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 $\frac{1}{\sqrt{2}}$.