# Simple Harmonic Motion

1. May 23, 2010

### fishingspree2

1. The problem statement, all variables and given/known data
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
Fs=k times x

3. 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: Fs=mg
kx = mg
so x = mg/k = 0.05*9.8 / 4 = 0.1225 meters

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

2. May 23, 2010

### Staff: Mentor

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?

3. May 23, 2010

### fishingspree2

Here's my thinking:

mg is always acting on the block. when we let go of the block, mg is acting on it and Fs is 0. As the blocks fall down, Fs 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 Fs is equal to mg. At t+dt, Fs is greater than mg and the block starts to go up.

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

4. May 23, 2010

### 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.

5. May 23, 2010

### Staff: Mentor

Would you agree that as the block falls, it moves faster? (Since there is a net downward force on it.)
There will certainly be a moment when Fs = 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 Fs > 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?

6. May 23, 2010

### 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?

7. May 23, 2010

### tomwilliam

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.

8. May 23, 2010

### Staff: Mentor

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.

9. May 23, 2010

### 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?

Last edited: May 23, 2010
10. May 23, 2010

### Staff: Mentor

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

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

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?

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

11. May 23, 2010

### fishingspree2

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

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

Is that correct?

Last edited by a moderator: May 4, 2017
12. May 23, 2010

### fishingspree2

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

13. May 23, 2010

### matonski

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 $\frac{1}{2} k(2A)^2$

14. May 24, 2010

### Staff: Mentor

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

15. May 24, 2010

### fishingspree2

I think im 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

16. May 24, 2010

### 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.

17. May 24, 2010

### Staff: Mentor

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

18. May 24, 2010