question_answer

The time period of a simple pendulum in a stationary train is T. The time period of a mass attached to a spring is also T. The train accelerates at the rate 5 m/s. If the new time periods of the pendulum and spring be Tp and T, respectively, then

A) Tp = T1                                  

B) Tp > Ts

C) Tp < Ts                                   

D) cannot be predicted

Correct Answer: C

Solution :

Time period of simple pendulum placed in a train accelerating at the rate of \[am/{{s}^{2}}\]is given by \[T=2\pi {{\left[ \frac{l}{{{g}^{2}}+{{a}^{2}}} \right]}^{\frac{1}{2}}}\] However, the time period of mass attached to the spring is  \[T=2\pi \sqrt{\left( \frac{m}{k} \right)}\] It is independent of g as well as a . Hence, when the train accelerates, the time period of the simple pendulum decreases and that of spring remains unchanged. Hence, \[{{T}_{p}}<T\]   and \[{{T}_{s}}=T\] ie,              \[{{T}_{p}}<{{T}_{s}}.\]