WBJEE · Physics · Oscillations
A particle is executing linear simple harmonic motion of amplitude A. At what displacement is the energy of the particle half potential and half kinetic?
- A \(\frac{\mathrm{A}}{4}\)
- B \(\frac{\mathrm{A}}{2}\)
- C \(\frac{\mathrm{A}}{\sqrt{2}}\)
- D \(\frac{\mathrm{A}}{\sqrt{3}}\)
Answer & Solution
Correct Answer
(C) \(\frac{\mathrm{A}}{\sqrt{2}}\)
Step-by-step Solution
Detailed explanation
Hints : Total Energy \((\mathrm{E})=\frac{1}{2} m \omega^2 \mathrm{~A}^2\) P.E. \(=\frac{1}{2} m \omega^2 x^2\) As P.E. \(=\frac{E}{2}\) Then, \(\frac{1}{2} m \omega^2 \mathrm{~A}^2 \times \frac{1}{2}=\frac{1}{2} m \omega^2 x^2\)…
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