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
\(\left[\sin 30^{\circ}=\cos 60^{\circ}=0 \cdot 5, \cos 30^{\circ}=\sin 60^{\circ}=\sqrt{3} / 2\right]\)
- A \(1: 2\)
- B \(1: 3\)
- C \(2: 1\)
- D \(3: 1\)
Answer & Solution
Correct Answer
(D) \(3: 1\)
Step-by-step Solution
Detailed explanation
\(\frac{PE}{KE} = \tan^2(\omega t)\) \(\omega t = \frac{2\pi}{T} \cdot \frac{T}{6} = \frac{\pi}{3}\)
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