A) \[2\sqrt{2}\,cm\]
B) \[2\,cm\]
C) \[\sqrt{2}\]
D) \[\frac{3}{\sqrt{2}}cm\]
Correct Answer: A
Solution :
Suppose at point\[x\] \[KE=PE\] \[\Rightarrow \] \[\frac{1}{2}m{{\omega }^{2}}({{A}^{2}}-{{x}^{2}})=\frac{1}{2}m{{\omega }^{2}}{{x}^{2}}\] \[2{{x}^{2}}={{A}^{2}}\] \[x=\frac{A}{\sqrt{2}}=\frac{4}{\sqrt{2}}\] \[x=2\sqrt{2}cm\]
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