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