Why Is My Calculation of Maximum Stretch in Simple Harmonic Motion Incorrect?

[fishingspree2]

Homework Statement

A 50 g block is hung from a vertical spring. The spring constant is 4 N/m. We let go of the block when the spring is not stretched.What is the maximal stretch of the spring?

2. Equations
F_s=k times x

The Attempt at a Solution

At the maximal stretch, the block is at equilibrium since it stops falling down. Gravity force is then equal to the spring force.

we have: F_s=mg
kx = mg
so x = mg/k = 0.05*9.8 / 4 = 0.1225 meters

This answer is wrong.

Can anyone help me? My reasoning looks logic to me :(

 

[Doc Al]

At the maximal stretch, the block is at equilibrium since it stops falling down.

This is incorrect. Note that they "let go" of the block, allowing it to fall. (As opposed to lowering it gently.) When it reaches the lowest point, does the block just sit there?

 

[fishingspree2]

This is incorrect. Note that they "let go" of the block, allowing it to fall. (As opposed to lowering it gently.) When it reaches the lowest point, does the block just sit there?

Hello, thank you for your reply.

Here's my thinking:

mg is always acting on the block. when we let go of the block, mg is acting on it and F_s is 0. As the blocks fall down, F_s grows and grows but it is still smaller than mg. Then, just as the spring reaches its maximum stretch, there is a time t at which F_s is equal to mg. At t+dt, F_s is greater than mg and the block starts to go up.

Can you please tell me what is wrong with that reasoning?

 

[tomwilliam]

That reasoning would work if the block was lowered very slowly and could not be said to have speed. At the point that kx is equal to mg, the forces will be balanced but the block will actually have a speed in the downward direction. What you want to find out is the extension at which the force has reduced the speed to zero.

 

[Doc Al]

mg is always acting on the block. when we let go of the block, mg is acting on it and F_s is 0. As the blocks fall down, F_s grows and grows but it is still smaller than mg.

Would you agree that as the block falls, it moves faster? (Since there is a net downward force on it.)

Then, just as the spring reaches its maximum stretch, there is a time t at which F_s is equal to mg. At t+dt, F_s is greater than mg and the block starts to go up.

There will certainly be a moment when F_s = mg (the equilibrium point), but by then the block is moving fast and there's no force slowing it down. As it continues to fall, it begins to slow, since now F_s > mg and there is a net upward force. It keeps falling, moving slower and slower, until reaching its lowest point. But there is an upward force acting, so the block begins rising again, repeating the cycle.

Make sense?

 

[fishingspree2]

Ok, it makes sense
We have:
$$x\left(t\right)=A\sin\left(\omega t+\phi\right)$$
Taking the derivative with respect to t, we have:
$$v\left(t\right)=A\omega\cos\left(\omega t+\phi\right)$$

We know that $$\omega=\sqrt\frac{k}{m}=8.944$$

so $$v\left(t\right)=8.944A\cos\left(8.944 t+\phi\right)$$

We don't know the time t at which it happens, we don't know phi and we are looking for A

I am stuck

EDIT: Can we use the fact that a period is equal to 2 Pi over omega and assume that the spring will be at its maximum stretch when t is equal to half a period?

 

[tomwilliam]

Try using energy considerations instead.

If you use the stored strain energy of the spring as 1/2kx^2 when the spring is fully extended, and mgh as the potential energy when it's in equilibrium, you can equate these and define the right coordinate system to find x.
Seems the simplest way to do it, to me.

 

[Doc Al]

We don't know the time t at which it happens, we don't know phi and we are looking for A

I am stuck

The simplest way to find the amplitude is to compare the initial position (where the block was released) to the equilibrium position. (You've done the needed calculations in your first post.)

Or, as tomwilliam says, you can use energy methods to solve for the lowest point. Compare the initial energy (at the release point) to the energy at the lowest point. (Hint: Measure gravitational PE from the lowest point.)

Do it both ways.

 

[fishingspree2]

Ok here's what I did

Before we let go of the block, it is at x=A. When the spring is at its maximum stretch, the block is at x = -A

gravitationnal potential energy = total energy of the system
= m*g*A

mgA = mg(A-A)+0.5*k*A^2

A = 2mg / k = 0.245

correct?

I am trying to understand how to do this... what are the conditions for equilibrium position? v=0?

The simplest way to find the amplitude is to compare the initial position (where the block was released) to the equilibrium position. (You've done the needed calculations in your first post.)

 

[Doc Al]

Ok here's what I did

Before we let go of the block, it is at x=A. When the spring is at its maximum stretch, the block is at x = -A

So A is the amplitude and the maximum stretch of the spring will be 2A.

gravitationnal potential energy = total energy of the system
= m*g*A

Looks like you are measuring gravitational PE from the equilibrium point. Be careful! (Or measure gravitational PE from the lowest point, like I suggested.)

mgA = mg(A-A)+0.5*k*A^2

A = 2mg / k = 0.245

correct?

Not exactly. What's the spring PE at the lowest point? (How much is the spring stretched?) What's the gravitational PE at the lowest point?

I am trying to understand how to do this... what are the conditions for equilibrium position? v=0?

No, the equilibrium position is where Fs = mg, just like you figured out in your first post. That will immediately give you the amplitude.

 

[fishingspree2]

[PLAIN]http://img580.imageshack.us/img580/9219/harmonic.jpg

$$U=2mgA$$
$$E_{tot}=2mgA=\frac{1}{2}kA^{2}$$
$$A=\frac{4mg}{k}$$

Is that correct?

 

[fishingspree2]

No it's not correct and I don't see why

 

[matonski]

$$U=2mgA$$
$$E_{tot}=2mgA=\frac{1}{2}kA^{2}$$
$$A=\frac{4mg}{k}$$

Is that correct?

If you are defining zero potential to be at full stretch with the block starting at 2A, then at full stretch, the spring will be stretched by 2A and not just by A. Thus, the potential energy will be $\frac{1}{2} k(2A)^2$

 

[Doc Al]

$$U=2mgA$$
$$E_{tot}=2mgA=\frac{1}{2}kA^{2}$$
$$A=\frac{4mg}{k}$$

Is that correct?

As matonski has already pointed out, the spring stretches by 2A, not A. Fix your expression for spring PE.

 

[fishingspree2]

I think I am missing something. I thought A was mesured from the center to one of the extremities on the x(t) graph like on this picture. So from A to 0

 

[matonski]

Yes, but at the top of the motion, the problem specifically says the spring is unstretched. Thus, at the bottom, it will be stretched by 2A. Starting from the beginning, the spring is unstretched at y = A. At y = 0, the gravitational force matches the spring force. This is the equilibrium point. Finally, y = -A is the lowest position of the object. You are getting confused because the position where the spring is unstretched and the equilibrium position of the oscillation are not at the same place.

 

[Doc Al]

I think I am missing something. I thought A was mesured from the center to one of the extremities on the x(t) graph like on this picture.

Exactly. So the amount that the spring is stretched is 2A, since it starts out at the top with zero stretch.

 

[EngineerHead]

we have: Fs=mg
kx = mg
so x = mg/k = 0.05*9.8 / 4 = 0.1225 meters

fs, what you're solving for here is where the object reaches terminal velocity, not where it stops. Where it stops occurs when all the PEnergy from the block is transferred to the spring (stretched), and therefore you would use energy, not F=ma. Remember, setting F=0 is finding when a=0, but not v=0, to find v=0 you would use energy (integration of F with respect to displacement is energy, which is in terms of velocity and not acceleration, therefore you can set velocity equal to zero).