Simple Harmonic Motion/Energy: Damped Oscillations and Energy Dissipation

AI Thread Summary
The discussion focuses on calculating the mechanical energy dissipated from a damped oscillating block-spring system. The initial total mechanical energy is determined to be 2.7 J, based on the spring constant and initial amplitude. After 12 seconds, the maximum amplitude is calculated using the damping time constant, resulting in a remaining mechanical energy of approximately 0.0067 J. By subtracting the remaining energy from the initial energy, the dissipated energy is found to be about 2.693 J. This calculation confirms the energy loss due to damping over the specified time period.
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Homework Statement



Problem:
A 2.0 kg block oscillates up and down on a spring with spring constant 240 N/m. Its initial amplitude is 15 cm. If the time constant ("tau") for damping of the oscillation is 4.0 s, how much mechanical energy has been dissipated from the block-spring system after 12 s?

Homework Equations



U_sp = 0.5kx^2
x_max(t) = Ae^(-t/tau)

The Attempt at a Solution



I only have one attempt at this problem left, and this is the best I could come up with, so I need this to be verified...

First I found the initial total mechanical energy (U_sp) within the system:
U_sp = (0.5)(240)(0.15)^2 = 2.7 J

Then I found the maximum amplitude (x_max) for the given time:
x_max(12) = 0.15e^(-12/4) = 0.007468060255 m

I plugged this value back into the spring's potential energy equation to find the remaining mechanical energy left in the system:
U_sp = (0.5)(240)(0.007468060255)^2 = 0.006692630877 J

I subtracted this final energy from its initial to find the dissipated energy from the system:
|deltaE| = 2.7 - 0.006692630877 = 2.69330736912 J
 
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Therefore, the amount of mechanical energy that has been dissipated from the block-spring system after 12 s is 2.69330736912 J.
 
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