A particle starts from mean position and performs S.H.M. with period 6 second. At what time its kinetic…

A particle starts from mean position and performs S.H.M. with period 6 second. At what time its kinetic energy is $50 \%$ of total energy? $\left(\cos 45^{\circ}=\frac{1}{\sqrt{2}}\right)$
  1. $0.75$ second
  2. 1 second
  3. $0.25$ second
  4. $0.50$ second

Solution

$T=6 \mathrm{~sec}$ $\frac{1}{2} m \omega^{2}\left(A^{2}-x^{2}\right)=\frac{1}{2}\left(\frac{1}{2} m \omega^{2} A^{2}\right.)$ $A^{2}-x^{2}=\frac{A^{2}}{2}$ $\therefore x^{2}=\frac{A^{2}}{2} \Rightarrow x=\frac{A}{\sqrt{2}}$ $\frac{A}{\sqrt{2}}=A \sin \omega t=A \sin \frac{2 \pi}{T} t$ $\frac{1}{\sqrt{2}}=\sin \frac{2 \pi}{6} t=\sin \frac{\pi}{3} t$ $\sin \frac{\pi}{4} t=\sin \frac{\pi}{3} t$ $\therefore t=\frac{3}{4}=0.75 \mathrm{sec}$

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

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