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/s2. If the new time periods of the pendulum and spring be Tp and Ts, respectively, then

Question

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/s2. If the new time periods of the pendulum and spring be Tp and Ts, respectively, then (A) Tp = Ts (B) Tp > Ts (C) Tp < Ts (D) cannot be predicted

Tardigrade Admin · Accepted Answer

Time period of simple pendulum placed in a train accelerating at the rate of a

m s^{- 2}

is given by

T = 2 π [\frac{1}{g ^{2} + a ^{2}}]^{\frac{1}{2}}

However, the time period of mass attached to the spring is

T = 2 π (\frac{m}{k})

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

i.e.,,

T_{p} < T_{s}

Q. 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 $5 m / s^{2}$ . If the new time periods of the pendulum and spring be $T_{p}$ and $T_{s}$ , respectively, then

$T_{p} = T_{s}$

$T_{p} > T_{s}$

$T_{p} < T_{s}$

cannot be predicted

Q. 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 5m/s2. If the new time periods of the pendulum and spring be Tp​ and Ts​, respectively, then

Tp​=Ts​

Tp​>Ts​

Tp​<Ts​

cannot be predicted

Q. 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 $5 m / s^{2}$ . If the new time periods of the pendulum and spring be $T_{p}$ and $T_{s}$ , respectively, then

$T_{p} = T_{s}$

$T_{p} > T_{s}$

$T_{p} < T_{s}$