitude up here. Change the amplitude, doesn't matter. Those effects offset. It only depends on the mass
and the spring constant. Again, I didn't derive this. If you're curious, watch
those videos that do derive it where we use calculus to show this. Something else that's important to note, this equation works even if
the mass is hanging vertically. So, if you have this mass
hanging from the ceiling, right, something like this, and this mass oscillates
vertically up and down, this equation would
still give you the period of a mass on a spring. You'd plug in the mass that
you had on the spring here. You'd plug in the spring
constant of the spring there. This would still give you the period of the mass on a spring. In other words, it does not depend on the gravitational constant, so little g doesn't show up in here. Little g would cause this
thing to hang downward at a lower equilibrium point, but it does not affect the
period of this mass on a spring, which is good news. This formula works for horizontal masses, works for vertical masses, gives you the period in both cases. So, recapping, the period
of a mass on a spring does not depend on the amplitude. You can change the amplitude, but it will not affect how
long it takes this mass to go through a whole cycle. And that's true for
horizontal masses on a spring and vertical masses on a spring. The period also does not depend on the gravitational acceleration, so if you took this mass on
a spring to Mars or the moon, hung it vertically, let it oscillate, if it's the same mass and the same spring, it would have the same period. It doesn't depend on what the
acceleration due to gravity is but the period is affected
by the mass on a spring. Bigger mass means you
would get more period because there's more inertia, and it's also affected
by the spring constant. Bigger spring constant
means you'd have less period because the force from the
spring would be larger.