1. Period of a mass-spring system

For a mass $m$ attached to a spring of spring constant $k$, the time period is

$t = 2\pi \sqrt{\frac{m}{k}}$

So for the two individual springs:

$t_1 = 2\pi \sqrt{\frac{m}{k_1}}$

$t_2 = 2\pi \sqrt{\frac{m}{k_2}}$

2. Equivalent spring constant for springs in series

If two springs of constants $k_1$ and $k_2$ are connected in series, then

$\frac{1}{k_{\text{eq}}} = \frac{1}{k_1} + \frac{1}{k_2}$

Hence,

$k_{\text{eq}} = \frac{k_1 k_2}{k_1 + k_2}$

3. Time period with the equivalent spring

The new period is

$T = 2\pi \sqrt{\frac{m}{k_{\text{eq}}}}$

Substituting $k_{\text{eq}}$:

$T = 2\pi \sqrt{\frac{m(k_1+k_2)}{k_1k_2}}$

Squaring,

$T^2 = 4\pi^2 m\left(\frac{1}{k_1} + \frac{1}{k_2}\right)$

4. Relate with $t_1$ and $t_2$

From the individual periods,

$t_1^2 = 4\pi^2 \frac{m}{k_1}, \qquad t_2^2 = 4\pi^2 \frac{m}{k_2}$

Therefore,

$t_1^2 + t_2^2 = 4\pi^2 m\left(\frac{1}{k_1} + \frac{1}{k_2}\right)$

Comparing with the expression for $T^2$,

$T^2 = t_1^2 + t_2^2$

5. Check options

• A: $T^{-1} = t_1^{-1} + t_2^{-1}$ ❌
• B: $T^2 = t_1^2 + t_2^2$ ✅
• C: $T = t_1 + t_2$ ❌
• D: $T^{-2} = t_1^{-2} + t_2^{-2}$ ❌

So the correct option is B.