Calculating Energy in Simple Harmonic Motion: Mass-Spring System

[doc1388]

Homework Statement

A mass "m" is attached to the free end of a light vertical spring (unstretched length L) of spring contant "k" and suspended from a ceiling. The spring streches by Delta L under the load and comes to equilibrium 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 the ground level, show that the total energy of the spring-mass system is (1/2)k{Delta L^2 + A^2} + mgh.

Homework Equations

x(t)= Asin(Omega t + phi)
v(t)= wAcos(Omega t +phi)
Total Energy = 1/2 k A^2
Delta L k = mg

Hint: Draw four sketches of the vertical free spring, spring-mass in equilibrium, and the two extreme positions of oscillation of the mass. Below each of the above sketches (except free spring) write equations for Kinetic Energy, Elastic Potential Energy, Gravitational Potential Energy and Total Energy using symbols k, Delta L, m, A, g, h and v_m_

The Attempt at a Solution

I am struggling trying to find where Delta L fits into this equation. I understand the change in energies at the top and bottom are both full PE and GravPE and while it passes through equilibrium the KE is max because velocity is max. I just don't know how to even go about this problem I guess. I've labeled my drawing with the differences in h where at equilibrium y=h and the top y=h+A and at the bottom y=h-A. I'm lost. HELP please. Thank you in advance.

 

[doc1388]

I have attempted a bit more to this and I believe at equilibrium the total energy of the system is split up into three parts: Kinetic Energy is 1/2 m v_max_^2 + Elastic Potential Energy which is 1/2 k Delta L^2 + mgh

This is the answer if I sub in Omega and such and cancel masses I end up with the answer. But this is only the answer for equilibrium and I can't prove it is true at the extrema.

Any ideas?

 

[nlightner]

I would approach this problem by first understanding the physical situation described in the problem and then using relevant equations and principles to solve it. In this case, we have a mass-spring system that is undergoing simple harmonic motion. This means that the mass attached to the spring is oscillating vertically with a certain amplitude and frequency.

To solve this problem, we can start with the equation for the potential energy of a spring, which is given by PE = 1/2 kx^2, where k is the spring constant and x is the displacement from the equilibrium position. In this case, the displacement is given by Delta L + A, since the spring is already stretched by Delta L and the mass is pushed up by A from its equilibrium position. So, the potential energy of the spring at the top of the oscillation (when the mass is at its maximum height h+A) is given by PE_top = 1/2 k(Delta L + A)^2.

Similarly, at the bottom of the oscillation (when the mass is at its minimum height h-A), the potential energy of the spring is given by PE_bottom = 1/2 k(Delta L - A)^2.

Now, let's consider the kinetic energy of the mass. At the top of the oscillation, the mass is at rest and hence, the kinetic energy is zero. At the bottom of the oscillation, the mass has maximum velocity and hence, the kinetic energy is given by KE_bottom = 1/2 mv_m^2, where v_m is the maximum velocity of the mass.

Next, we need to consider the gravitational potential energy of the mass. At the ground level (y=0), the gravitational potential energy is zero. At the top of the oscillation, the mass is at a height h+A above the ground level and hence, the gravitational potential energy is given by GravPE_top = mgh + mgh = 2mgh (since the mass travels twice the distance h+A).

Using the equations for potential energy and kinetic energy at the top and bottom of the oscillation, we can write the total energy of the system as:

Total Energy = PE_top + KE_bottom + GravPE_top
= 1/2 k(Delta L + A)^2 + 1/2 mv_m^2 + 2mgh

Simplifying this expression, we get:

Total Energy = 1/2