A body executes simple harmonic motion with an amplitude \(A\). At what displacement, from the mean position…
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?
\(\frac{A}{4}\)
\(\frac{A}{2}\)
\(\frac{3 A}{4}\)
\(3 A\)
Solution
According to question,
In simple harmonic motion, potential energy \(=\frac{1}{4}\) (total energy)
\(\Rightarrow \frac{1}{2} m \omega^2 y^2=\frac{1}{4}\left[\frac{1}{2} m \omega^2 A^2\right] \Rightarrow y^2=\frac{A^2}{4} \Rightarrow y=\frac{A}{2}\)