The time taken for the amplitude of vibrations of a damped oscillator to drop to $25 \%$ of its initial…
The time taken for the amplitude of vibrations of a damped oscillator to drop to $25 \%$ of its initial value is $t$. The time taken for its mechanical energy to drop to $50 \%$ of its initial mechanical energy is
$\mathrm{t}$
$\frac{\mathrm{t}}{2}$
$\frac{t}{4}$
$\frac{t}{8}$
Solution
$\begin{aligned} & A=A_0 e^{-\frac{b t}{2 m}} \\ & A=25 \% \text { of } A_0=\frac{A_0}{4} \\ & \frac{A_0}{4}=A_0 e^{-\frac{b t}{2 m}} \\ & (2)^{-2}=e^{-\frac{b t}{2 m}} \\ & \frac{b t}{2 m}=2 \ln 2 \Rightarrow t=\frac{42 m \ell n 2}{b} ... (1) \\ & E(t)=E(0) e^{-\frac{b t}{2 m}} \\ & E(t)=50 \% \text { of } E_0 \\ & \frac{E_0}{2}=E_0 e^{-\frac{b t^{\prime}}{2 m}}\end{aligned}$
$\begin{aligned}
& (2)^{-1}=e^{-\frac{b t^{\prime}}{2 m}} \\
& \frac{b t^{\prime}}{2 m}=\ln 2 \\
& t^{\prime}=\frac{2 m \ln 2}{b}
\end{aligned}$
Divide equation (2) by (1), we have
$\frac{t^{\prime}}{t}=\frac{2 m \ln 2}{b} \times \frac{b}{4 m \ln 2} \quad t^{\prime}=\frac{t}{2}$