A body performs linear S.H.M. with amplitude 'a'. When it is at a distance $\frac{\mathrm{a}}{3}$ from…

A body performs linear S.H.M. with amplitude 'a'. When it is at a distance $\frac{\mathrm{a}}{3}$ from extreme position, the magnitude of velocity is $\frac{1}{3}$ times the magnitude of acceleration. The period of S.H.M. is
  1. $\frac{3 \pi}{2 \sqrt{5}}\mathrm{~s}$
  2. $\frac{5 \pi}{3 \sqrt{5}}\mathrm{~s}$
  3. $\frac{4 \pi}{3 \sqrt{5}}\mathrm{~s}$
  4. $\frac{\pi}{3 \sqrt{5}}\mathrm{~s}$

Solution

$a=$ amplitude $x=a-\frac{a}{3}=\frac{2 a}{3}$ $a_{p}=\omega^{2} x=\omega^{2} \frac{2 a}{3}$ $v_{p}=\omega \sqrt{a^{2}-x^{2}}=\omega \sqrt{a^{2}-\frac{4 a^{2}}{9}}=\omega \sqrt{\frac{5 a^{2}}{9}}$ $=\omega \sqrt{5} \frac{\mathrm{a}}{3}$ $3 v_{p}=a_{p}$ $\frac{v_{p}}{a_{p}}=\frac{1}{3}=\frac{\omega \sqrt{5} a / 3}{\omega^{2} \frac{2 a}{3}}=\frac{\sqrt{5}}{2 \omega}$ $3=\frac{2 \omega}{\sqrt{5}}=\frac{2}{\sqrt{5}} \times \frac{2 \pi}{T}$ $\therefore \mathrm{T}=\frac{4 \pi}{3 \sqrt{5}}$

Asked in: MHT CET 2020 (20 Oct Shift 2)

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