A) \[\frac{A}{4}\]
B) \[\frac{A}{2}\]
C) \[\frac{A}{\sqrt{2}}\]
D) \[\frac{A}{\sqrt{3}}\]
Correct Answer: C
Solution :
The total energy a particle executing SHM\[=\frac{1}{2}m{{\omega }^{2}}{{A}^{2}}\] The PE of the particle at a distance x from the equilibrium position\[=\frac{1}{2}m{{\omega }^{2}}{{x}^{2}}\] From the question, \[\frac{1}{2}m{{\omega }^{2}}{{x}^{2}}=\frac{1}{2}\left( \frac{1}{2}m{{\omega }^{2}}{{A}^{2}} \right)\] \[\Rightarrow \] \[{{x}^{2}}=\frac{{{A}^{2}}}{2}\Rightarrow x=\frac{A}{\sqrt{2}}\]
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