A body executes simple harmonic motion under the action of a force $F_1$ with a time period $\frac{4}{5}s.$. If  the force is changed to $F_2$ it executes S.H.M. with time period $\frac{3}{5}s.$. If both the forces $F_1$ and $F_2$  act, simultaneously in the same direction on the body, its time period in second will be 

Let the body be displaced by a distance x. If the restoring force is $F_1$, then the angular frequency of the resulting simple harmonic motion is given by

${\omega }_1^2=\frac{K}{m}=\frac{Kx}{mx}=\frac{F_1}{mx}$       (i)

where $m$ ·is the mass ,of the body. For force· $F_2$, we have

${\omega }_2^2=\frac{F_2}{mx}$                                    (ii)

if $F_1$ and $F_2$ act simultaneously, then

${\omega }^2=\frac{F_1+F_2}{mx}$                             (iii)

From (i), (ii) and (iii) we get

$$\begin{gathered} {\omega }^2={\omega }_1^2+{\omega }_2^2 \\ \text{ or }{\left(\frac{2\pi }{T}\right)}^2={\left(\frac{2\pi }{T_1}\right)}^2+{\left(\frac{2\pi }{T_2}\right)}^2 \\ \text{ or }\frac{1}{T^2}=\frac{1}{T_1^2}+\frac{1}{T_2^2}=\frac{1}{{\left(\frac{4}{5}\right)}^2}+\frac{1}{{\left(\frac{3}{5}\right)}^2} \\ \text{ or }\frac{1}{T^2}=\frac{25}{16}+\frac{25}{9}=\frac{25\times 25}{16\times 9}=\frac{25\times 25}{144} \\ \text{ or }T=\sqrt{\frac{144}{25\times 25}}=\frac{12}{25}\mathrm{s} \end{gathered}$$