# Simple harmonic motion - Find amplitude given period

BobRoss

## Homework Statement

A block rides on a piston that is moving vertically with simple harmonic motion. (a) If the SHM has period 2.65 s, at what amplitude of motion will the block and piston separate?

## Homework Equations

x(t)=xmcos(ωt + ∅)
a(t)=-w2xmcos(ωt + ∅)

Where xm is the amplitude.

## The Attempt at a Solution

So from the Period T=2.65 s I found the angular frequency ω=2.37 rad/s. I assume that the block and piston separate when the acceleration is at a maximum. How do I find the amplitude from this information?

Homework Helper
The block will separate from the cylinder when the acceleration of the cylinder is greater than the weight of the block.

BobRoss
Hmm, I found a thread on here with a very similar question to this one that said they separate when the acceleration is at a maximum. Then it said that the ω2xm of the equation a(t)=-ω2xmcos(ωt + ∅) gave the maximum acceleration, although it never explained how to find xm. Is that not correct?

Homework Helper
Hmm, I found a thread on here with a very similar question to this one that said they separate when the acceleration is at a maximum. Then it said that the ω2xm of the equation a(t)=-ω2xmcos(ωt + ∅) gave the maximum acceleration, although it never explained how to find xm. Is that not correct?

Xm is the maximum amplitude when cos( ωt + ∅ ) is equal to 1

Homework Helper
The forces acting on the particle are gravity, (mg, downward) and normal force from the piston (N, upward) The particle is not fixed to the piston, the piston can not pull it, N is upward.
The particle moves together with the piston, and accelerates with a(t).
Setting "upward" as positive, the acceleration of the particle is a=(N-mg)/m, and that is equal to the acceleration of the piston, a(t).
So N-mg=m a(t): N=m(a(t)+g). The particle is separated from the piston if N=0. When does that happen? You need the minimum amplitude when that occurs, so the separation happens at maximum downward acceleration. What is cos(ωt + ∅) then?

ehild

BobRoss
Ok I'm following everything you are saying but I still don't understand how to find minimum amplitude where that occurs. And I don't know how to find cos(ωt+∅) either.

BobRoss
Wait, I got it. I found the max downward acceleration to be 9.81 m/s2 and I divided that by ω2 which gave me 1.747 m. Although I am having a bit of trouble understanding the second part to the question.

b.)If the piston has an amplitude of 6.05 cm, what is the maximum frequency (in Hz) for which the block and piston will be in contact continuously?

I don't know where to really start with this part now either though.

Homework Helper
You have figured out the relation between amplitude and angular frequency when the block loses contact from the piston: Aω2≥mg. NOT to be separated, Aω2<mg must hold. Given the amplitude, what is true for ω? And how do you get the frequency from ω?

ehild

BobRoss
Sorry I'm not getting it. I know that f=ω/2π but I'm not sure what amplitude has to do with it or even what it means by the max frequency that the block and piston will be in contact for.

Homework Helper
b.)If the piston has an amplitude of 6.05 cm, what is the maximum frequency (in Hz) for which the block and piston will be in contact continuously?
Being in contact continuously means that the block does not separate from the piston.

ehild

BobRoss
Ahh okay I got it. I set ω2xm=9.81 then I solved for ω and found the frequency to be 2.03 Hz from that. Thanks for the help on this one! I have one more spring question that I am struggling with and am hoping you can help there too. I thought I knew what I was doing for it but I only got 1/4 parts correct.

A block weighing 21 N oscillates at one end of a vertical spring for which k = 130 N/m; the other end of the spring is attached to a ceiling. At a certain instant the spring is stretched 0.25 m beyond its relaxed length (the length when no object is attached) and the block has zero velocity. (a) What is the net force on the block at this instant? What are the (b) amplitude and (c) period of the resulting simple harmonic motion? (d) What is the maximum kinetic energy of the block as it oscillates?

So first I thought that if the block has zero velocity when it is 0.25 m below the relaxed length of the spring, it must be at it's maximum displacement. So the amplitude of the spring from that would be 0.25/2=0.125 m. But when I input that amplitude for part b.) it says it is incorrect.

For a.) I know that Fnet=kx - mg. I'm not really sure which value to use for x here though. Do I use 0.25 m or 0.125 m (which I think is the amplitude)? When I use 0.125 m I get

Fnet=(130N/m)(0.125m) - (2.141kg)(9.81m/s2)
= -4.75N

But that isn't correct.

For part c.) I correctly found the period to be 0.81 s.

For part d.) I know that the max kinetic energy occurs when the displacement is 0, at which point the potential energy is 0. So if I find the potential energy at it's max displacement, that is equal to the max kinetic energy.

Ep=1/2kx2
=1/2(130N/m)(0.125m)2
=1.02 J

But that answer is also incorrect.

I am assuming I went wrong somewhere with the amplitude and that I am using the wrong values for x in my equations, but I'm not sure. What have I done wrong here and how do I fix it?

Homework Helper
The amplitude of the oscillation is 0.25 m.

BobRoss
Hmm, that is what I had initially entered as my answer for b.) the first time I did the question but when I submitted it, it came back as incorrect...

