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 `ms^(-2)`. If the new time periods of the pendulum and spring be `T_(p) and T_(s)`, respectively. Then,

Options

A. `T_(p) = T_(s)`
B. `T_(p) gt T_(s)`
C. `T_(p) lt T_(s)`
D. Cannot be predicted

Answer

C

Explanation

Time period of simple pendulum placed in a train accelerating at the rate of a `ms^(-2)` is given by

`T = 2pi [(I)/(sqrt(g^(2)+a^(2)))]^((1)/(2))`
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) lt T`
and `T_(s)="T i.e." T_(p) lt T_(s)`