In a time of 2 s , the amplitude of a damped oscillator becomes $\frac{1}{e}$ times its initial amplitude A.…

In a time of 2 s , the amplitude of a damped oscillator becomes $\frac{1}{e}$ times its initial amplitude A. In the next two seconds, the amplitude of the oscillator is
  1. $\frac{1}{2 e}$
  2. $\frac{2}{e}$
  3. $\frac{1}{e^2}$
  4. $\frac{2}{e^2}$

Solution

$\mathrm{t}=2 \mathrm{~s}, \mathrm{~A}^{\prime}=\frac{\mathrm{A}}{\mathrm{e}}$ In damped oscillation, $\mathrm{A}^{\prime}=\mathrm{A} \mathrm{e}^{-\left(\frac{\mathrm{b}}{2 \mathrm{M}}\right) \mathrm{t}}$ $\Rightarrow \frac{\mathrm{A}}{\mathrm{e}}=\mathrm{Ae} \mathrm{e}^{-\left(\frac{\mathrm{b}}{2 \mathrm{~m}}\right) \times 2} \Rightarrow \frac{\mathrm{~b}}{2 \mathrm{M}}=\frac{1}{2}$ $\therefore \quad \mathrm{A}^{\prime}=\mathrm{A}^{-\mathrm{t} / 2}$ $\therefore \quad$ At. $t=4 \mathrm{~s}, \mathrm{~A}^{\prime}=\mathrm{Ae}^{-4 / 2}=\frac{\mathrm{A}}{\mathrm{e}^2}$

Asked in: AP EAMCET 2024 (20 May Shift 1)

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