question_answer

A simple pendulum is suspended from the roof of a trolley which moves in a horizontal direction with an acceleration a, then the time period is given by \[T=2\pi \sqrt{\frac{l}{{{g}'}}}\],  where \[{g}'\] is equal to                                      [BHU 1997]

A)            g    

B)            \[g-a\]

C)            \[g+a\]                                   

D)            \[\sqrt{{{g}^{2}}+{{a}^{2}}}\]

Correct Answer: D

Solution :

                   \[{g}'=\sqrt{{{g}^{2}}+{{a}^{2}}}\]