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?
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