A) \[2\sqrt{2}cm\]
B) \[\sqrt{2}cm\]
C) \[2\,cm\]
D) \[1\,cm\]
Correct Answer: A
Solution :
In simple harmonic motion total energy is \[E=\frac{1}{2}m{{a}^{2}}{{\omega }^{2}}\] where a is amplitude, \[\omega \] the angular velocity, m the mass. Potential energy is \[U=\frac{1}{2}m{{y}^{2}}{{\omega }^{2}}\] where y is displacement. Given, \[U=\frac{1}{2}E\] \[\therefore \] \[\frac{1}{2}m{{y}^{2}}{{\omega }^{2}}=\frac{1}{2}\left( \frac{1}{2}m{{a}^{2}}{{\omega }^{2}} \right)\] \[\Rightarrow \] \[{{y}^{2}}=\frac{{{a}^{2}}}{2}=\frac{{{4}^{2}}}{2}=8\] \[\Rightarrow \] \[y=2\sqrt{2}cm\]
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