question_answer

A man measures time period of a pendulum (T) in stationary lift. If the lift moves upward with p acceleration \[\frac{g}{4}\], then new time period will be :

A) \[\sqrt{\frac{2}{5}T}\]                                   

B)  \[\sqrt{\frac{5}{2}T}\]

C)  \[\frac{\sqrt{5}\,T}{2}\]                                               

D)  \[\frac{2T}{\sqrt{5}}\]

Correct Answer: D

Solution :

Key Idea: when the lift moves upwards effective acceleration increases. When the lift is moving upwards, then from Newton?s second law of motion net force on him is \[F=ma\]                                (upwards) Or           \[F=-ma\]            (downwards) Therefore,                                                 \[-ma=mg-R\] \[\Rightarrow \]                               \[R=mg+ma\] \[\therefore \]                  \[R=m\left( g+a \right)\] Time Period,    \[T=2\pi \sqrt{\frac{l}{g}}\,is\]                                 \[T=2\pi \sqrt{\frac{l}{g+a}}\,\]                 ?(1) When lift is stationary time period is                                 \[T=2\pi \sqrt{\frac{l}{g}}\,\]                                      ?(2) Also, \[g'=g+a=g+\frac{g}{4}=\frac{5g}{4}\] Dividing Eq. (2) by (1), we get                                 \[\frac{T}{T'}=\sqrt{\frac{5/4g}{g}}=\frac{\sqrt{5}}{2}\]                                 \[T'=\frac{2}{\sqrt{5}}T\]