MHT CET · Physics · Oscillations
A particle is executing linear S.H.M. starting from mean position. The ratio of the kinetic energy to the potential energy of the particle at a point of half the amplitude is
- A \(2: 1\)
- B \(3: 1\)
- C \(4: 1\)
- D \(8: 1\)
Answer & Solution
Correct Answer
(B) \(3: 1\)
Step-by-step Solution
Detailed explanation
\(PE = \frac{1}{2} m \omega^2 x^2 = \frac{1}{2} m \omega^2 (\frac{A}{2})^2 = \frac{1}{8} m \omega^2 A^2\) \(KE = \frac{1}{2} m \omega^2 (A^2 - x^2) = \frac{1}{2} m \omega^2 (A^2 - (\frac{A}{2})^2) = \frac{1}{2} m \omega^2 (A^2 - \frac{A^2}{4}) = \frac{3}{8} m \omega^2 A^2\)
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