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A body executes simple harmonic motion with an amplitude \(A\). At what displacement, from the mean position, is the potential energy of the body one fourth of its total energy?
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The correct answer is:
\(\frac{A}{2}\)
According to question,
In simple harmonic motion, potential energy \(=\frac{1}{4}\) (total energy)
\(\Rightarrow \frac{1}{2} m \omega^2 y^2=\frac{1}{4}\left[\frac{1}{2} m \omega^2 A^2\right] \Rightarrow y^2=\frac{A^2}{4} \Rightarrow y=\frac{A}{2}\)
In simple harmonic motion, potential energy \(=\frac{1}{4}\) (total energy)
\(\Rightarrow \frac{1}{2} m \omega^2 y^2=\frac{1}{4}\left[\frac{1}{2} m \omega^2 A^2\right] \Rightarrow y^2=\frac{A^2}{4} \Rightarrow y=\frac{A}{2}\)
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