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Finding the maximum kinetic energy of a harmonic oscillator

  • Thread starter apbuiii
  • Start date
  • #1
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Homework Statement


A harmonic oscillator with a vertical mass on a string has a hanging mass of 2m and a spring constant of K. It oscillates with an amplitude of Z. When its position is at a distance Z/2 of the equilibrium point, its potential energy is Ui. What is the maximum kinetic energy Km?


Homework Equations


E=1/2mV2 + 1/2mA2


The Attempt at a Solution


I realized that the max kinetic energy must be equal to the to the max potential energy because total energy is constant. I just don't know where to put the equations to make it fit.

Can someone show me the set up? The answer give was Km = 4Ui

Thanks in advance!
 

Answers and Replies

  • #2
1,506
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Can you see that ω^2 = k/2m ?
Also max velocity = ωZ
So you can get an expression for the max KE
The velocity at a displacement x is given by v^2 = ω^2(A^2-x^2)
can you use this to get the KE at a displacement Z/2
If you combine this lot you should get the answer
Hope this helps..... I will follow this up if you have trouble
 
  • #3
14
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I don't think I used your method, but I did end up figuring it out. Thanks for your help though.
 

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