A) \[U=K{{X}^{2}}\]
B) \[\sqrt{2gl(1+\cos \theta )}\]
C) \[U=KX\]
D) \[\frac{a}{2}\]
Correct Answer: C
Solution :
If suppose bob rises up to a height h as shown then after releasing potential energy at extreme position becomes kinetic energy of mean position \[\Rightarrow mgh=\frac{1}{2}mv_{\max }^{2}\]\[\Rightarrow {{v}_{\max }}=\sqrt{2gh}\] Also, from figure \[\cos \theta =\frac{l-h}{l}\] \[\Rightarrow h=l(1-\cos \theta )\] So, \[{{v}_{\max }}=\sqrt{2gl(1-\cos \theta )}\]You need to login to perform this action.
You will be redirected in
3 sec