Q121 / 160 · AP EAMCET 2020 (21 Sep Shift 2) · Physics

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?

  • A
    \(\frac{A}{4}\)
  • Correct answer
    B
    \(\frac{A}{2}\)
  • C
    \(\frac{3 A}{4}\)
  • D
    \(3 A\)

Solution

According to question, In simple harmonic motion, potential energy \(=\frac{1}{4}\) (total energy) \(\Rightarrow \frac{1}{2} m \omega^2 y^2=\frac{1}{4}\left[\frac{1}{2} m \omega^2 A^2\right] \Rightarrow y^2=\frac{A^2}{4} \Rightarrow y=\frac{A}{2}\)

Hence, the correct answer is option: 'B'.

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