A mass is suspended separately by two springs and the time periods in the two cases are $T_1$ and $T_2.$ If the same mass be suspended by connecting the two springs in parallel then the time period of oscillations is $T_p$ and when the two springs are connected in series then the time period of oscillations is $T_s$. Then select the correct relation(s).

The correct option is C $T_p^{-2}=T_1^{-2}+T_2^{-2}$

We know that, time period of oscillations, $$T=2\pi \sqrt{\frac{m}{K}}$$Here,
$m=$suspended mass,
$K=$ spring constant,

Since here mass is kept same, so
$2\pi \sqrt{m}=$ constant$=\alpha$ (assume)

$\therefore T=\frac{\alpha }{\sqrt{K}}$

$\Rightarrow K=\frac{{\alpha }^2}{T^2}......\left(1\right)$

$\Rightarrow \frac{1}{K}=\frac{T^2}{{\alpha }^2}......\left(2\right)$

Let, the spring constant for first spring and second spring are $K_1$ and $K_2$ respectively.

So, for the first case where spring are connected in parallel:

$K=K_1+K_2$

Using equation $\left(1\right)$,

$\frac{{\alpha }^2}{T_p^2}=\frac{{\alpha }^2}{T_1^2}+\frac{{\alpha }^2}{T_2^2}$

$\therefore T_p^{-2}=T_1^{-2}+T_2^{-2}$

This is the desired relation.

For the second case where spring are connected in series,

$\frac{1}{K}=\frac{1}{K_1}+\frac{1}{K_2}$

Now, using equation $\left(2\right)$ we have,

$\frac{T_s^2}{{\alpha }^2}=\frac{T_1^2}{{\alpha }^2}+\frac{T_2^2}{{\alpha }^2}$

$\therefore T_s^2=T_1^2+T_2^2$

This is the desired relation in this case.

Hence, option (c) and (d) are correct answers.