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

$\therefore$ ${t_1} = 2\pi \sqrt {m/{k_1}}$   ...(i)
and ${t_2} = 2\pi \sqrt {m/{k_2}}$   ...(ii)


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 = k₁ + k₂.
[Since mg = k₁x + k₂x = k_effx]

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

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

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

From (ii), ${1 \over {t_0^2}} = {1 \over {4{\pi ^2}}} \times {{{k_1} + {k_2}} \over m}$   ...(vi)

Now (iv) + (v)

${1 \over {t_1^2}} + {1 \over {t_2^2}} = {1 \over {4{\pi ^2}m}}\left( {{k_1} + {k_2}} \right) = {1 \over {t_0^2}}$

$\therefore$ $t_0^{ - 2} = t_1^{ - 2} + t_2^{ - 2}$