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
- A \(2\)
- B \(3\)
- C \(4\)
- D \(6\)
Answer & Solution
Correct Answer
(C) \(4\)
Step-by-step Solution
Detailed explanation
\(E = n \cdot K\) \(K = \frac{1}{2} m \omega^2 (A^2 - x^2)\)
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