A vertical spring mass system has the same time period as simple pendulum undergoing small oscillations. Now, both of them are put in an elevator going downwards with an acceleration $5 \mathrm{~m} / \mathrm{s}^2$. The ratio of time period of the spring mass system to the time period of the pendulum is (Assume, acceleration due to gravity, $g=10 \mathrm{~m} / \mathrm{s}^2$ )

We know that time period of a spring mass system, $$ \mathrm{T}_{\mathrm{l}}=2 \pi \sqrt{\frac{m}{k}} $$ where, $m=$ mass of body and $k=$ force constant of the spring Time period of simple pendulum, $$ \mathrm{r}_2=2 \pi \sqrt{\frac{l}{g}} $$ According to the question, initially time period of a spring mass system, $T_1=$ time period of simple pendulum, $\mathrm{T}_2$ $$ \begin{aligned} 2 \pi \sqrt{\frac{m}{k}} & =2 \pi \sqrt{\frac{l}{g}} \\ \sqrt{\frac{m}{k}} & =\sqrt{\frac{l}{10}} \end{aligned} $$ $\ldots$ (i) $\left[\because g=10 \mathrm{~m} / \mathrm{s}^2\right]$ When both of them are put in an elevator going downwards with an acceleration $5 \mathrm{~m} / \mathrm{s}^2$, then no effect occurs on the time period of spring mass system. i.e., $\quad T_1^{\prime}=T_1=2 \pi \sqrt{\frac{m}{k}}$ But time period of simple pendulum changes in elevator and is given by $T_2^{\prime}=2 \pi \sqrt{\frac{l}{g_{\text {eff }}}}$, where $g_{\text {eff }}=g-a=$ effective acceleration of the pendulum when elevator is accelerating downwards with a $\mathrm{m} / \mathrm{s}^2$. $$ \begin{array}{rlrl} T_2^{\prime} & =2 \pi \sqrt{\frac{l}{g-5}}=2 \pi \sqrt{\frac{l}{5}} \\ \therefore \quad & \frac{T_1^{\prime}}{T_2^{\prime}} & =\frac{2 \pi \sqrt{\frac{m}{k}}}{2 \pi \sqrt{\frac{l}{5}}}=\frac{\sqrt{l / 10}}{\sqrt{l / 5}} & \text { [From Eq. (i)] } \\ \therefore \quad & \frac{T_1^{\prime}}{T_2^{\prime}} & =\frac{1}{\sqrt{2}} \end{array} $$ [From Eq. (i)] Hence, the ratio of time period of the spring mass system to the time period of the pendulum is $\frac{1}{\sqrt{2}}$.