A) \[2\sqrt{2}cm\]
B) \[\sqrt{2}cm\]
C) 3 cm
D) 1 cm
Correct Answer: A
Solution :
For a particle executing SHM of amplitude a, angular velocity o, displacement\[x\]then Kinetic energy \[(KE)=\frac{1}{2}m{{\omega }^{2}}({{a}^{2}}-{{x}^{2}})\] Potential energy\[(PE)=\frac{1}{2}m{{\omega }^{2}}{{x}^{2}}\] Given, \[KE=PE\] \[\frac{1}{2}m{{\omega }^{2}}{{a}^{2}}-\frac{1}{2}m{{\omega }^{2}}{{x}^{2}}=\frac{1}{2}m{{\omega }^{2}}{{x}^{2}}\] \[\frac{1}{2}m{{\omega }^{2}}{{a}^{2}}=m{{\omega }^{2}}{{x}^{2}}\] \[\Rightarrow \] \[x=\frac{a}{\sqrt{2}}=\frac{4}{\sqrt{2}}=2\sqrt{2}\,cm\]
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