What is the Total Energy of a Spring-Mass System in Vertical Oscillations?

[stryker123]

Homework Statement

A mass 'm' is attached to the free end of a spring (unstretched = l) of spring constant 'k' and suspended vertically from ceiling. The spring stretches by Δl under the load andd comes to equilibrium position. The mass is pushed up vertically by "A" from its equilibrium position and released from rest. The mass-spring executes vertical oscillations. Show from energy considerations that the total energy of the spring-mass system is (1/2)k[Δl^2 + A^2] assuming the gravitational potential energy is zero at the equilirium position of the mass 'm'

Homework Equations

PE= (1/2)kx^2
KE= (1/2)mv2
Total Energy = PE + KE
Oscillating systems = x=Asinωt
v= Aωcosωt

The Attempt at a Solution

Since energy is conserved, I should get (kx^2/2) + (mv^2/2) = E. Therefore, I can substitute in (kA^2sin^2ωt)/2 + (mA^2ω^2cos^2ωt)/2 = E

Since ω^2 = k/m, I get (kA^2/2)sin^2ωt + (kA^2/2)cos^2ωt = E.
By using trig identities, I can reduce this to E = (KA^2)/2.

This is close to what I'm supposed to get, but I'm not quite there yet. Can anyone help me?

 

[stryker123]

so, I got the equation to equal E = (KA^2)/2 yet I do not see where the 'Delta' l comes in. Is this something that I can just add on?

 

[Dick]

The initial displacement of the spring by dl to equilibrium tells you the relation between mg and k*dl. Then figure out total energy when the spring is pushed up by A. It's (1/2)*k*(A-dl)^2+mgA, right? Now use your relation between mg and k*dl to get that to the final form. I think that is all you need to do is to compute the energy at that one point. You know energy is conserved after you release the spring, right?