Fractional energy in a damped harmonic oscillator

AI Thread Summary
The discussion focuses on deriving the fractional energy lost per period in a damped harmonic oscillator, expressed as ΔE/E = 2πb/(mω₀) = 2π/Q. The attempt at a solution involves calculating the change in energy over one cycle and applying the Taylor series to simplify the exponential term. The negative sign in the energy change is clarified, emphasizing that while the change in energy is negative, the fractional energy lost is a positive quantity. The importance of assuming small damping (b << mω₀) is noted to justify dropping higher-order terms in the Taylor expansion. The conversation concludes with a clear distinction between the change in energy and the actual energy lost during one cycle.
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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?
 
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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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