The amplitude of a damped oscillator becomes $\left(\frac{1}{3} ight)^{ ext {rd }}$ in 2 seconds.

Question

The amplitude of a damped oscillator becomes $\left(\frac{1}{3}ight)^{	ext {rd }}$ in 2 seconds. If its amplitude after 6 seconds. If its amplitude after 6 seconds is $\frac{1}{\mathrm{n}}$ times the original amplitude, the value of n is\

Professor PC Thomas Classes · Answer

Amplitude of a damped oscillator at any instant $t$ is given by\ $\mathrm{A}=\mathrm{A}_{0} \mathrm{e}^{-\mathrm{h} / 2 \mathrm{~m}}$\ Where $\mathrm{A}_{0}$ is the original amplitude\ When $t=2 s, A=\frac{A_{0}}{3}$\ $\frac{\mathrm{A}_{0}}{3}=\mathrm{A}_{0} \mathrm{e}^{-2 \mathrm{~h} / 2 \mathrm{~m}}$ $$ \begin{equation*} \frac{1}{3}=\mathrm{e}^{-\mathrm{b} / \mathrm{m}} 	ag{i} \end{equation*} $$when $\mathrm{t}=6 \mathrm{~s}, \mathrm{~A}=\frac{\mathrm{A}_{0}}{\mathrm{n}}$\ $	herefore \frac{\mathrm{A}_{0}}{\mathrm{n}}=\mathrm{A}_{0} \mathrm{e}^{-6 \mathrm{~b} / 2 \mathrm{~m}}$\ $\frac{1}{\mathrm{n}}=\mathrm{e}^{-3 \mathrm{~b} / \mathrm{m}}=\mathrm{e}^{(-\mathrm{b} / \mathrm{m})^{3}}$\ $\frac{1}{\mathrm{n}}=\left(\frac{1}{3}ight)^{3}$\ (Using (i))\ $	herefore \mathrm{n}=3^{3}$\