question_answer

The period of oscillation of a simple pendulum is T in a stationary lift. If the lift moves upwards with acceleration of 8g, the period will:

A) remain the same

B) decreases by \[\frac{T}{2}\]

C) increase by \[\frac{T}{3}\]

D) none of the above

Correct Answer: C

Solution :

Key Idea: When lift moves upwards net acceleration of lift increases. When lift moves upwards that net force on it is in the   upward direction. Therefore, from Newton's second law, the net force                    \[F-mg=ma\]                 \[\Rightarrow \]   \[F=m(g+a)\]                 Given, \[a=8g\] \[\therefore \]    \[g'=g+8g=9g\] Also time period \[(T)=2\pi \sqrt{\frac{l}{g}}\] \[\Rightarrow \]    \[T\propto \frac{1}{\sqrt{g}}\] Since \[{{T}^{2}}g=\text{constant}\] \[\therefore \]    \[T_{1}^{2}g=T_{2}^{2}\times 9g\] \[\Rightarrow \]    \[{{T}_{2}}=\frac{{{T}_{1}}}{3}=\frac{T}{3}\]