BobRoss
Okay so for a.) I just did the equation Fnet=kx-mg using x=0.25 and I got Fnet=11.5 N, and that came out correct. I entered 0.25 for the amplitude for b.) and it tells me that that is incorrect. And I used x=0.25 in my equation for d.) to find the maximum kinetic energy and got 4.06 J and it tells me that is incorrect too. How do I find the correct amplitude and max kinetic energy here?

nasu
The amplitude is measured from the equilibrium position of the system.
This is not the position of the spring when no object is attached. It is the equilibrium position with the 21N object attached. The maximum displacement from this position is the amplitude.

BobRoss
Ya I knew that. But it says "At a certain instant the spring is stretched 0.25 m beyond its relaxed length (the length when no object is attached) and the block has zero velocity".

So if the block is released and reaches a point where it has zero velocity, that means that it is at it's maximum displacement at that point, right? And if that point is 0.25m below the original relaxed equilibrium point, wouldn't the new equilibrium point with the block attached be at 0.25/2=0.125m from the original relaxed positions equilibrium? And that would make the amplitude with the block attached 0.125m wouldn't it?

Homework Helper
When the block is attached to the spring , it is stretched through a distance given by x = N / k before it attains the equilibrium position. So the amplitude of the oscillation is 0.25 - x .

Homework Helper
So if the block is released and reaches a point where it has zero velocity, that means that it is at it's maximum displacement at that point, right?

That is right. But: When a block hangs on a spring the spring is stretched by xo in equilibrium when the weight is balanced by the spring force, that is kxo=Weight.

If you move out the block from equilibrium it will oscillate about xo. The stretching of the spring varies with time as x=xo+Acos(ωt+θ). At maximum displacement, x=xo+A.

ehild

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BobRoss
Okay, I think maybe I was initially interpreting the question wrong. When it says "At a certain instant the spring is stretched 0.25 m beyond its relaxed length (the length when no object is attached) and the block has zero velocity", this just means that they've attached the block and it is at rest (it's new equilibrium point) 0.25m below the original equilibrium point, and it isn't currently oscillating? As in it hasn't even been set in motion at this point? I was thinking that they attached the block and it was already oscillating, and the 0.25 m below the original equilibrium was the maximum displacement of the oscillation, before it begins to move back upwards. That is incorrect isn't it?

nasu
Okay, I think maybe I was initially interpreting the question wrong. When it says "At a certain instant the spring is stretched 0.25 m beyond its relaxed length (the length when no object is attached) and the block has zero velocity", this just means that they've attached the block and it is at rest (it's new equilibrium point) 0.25m below the original equilibrium point, and it isn't currently oscillating? As in it hasn't even been set in motion at this point? I was thinking that they attached the block and it was already oscillating, and the 0.25 m below the original equilibrium was the maximum displacement of the oscillation, before it begins to move back upwards. That is incorrect isn't it?

No, the equilibrium position with block attached is not 0.25 m below the end of the spring without block. You have to calculate this from the given data. (Hint: it will be less than 0.25 m). Now imagine that the block is in the equilibrium position (the one you have to calculate) and then you pull it down some more, so it will be at the 0.25 m position, at rest.
When you release the block, it will go up, pass through the equilibrium position and so on.

If the equilibrium position with the block were right there, at 0.25 m, and there is no initial speed, why would the block start to oscillate?

Homework Helper
A block weighing 21 N oscillates at one end of a vertical spring for which k = 130 N/m; the other end of the spring is attached to a ceiling. At a certain instant the spring is stretched 0.25 m beyond its relaxed length (the length when no object is attached) and the block has zero velocity. (a) What is the net force on the block at this instant? What are the (b) amplitude and (c) period of the resulting simple harmonic motion? (d) What is the maximum kinetic energy of the block as it oscillates?
So the block is oscillating and it is below the original relaxed length by 0.25 m. See attachment.

If you hang the block on the spring and let it occupy the equilibrium position slowly, so it is not oscillating, the spring stretches by ΔLo=mg/k. Moving the block out from equilibrium, it will oscillate about the equilibrium position, so the length of the spring changes between ΔLo+A and ΔLo-A.

ehild

Edit: figure replaced with a corrected one.

#### Attachments

• springblock.JPG
11 KB · Views: 496
Last edited:
nasu
@ehild
The block does not have to reach the position of the relaxed spring without block.
2A does not have to be 0.25 m. And is not, in this problem. (ΔLo is not 0.125 m.)

Homework Helper
@ehild
The block does not have to reach the position of the relaxed spring without block.
2A does not have to be 0.25 m. And is not, in this problem. (ΔLo is not 0.125 m.)

@nasu,

You are right, thank you for correcting me.
(I have drawn the same figure too many times, which illustrated the case when the block was hanged on the relaxed spring and released, so it started oscillation from the relaxed length. The drawing was from my memory... )

ehild

nasu
Yes, the new figure is OK.
I hope it will help the OP to understand the problem.

BobRoss
So if I find

ΔLo=mg/k
ΔLo=21/130
ΔLo=0.162m

Then the ΔLo+A=0.25m

Is that correct?

So if I solve for A we get 0.088m. Is that the amplitude here?

Homework Helper
Yes, it is correct.

ehild