# Homework Help: [CONFUSED]Question about Hooke's Law

1. Nov 22, 2009

### qazxsw11111

If I have a spring with a load and I oscillate it freely, applying hooke's law,

TOP: GPE+EPE(Given by area under F-x graph)
Equilibrium:KE (No EPE since x=0)
Bottom:EPE Only

Since energy is conserved, I have to assume that the extension (actually compression) at the top is less than extension at bottom as there is also GPE also at the top.

Is this reasoning correct?

But what about an oscillating spring being a example of simple harmonic motion? Isnt the amplitude from the equilibrium position supposed to be the same?

Anyone help?

2. Nov 22, 2009

### tiny-tim

Hi qazxsw11111!

Everything is shifted slightly.

The equation of motion is x'' = -kx - mg,

so put y = x + mg/k, then y'' = -ky, which is SHM, with equal amplitude either side of the equilibrium position, but with the equilibrium position where the mass would remain at rest.

Last edited: Nov 22, 2009
3. Nov 22, 2009

### qazxsw11111

Hi Tiny-Tim =),

How do you get a = -kx - mg ?

What is actually the loophole in my reasoning:

Quoted: "Since energy is conserved, I have to assume that the extension (actually compression) at the top is less than extension at bottom as there is also GPE also at the top."

At the top of the oscillation, both GPE and EPE are present right?

I appreciate the help but still I can't help feeling confused.

4. Nov 22, 2009

### tiny-tim

(ooh, i left out an m … it should be mx'' = -kx - mg and my'' = - ky )
Hi qazxsw11111!

Good ol' Newton's second law … F = ma

the F is -kx for the spring and -mg for gravity.
Your reasoning is correct, but it doesn't contradict the actual result …

everything is displaced slightly, by a distance mg/k.

The extension (actually compression) at the top is less than the extension at bottom, if you measure it from the equilibrium-position-without-gravity (eg, if the spring is horizontal), but they are the same if you measure from a position mg/k lower.