question_answer

A lift is ascending with an acceleration equal to g/3. What will be the time period of a simple pendulum suspended from ceiling if its time period in stationary lift is T?

A) T/4                                        

B) \[\left( \frac{\sqrt{3}}{4} \right)T\]

C) \[\left( \frac{\sqrt{3}}{2} \right)T\]                                         

D) \[\frac{T}{2}\]

Correct Answer: C

Solution :

The periodic time of pendulum in SHM is \[T=2\pi \sqrt{\frac{l}{g}}\]       ?(i) where, \[l\] is length of pendulum and g is gravity. Effective gravity when lift    accelerates upwards is          \[g'=g+\frac{g}{3}=\frac{4}{3}g\] \[\therefore \]    \[T'=2\pi \sqrt{\frac{l}{(4/3)g}}\]            ?(ii) Dividing Eq. (i) by (ii),we get \[\frac{T}{T'}=\sqrt{\frac{\frac{4}{3}g}{g}}=\sqrt{\frac{4}{3}}\] \[\therefore \]     \[T'=\sqrt{\frac{3}{4}}T=\frac{\sqrt{3}}{2}T\]