MHT CET · Physics · Oscillations
A particle is executing S.H.M. of amplitude 'A'. When the potential energy of the particle is half of its maximum value during the oscillation, its displacement from the equilibrium position is
- A \(\pm \frac{\mathrm{A}}{4}\)
- B \(\pm \frac{\mathrm{A}}{2}\)
- C \(\pm \frac{\mathrm{A}}{\sqrt{3}}\)
- D \(\pm \frac{\mathrm{A}}{\sqrt{2}}\)
Answer & Solution
Correct Answer
(D) \(\pm \frac{\mathrm{A}}{\sqrt{2}}\)
Step-by-step Solution
Detailed explanation
\(PE = \frac{1}{2}PE_{max}\) \(\frac{1}{2}kx^2 = \frac{1}{2}\left(\frac{1}{2}kA^2\right)\)
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