Answer:
Option C
Explanation:
We know that time period of a spring mass system
$T_{1}=2\pi\sqrt{\frac{m}{k}}$
where, m= mass of body
and k= force constant of the spring
Time period of simple pendulum
$T_{2}=2\pi\sqrt{\frac{l}{g}}$
According to the question, initially time period of a spring mass system. $T_{1}$= time period of simple pendulum , $T_{2}$
$2\pi\sqrt{\frac{m}{k}}=2\pi\sqrt{\frac{l}{g}}$
$\sqrt{\frac{m}{k}}=\sqrt{\frac{l}{10}}$ .........(i) [ $\because g=10 m/s^{2}$]
when both of them are put in an elevator going downwards with an acceleration 5 m/s2 , then no effect occurs on the time period of spring mass system.
i.e, $T_{1}'=T_{2}=2\pi\sqrt{\frac{m}{k}}$
But time period of simple pendulum changes in elevator and is given by
$T_{2}'=2\pi\sqrt{\frac{l}{g_{eff}}}$ , where $g_{eff}$= g-a = effective
acceleration of the pendulum when elevator is accelerating downwards with a m/s2
$T_{2}'=2\pi\sqrt{\frac{l}{g-5}}=2\pi \sqrt{\frac{l}{5}}$
$\therefore$ $\frac{T_{1}'}{T_{2}'}=\frac{2\pi\sqrt{\frac{m}{k}}}{2\pi\sqrt{\frac{l}{5}}}=\frac{\sqrt{l/10}}{\sqrt{l/5}}$ [From Eq. (i)]
$\therefore$ $\frac{T_{1}'}{T_{2}'}=\frac{1}{\sqrt{2}}$
Hence, the ratio of time period of the spring mass system to the time period of the pendulum is $\frac{1}{\sqrt{2}}$
