A) \[{{\cos }^{1}}\left[ \frac{L\,\cos a+l}{L+l} \right]\]
B) \[{{\cos }^{1}}\left[ \frac{L\,\cos a+l}{L-l} \right]\]
C) \[{{\cos }^{1}}\left[ \frac{L\,\cos a-l}{L-l} \right]\]
D) \[{{\cos }^{1}}\left[ \frac{L\,\cos a-l}{L+l} \right]\]
Correct Answer: C
Solution :
[c] Since the pendulum started with no kinetic energy. Conservation of energy, implies that the potential energy at \[{{Q}_{\max }}\]must be equal to the original potential energy, i.e., the vertical position will be same. Therefore, \[L\cos \alpha =l+(L-1)cos\theta \] \[\Rightarrow \cos \theta =\frac{L\cos \alpha -l}{L-1}\] \[\Rightarrow \theta ={{\cos }^{-1}}\left[ \frac{L\cos \alpha -l}{L-l} \right]\]
You need to login to perform this action.
You will be redirected in
3 sec
