A) \[a\text{/4}\]
B) \[a\text{/3}\]
C) \[a\text{/2}\]
D) \[2a\text{/3}\]
Correct Answer: C
Solution :
Potential energy of particle executing SHM is given by \[PE=\frac{1}{2}m{{\omega }^{2}}{{y}^{2}}\] maximum \[PE({{U}_{\max }})=\frac{1}{2}m{{\omega }^{2}}{{a}^{2}}\] Given \[PE(U)=\frac{1}{4}{{U}_{\max }}\] \[\therefore \] \[\frac{1}{2}m{{\omega }^{2}}{{y}^{2}}=\frac{1}{4}\frac{1}{2}m{{\omega }^{2}}{{a}^{2}}\] or \[{{y}^{2}}={{a}^{2}}/4\] \[\Rightarrow \] \[y=\pm \frac{a}{2}\]
You need to login to perform this action.
You will be redirected in
3 sec
