A mass is suspended separately by two different springs in successive order then time periods is $t_1$ and $t_2$ respectively. If it is connected by both spring as shown in figure then time period is $t_0$, the correct relation is

The correct option is D $t_0^{-2}=t_1^{-2}+t_2^{-2}$

The time period of a spring mass system as shown in figure 1 is given by
$T=2\pi \sqrt{\frac{m}{k}}$,
where k is the spring constant.

$\therefore t_1=2\pi \sqrt{\frac{m}{k_1}}....\left(i\right)$ and

$t_2=2\pi \sqrt{\frac{m}{k_2}}...\left(ii\right)$



Now, when they are connected in parallel as shown in figure 2(a), the system can be replaced by a single spring of spring constant, $k_{eff}=k_1+k_2$, as shown in figure 2(b).

Since $mg=k_{1}x+k_{2}x=k_{eff}x$

$\therefore t_0=2\pi \sqrt{\frac{m}{k_{eff}}}=2\pi \sqrt{\frac{m}{(k_1+k_2)}}...\left(iii\right)$,

From $\left(i\right)\frac{1}{t_1^2}=\frac{1}{4{\pi }^2}\times \frac{k_1}{m}....\left(iv\right)$,

From (ii), $\frac{1}{t_2^2}=\frac{1}{4{\pi }^2}\times \frac{k_2}{m}....\left(v\right)$,

From (iii), $\frac{1}{t_0^2}=\frac{1}{4{\pi }^2}\times \frac{k_1+k_2}{m}....\left(vi\right)$,

From equations (iv) , (v) and (vi)

$\frac{1}{t_1^2}+\frac{1}{t_2^2}=\frac{1}{t_0^2};\therefore t_0^{-2}=t_1^{-2}+t_2^{-2}$
Hence, option $\left(B\right)$ is correct.