A vertical spring mass system has the same time-period as a simple pendulum undergoing small oscillations. Now both of them are put in an elevator going downwards with an acceleration 5 m/s². The ratio of time period of the spring mass system to the time period of the pendulum is (Assume g = 10m/s²)  
 

We know that of the spring-mass system's time period,

$T_i=2\mathrm{\pi }\sqrt{\frac{\mathrm{m}}{\mathrm{k}}}$

The time period of the simple pendulum,

$T_2=2\mathrm{\pi }\sqrt{\frac{l}{\mathrm{g}}}$

The initial time period of a spring mass system, $T_1$, is equal to the starting time period of a simple pendulum, $T_2$, the question states.

$2\mathrm{\pi }\sqrt{\frac{\mathrm{m}}{\mathrm{k}}}=2\mathrm{\pi }\sqrt{\frac{l}{\mathrm{g}}}$

$\sqrt{\frac{m}{k}}=\sqrt{\frac{l}{10}}\_\_\_\_\_\_\_\left(1\right)$

The duration of the spring mass system is unaffected when they are both placed in an elevator moving downward at an acceleration of $5m/s^2.$

$T_1\text{'}=T_1=2\mathrm{\pi }\sqrt{\frac{\mathrm{m}}{\mathrm{k}}}$

But still, the elevator's simple pendulum's changing period is given by

$T{\text{'}}_2=2\mathrm{\pi }\sqrt{\frac{l}{{\mathrm{g}}_{\mathrm{eff}}}} ({\mathrm{g}}_{\mathrm{eff}}=\mathrm{g}-\mathrm{a})$

$T{\text{'}}_2=2\mathrm{\pi }\sqrt{\frac{l}{\mathrm{g}-5}}=2\mathrm{\pi }\sqrt{\frac{l}{5}}$

$\frac{T_1\text{'}}{T_2\text{'}}=\frac{2\mathrm{\pi }\sqrt{{\displaystyle \frac{\mathrm{m}}{\mathrm{k}}}}}{2\mathrm{\pi }\sqrt{{\displaystyle \frac{l}{5}}}}=\frac{\sqrt{{\displaystyle \frac{l}{10}}}}{\sqrt{{\displaystyle \frac{l}{5}}}}(from equation 1)$

$\frac{T_1\text{'}}{T_2\text{'}}=\frac{1}{\sqrt{2}}$

Hence the correct answer is $\frac{1}{\sqrt{2}}.$