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An undamped oscillator has period ro = 1.000 s, but I now
Chapter 5, Problem 5.26(choose chapter or problem)
An undamped oscillator has period \(\tau_{\mathrm{o}}=1.000 \mathrm{~s}\), but I now add a little damping so that its period changes to \(\tau_{1}=1.001 \mathrm{~s}\). What is the damping factor \(\beta\)? By what factor will the amplitude of oscillation decrease after 10 cycles? Which effect of damping would be more noticeable, the change of period or the decrease of the amplitude?
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QUESTION:
An undamped oscillator has period \(\tau_{\mathrm{o}}=1.000 \mathrm{~s}\), but I now add a little damping so that its period changes to \(\tau_{1}=1.001 \mathrm{~s}\). What is the damping factor \(\beta\)? By what factor will the amplitude of oscillation decrease after 10 cycles? Which effect of damping would be more noticeable, the change of period or the decrease of the amplitude?
ANSWER:Step 1 of 4
The angular frequency change due to damping can be expressed as,
\(\begin{aligned} \omega_{1} & =\sqrt{\omega_{0}^{2}-\beta^{2}} \\ \omega_{1}^{2} & =\omega_{0}^{2}-\beta^{2} \\ \beta^{2} & =\omega_{0}^{2}-\omega_{1}^{2} \\ \beta & =\omega_{0} \sqrt{1-\frac{\omega_{1}^{2}}{\omega_{0}^{2}}} \end{aligned}\) ….. (1)
The ratio of periods of undamped oscillator to the little damping (damped oscillator) can be expressed as,
\(\frac{\omega_{1}^{2}}{\omega_{0}^{2}}=\frac{\tau_{0}^{2}}{\tau_{1}^{2}}\)
For \(\tau_{0}=1.000 \mathrm{~s}\), \(\tau_{1}=1.001 \mathrm{~s}\),
\(\begin{array}{l} \frac{\omega_{1}^{2}}{\omega_{0}^{2}}=\frac{(1.000)^{2}}{(1.001)^{2}} \\ \frac{\omega_{1}^{2}}{\omega_{0}^{2}}=0.998 \\ \omega_{1}^{2}=0.998 \omega_{0}^{2} \end{array}\)
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