Solving for Total Energy of a Spring Mass System: Simple Harmonic Oscillations

[mathilin]

Homework Statement

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.

Homework Equations

EP = 1/2kx^2
K = 1/2kx^2
GE = mgh

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.

 

[LowlyPion]

Welcome to PF.

What is A?

 

[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.

 

[LowlyPion]

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

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

 

[mathilin]

Yes exactly, I just phrased it poorly.

 

[LowlyPion]

Yes exactly, I just phrased it poorly.

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

 

[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.

 

[LowlyPion]

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.

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?

 

[mathilin]

But what is the term 1/2*k*(-2AΔL)

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)

 

[LowlyPion]

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)

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)

 

[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!

 

[PhysicsDaoist]

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

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

 

[LowlyPion]

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

The m*g*h is there because the problem states

Assuming that gravitational potential energy of the mass "m" is zero at ground level, ...

... and they do want the total energy in the system.

 

[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.