question_answer

A pendulum bob has a speed of 3 m/s at its lowest position. The pendulum is 0.5 m long. The speed of the bob, when the length makes an angle of \[{{60}^{o}}\] to the vertical, will be (If  \[g=10m/{{s}^{2}}\])                                           [MP PET 1996]

A)            \[\frac{E}{4}\]                      

B)            \[\frac{3E}{4}\]

C)            \[\frac{\sqrt{3}}{4}E\]      

D)            \[{{K}_{1}}\]

Correct Answer: D

Solution :

                   Let bob velocity be v at point B where it makes an angle of 60o with the vertical, then using conservation of mechanical energy  \[K{{E}_{A}}+P{{E}_{A}}\]\[=K{{E}_{B}}+P{{E}_{B}}\] Þ \[\frac{1}{2}m\times {{3}^{2}}=\frac{1}{2}m{{v}^{2}}+mgl(1-\cos \theta )\] \[\therefore \frac{{{T}_{2}}}{{{T}_{1}}}=\sqrt{\frac{{{M}_{2}}}{{{M}_{1}}}}=\sqrt{\frac{4M}{M}}=2\]