Simple Harmonic Motion/Energy: Damped Oscillations and Energy Dissipation

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SUMMARY

The discussion focuses on calculating the mechanical energy dissipated from a damped oscillating block-spring system. A 2.0 kg block with a spring constant of 240 N/m and an initial amplitude of 15 cm experiences damping characterized by a time constant of 4.0 s. After 12 seconds, the total mechanical energy dissipated is calculated to be approximately 2.693 J, derived from the initial potential energy of 2.7 J and the remaining energy of 0.0067 J.

PREREQUISITES
  • Understanding of simple harmonic motion principles
  • Familiarity with potential energy equations, specifically U_sp = 0.5kx^2
  • Knowledge of exponential decay in oscillatory systems
  • Ability to perform calculations involving damping and energy dissipation
NEXT STEPS
  • Study the effects of varying the spring constant on energy dissipation
  • Learn about the role of mass in damped oscillations
  • Explore advanced topics in oscillatory motion, such as forced oscillations
  • Investigate real-world applications of damped harmonic motion in engineering
USEFUL FOR

Students studying physics, particularly those focusing on mechanics and oscillatory motion, as well as educators seeking to enhance their understanding of energy dissipation in damped systems.

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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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