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 acceleration at the rate of a `ms^(-2)` is lk given by

`T = 2pi [(1)/(sqrt(g^(2) + a^(2)))]^(1/2)`
However, the time period of mass attached to the spring
`T = 2pi sqrt(((m)/(k)))`