Fractional energy in a damped harmonic oscillator

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SUMMARY

The fractional energy lost per period in a damped harmonic oscillator is defined by the equation \(\frac{\Delta E}{E} = \frac{2\pi b}{m\omega_0}\), where \(\omega_0 = \sqrt{\frac{k}{m}}\) and \(Q = \frac{m\omega_0}{b}\). The energy change over one cycle is calculated as \(\Delta E = \frac{1}{2} k A^2 e^{-(b/m)(t + T)} - \frac{1}{2} k A^2 e^{-(b/m)t}\), leading to \(\frac{\Delta E}{E} = e^{-\frac{2\pi b}{m\omega_0}} - 1\). The negative sign in the energy change indicates a loss, aligning with the requirement to express energy lost as a positive quantity.

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  • Understanding of damped harmonic motion
  • Familiarity with the Taylor series expansion
  • Knowledge of the concepts of energy in oscillatory systems
  • Basic proficiency in calculus and differential equations
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Homework Statement


Show that the fractional energy lost per period is
\frac{\Delta E}{E} = \frac{2\pi b}{m\omega_0} = \frac{2\pi}{Q}
where \omega_0 = \srqt{k/m} and Q = m\omega_0 / b

Homework Equations


E = 1/2 k A^2 e^{-(b/m)t} = E_0 e^{-(b/m)t}

The Attempt at a Solution


\Delta E = 1/2 k A^2 e^{-(b/m)(t + T)} - 1/2 k A^2 e^{-(b/m)t} where T = 2\pi / \omega_0
\frac{\Delta E}{E} = e^{-(b/m)T} - 1
\frac{\Delta E}{E} = e^{-\frac{2\pi b}{m\omega_0}} - 1
What should I do now?
 
Last edited:
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Using the Taylor series of e^x,
e^{-(2\pi b)/(m\omega_0)} = 1 + -\frac{2\pi b}{m\omega_0}
Can you explain why I would drop the rest of the terms?
So
\frac{\Delta E}{E} = -\frac{2\pi b}{m\omega_0}
first of all, is this right?
second, how do i account for the negative sign?
 
All this assumes the damping is small, i.e. b << m \omega_0, so you can drop the higher terms of the Taylor series.

Re the negative sign, the question asks for the fractional energy lost.

Energy lost in 1 cycle (a positive number) = Initial energy - final energy.

Your equation found the change in energy as

Change in energy in 1 cycle (a negative number) = final energy - initial energy.
 

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