Finding the maximum kinetic energy of a harmonic oscillator

AI Thread Summary
The discussion centers on calculating the maximum kinetic energy (Km) of a harmonic oscillator with a mass on a string, given a hanging mass of 2m, spring constant K, and amplitude Z. It is established that the maximum kinetic energy equals the maximum potential energy due to the conservation of total energy. The relationship between angular frequency (ω), spring constant (k), and mass (m) is noted, with ω^2 = k/2m. The maximum velocity is expressed as ωZ, and the velocity at a displacement of Z/2 is derived using v^2 = ω^2(A^2-x^2). Ultimately, the solution reveals that Km equals 4Ui, confirming the calculations align with the principles of harmonic motion.
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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!
 
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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
 
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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