Energy loss of damped oscillator

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SUMMARY

The discussion focuses on deriving the energy loss of a damped oscillator, represented by the equation X(t) = A exp(-Beta*t)cos(wt-delta). The user initially struggles with the complexity of the energy expression, E = K + U, which includes terms involving mass (m), amplitude (A), and damping coefficient (Beta). The key insight provided is that in the underdamped regime, the energy loss can be simplified to an exponential decay, with the loss rate being proportional to the square of the velocity (v^2). This allows for a clearer understanding of energy loss over time.

PREREQUISITES
  • Understanding of damped harmonic motion
  • Familiarity with calculus, specifically differentiation and integration
  • Knowledge of energy conservation principles in oscillatory systems
  • Basic grasp of exponential functions and their properties
NEXT STEPS
  • Study the derivation of energy expressions for damped oscillators
  • Learn about the underdamped regime in oscillatory systems
  • Explore the mathematical techniques for integrating velocity squared (v^2) over time
  • Investigate the effects of damping coefficients on oscillation behavior
USEFUL FOR

Students and professionals in physics, particularly those studying mechanics and oscillatory motion, as well as engineers working with systems involving damped oscillations.

Bill12
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Hi,

I do not know how to drive an experession for energy loss of damped oscillator.I know that:

X(t)=A exp(-Beta*t)cos(wt-delta)
and:
v=dx/dt...
I found E=K+U
but it seems to be so messy. It is like:

E=(1/2)*m*(A^2)*exp(-2*beta*t)[ beta^2 (cos(wt-delta))^2)+beta*

sin 2(wt-delta)+w^2 ]

I do not know if it is right or not but also I do not know how to get the energy loss from it.

I will thank for help.
 
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Do you want loss, or loss rate? Loss rate is easy. dE/dt is proportional to v^2. Integrating results indeed in a complicated expression. But if you are in an underdamped regime (many oscillations before the movement decays away), the expression represents an exponential decay with a small modulation on top of it. If you are not interested in the small modulation, the expression is very simple. You can find it for example by finding 1/2 mv^2 at those times when x goes through zero.
 
Finally solved it.
Thanks for help.
 

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