# Simple Harmonic Oscillations

1. Apr 7, 2009

### mathilin

1. The problem statement, all variables and given/known data
I have to prove that the total energy of a spring mass system is equal to (1/2)k(delta L^2 + A^2)
The spring is in three sates, equilibrium (I proved that already), maximally stretched, and maximally compressed. The spring is at equilibrium at a height h above ground level. delta L is the amount that the string stretches after the weight is added to the free spring.

2. Relevant equations
EP = 1/2kx^2
K = 1/2kx^2
GE = mgh

3. The attempt at a solution
For the spring maximally compressed : I have KE = 0 becuase it is not moving. EP = (1/2)k*(A-delta L)^2. when that is extended EP = 1/2*k*(A^2-2AdeltaL+deltaL^2).
GE = m*g*(h-a). When all of these are added together it does NOT equal (1/2)k(delta L + A^2). I can't seem to find my error.

2. Apr 7, 2009

### LowlyPion

Welcome to PF.

What is A?

3. Apr 7, 2009

### mathilin

A is just the distance that it moves up or down in either dirrection. For example, at its max height, h is h+A.

4. Apr 7, 2009

### LowlyPion

So A is the amplitude and ΔL is the static displacement of an added weight?

5. Apr 7, 2009

### mathilin

Yes exactly, I just phrased it poorly.

6. Apr 7, 2009

### LowlyPion

Do you have the actual problem statement? Maybe that would mean fewer questions?

7. Apr 7, 2009

### mathilin

Yea here it is verbatim.

A mass "m" is attached to the free end of a light vertical spring (unstretched length l) of spring constant "k" and suspended form a ceiling. The spring stetches delta L under the load and comes to equlibrium at a height "h" above the ground level (y=0). The mass is pushed up vertically by "A" from its equilibrium position and released from rest. The mass-spring executes vertical oscillations. Assuming that gravitational potential energy of the mass "m" is zero at ground level, show that the total energy of the spring-mass system is (1/2)*k*(delta L2 + A2) + mgh.

I already did this for equilibrium, I can't quite figure it out for when it is at its max and min positions.

8. Apr 7, 2009

### LowlyPion

That makes a big difference having the m*g*h.

Consider the point at which you release the weight after pushing it up. Examine the sum of the Gravitational potential energy and the spring potential energy.

PE = m*g*(h + A)

SPE = 1/2*k*(A - ΔL)2 = 1/2*k*(A2 -2AΔL + ΔL2)

But what is the term 1/2*k*(-2AΔL) ?

Does that look familiar?

Does the expression mg = kΔL hold true? The weight = the displacement times spring constant?

9. Apr 7, 2009

### mathilin

It might by the -F of the spring, because F=-kx. If this were true, I don't see how this would help though.
Assuming that is correct the equation would reduce to.

(1/2)*k*(A^2 + F + ΔL^2) + mg(h+A)

Last edited: Apr 7, 2009
10. Apr 7, 2009

### LowlyPion

Sorry, I think you missed my point.

Total Energy = m*g*(h + A) + 1/2*k*(A2 -2AΔL + ΔL2)

Total Energy = m*g*(h + A) + 1/2*k*(A2 + ΔL2) - 1/2*k*2AΔL

Noting that mg = k*ΔL you get ...

Total Energy = m*g*(h + A) + 1/2*k*(A2 + ΔL2) - A*kΔL = m*g*(h + A) + 1/2*k*(A2 + ΔL2) - A*mg

Total Energy = m*g*h + 1/2*k*(A2 + ΔL2)

11. Apr 7, 2009

### mathilin

ohh, it makes so much sense once I see it mapped out. Thanks for all of your help, I'll do the other state now!

12. Apr 14, 2009

### PhysicsDaoist

Why is the mgh term still there? I thought the original question was asking one to show
Total Energy = 1/2*k*(A2 + ΔL2)

13. Apr 14, 2009

### LowlyPion

The m*g*h is there because the problem states
... and they do want the total energy in the system.

14. Apr 15, 2009

### PhysicsDaoist

You are absolutely right, I missed the Mathilin's verbatim restating of the original problem.
The frame of reference is at ground level and mgh is definitely needed.