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MHT CET · Physics · Oscillations

For a particle performing S.H.M., the total energy is ' n ' times the kinetic energy, when the displacement of a particle from mean position is \(\frac{\sqrt{3}}{2}\) A, where \(A\) is the amplitude of S.H.M. The value of ' \(n\) ' is

  1. A \(2\)
  2. B \(3\)
  3. C \(4\)
  4. D \(6\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(4\)

Step-by-step Solution

Detailed explanation

\(E = n \cdot K\) \(K = \frac{1}{2} m \omega^2 (A^2 - x^2)\)