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MHT CET · Physics · Oscillations

A particle starts oscillating simple harmonically from its mean position with time period ' \(T\) '. At time \(t=\frac{T}{6}\), the ratio of the potential energy to kinetic energy of the particle is
\(\left[\sin 30^{\circ}=\cos 60^{\circ}=0 \cdot 5, \cos 30^{\circ}=\sin 60^{\circ}=\sqrt{3} / 2\right]\)

  1. A \(1: 2\)
  2. B \(1: 3\)
  3. C \(2: 1\)
  4. D \(3: 1\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(3: 1\)

Step-by-step Solution

Detailed explanation

\(\frac{PE}{KE} = \tan^2(\omega t)\) \(\omega t = \frac{2\pi}{T} \cdot \frac{T}{6} = \frac{\pi}{3}\)