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, co the angular velocity, m the mass. Potential energy is \[U=\frac{1}{2}m{{y}^{2}}{{\omega }^{2}}\] where y is displacement. Given, \[U=\frac{1}{2}E\] \[\therefore \] \[\frac{1}{2}m{{y}^{2}}{{\omega }^{2}}=\frac{1}{2}\left( \frac{1}{2}m{{a}^{2}}{{\omega }^{2}} \right)\] \[\Rightarrow \] \[{{y}^{2}}=\frac{{{a}^{2}}}{2}=\frac{{{4}^{2}}}{2}=8\] \[\Rightarrow \] \[y=2\sqrt{2}cm\]
You need to login to perform this action.
You will be redirected in
3 sec
