Oscillations - Result Question 32

35. A body executes S.H.M with an amplitude A. At what displacement from the mean position is the potential energy of the body is one fourth of its total energy?

[1993]

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Answer:

Correct Answer: 35. (b)

Solution:

(b) $P \cdot E=\frac{1}{2} m \omega^{2} x^{2}$.

Total energy $E=\frac{1}{2} m \omega^{2} A^{2}$

P.E. $=\frac{1}{4} E \Rightarrow \frac{1}{2} m \omega^{2} x^{2}=\frac{1}{8} m \omega^{2} A^{2}$

$\therefore x=\frac{1}{2} A$.