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
AFMCAFMC 2007Oscillations
Solution:
Time period of simple pendulum placed in a train accelerating at the rate of a $ ms ^{-2}$ is given by
$T=2 \pi\left[\frac{1}{\sqrt{g^{2}+a^{2}}}\right]^{\frac{1}{2}}$
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 $T_{s}=T$
i.e.,, $T_{p} < T_{s}$
