A) \[2\,a\]
B) a
C) \[a\sqrt{2}\]
D) \[\frac{a}{\sqrt{2}}\]
Correct Answer: D
Solution :
In SHM, KE=\[=\frac{1}{2}m\,({{a}^{2}}-{{y}^{2}}){{\omega }^{2}}\] \[PE=\frac{1}{2}m{{\omega }^{2}}{{y}^{2}}\] They are equal if \[\frac{1}{2}m{{\omega }^{2}}({{a}^{2}}-{{y}^{2}})=\frac{1}{2}m{{\omega }^{2}}{{y}^{2}}\] \[\Rightarrow \] \[{{a}^{2}}-{{y}^{2}}={{y}^{2}}\] \[\Rightarrow \] \[2{{y}^{2}}={{a}^{2}}\] \[\therefore \] \[y=\frac{a}{\sqrt{2}}\]
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