Energy of an damped/undriven oscillator in terms of time?

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SUMMARY

The discussion focuses on deriving the time rate of change of mechanical energy for a damped, undriven oscillator, specifically showing that dE/dt = -bV². The total energy equation is given as E = (1/2)mV² + (1/2)kx². Participants suggest substituting the position equation x(t) = (Ae^(-βt))cos(ωt - δ) into the energy equation and then differentiating to find the desired form. The challenge lies in expressing E in terms of time, E(t), and correctly applying derivatives.

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  • Understanding of mechanical energy concepts in oscillatory systems
  • Familiarity with differential calculus and derivatives
  • Knowledge of damped oscillators and their equations of motion
  • Experience with trigonometric functions and exponential decay
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Homework Statement



The Q asks to show that the time rate of change in mechanical energy for a damped, undriven oscillator is dE/dt=-bV^2.

Homework Equations



I assume you take the derivative of the total E eq, E=(1/2)mV^2 + (1/2)kx^2 but I'm unsure how to put the E eq into terms of t, like E(t).

The Attempt at a Solution



Would you have to punch in the pos eq [x(t)=(Ae^(-βt))cos(ωt-δ)] in for x, then its derivative in for V? and then takes that entire eq's derivative? Seems like it would be too much work and not enough concept.
 
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i can't seem to get it in the form dE/dt=-bV^2.
 

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