A) \[\frac{A}{4}\]
B) \[\frac{A}{3}\]
C) \[\frac{A}{2}\]
D) \[\frac{A}{\sqrt{2}}\]
Correct Answer: A
Solution :
Potential energy of particle \[U=\frac{1}{2}m{{\omega }^{2}}{{A}^{2}}\] Maximum potential energy of particle \[E=\frac{1}{2}m{{\omega }^{2}}{{y}^{2}}\] According to given position The potential energy\[U=\frac{E}{2}\] or \[\frac{1}{2}m{{\omega }^{2}}{{y}^{2}}=\frac{1}{2}\times \frac{1}{2}m{{\omega }^{2}}{{A}^{2}}\] \[{{y}^{2}}=\frac{{{A}^{2}}}{2}y=\frac{A}{\sqrt{2}}\]
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