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(C) The amplitude of a damped oscillator is given by $A(t) = A_0 e^{-(b/2m)t}$.Given that after $n = 100$ oscillations, the amplitude becomes $A_0/e$,

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$\frac{\pi^2}{600} \%$

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(C) The amplitude of a damped oscillator is given by $A(t) = A_0 e^{-(b/2m)t}$.Given that after $n = 100$ oscillations, the amplitude becomes $A_0/e$, so $e^{-(b/2m)T \cdot n} = e^{-1}$, where $T$ is the time period.Thus, $\frac{b}{2m} T \cdot n = 1$. Since $T = \frac{2\pi}{\omega}$, we have $\frac{b}{2m} \cdot \frac{2\pi}{\omega} \cdot 100 = 1$, which implies $\frac{b}{2m\omega} = \frac{1}{200\pi}$.The damped frequency is $\omega^{\prime} = \omega \sqrt{1 - \frac{b^2}{4m^2\omega^2}}$.Using the binomial approximation $(1-x)^{1/2} \approx 1 - x/2$ for small $x$, we get $\omega^{\prime} \approx \omega \left(1 - \frac{b^2}{8m^2\omega^2}ight)$.Therefore, $\frac{\omega - \omega^{\prime}}{\omega} = \frac{b^2}{8m^2\omega^2} = \frac{1}{2} \left(\frac{b}{2m\omega}ight)^2$.Substituting the value, $\frac{\omega - \omega^{\prime}}{\omega} = \frac{1}{2} \left(\frac{1}{200\pi}ight)^2 = \frac{1}{2} \cdot \frac{1}{40000\pi^2} = \frac{1}{80000\pi^2}$.Converting to percentage: $\frac{1}{80000\pi^2} 	imes 100 = \frac{1}{800\pi^2} \%$.

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