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MHT CET · Physics · Oscillations

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. A \(0.75\) second
  2. B 1 second
  3. C \(0.25\) second
  4. D \(0.50\) second
Verified Solution

Answer & Solution

Correct Answer

(A) \(0.75\) second

Step-by-step Solution

Detailed explanation

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