Consider a swing whose one end is fixed above the other rope by “b”. The distance between the poles of the swing is $\alpha$ . The lengths $l_1$ and $l_2$ are such that $l_1^2+l_2^2=a^2+b^2$ as shown. The period of the small oscillations of the swing, by neglecting the height of the swinging person, is....

From the given information, it is inferred that two ropes are perpendicular to each other:

$$\begin{gathered} \therefore T_1\cos \theta +T_2\sin \alpha =mg\dots \dots \dots \dots ..\left(1\right) \\ {T}_1\sin \alpha =T_2\cos \alpha . . . . . . . . .\left(2\right) \\ \Rightarrow T_1=mg\cos \alpha \text{ and }{T}_2=mg\sin \alpha \end{gathered}$$

When the swing is given a displacement, x , along outward normal to plane of the paper such that the strings 1 and 2 make angles ${\theta }_1 and {\theta }_2 with their respective initial positions , then$

$F_{\text{restoring }}=-(T_1 \sin {\theta }_1+T_2 \sin {\theta }_2) =-(T_1\frac{x}{l_1}+T_2\frac{x}{l_2})=-\frac{mga}{l_1{l}_2}x$

So the time period T  = 2$\mathrm{\pi }\sqrt{\frac{{\mathrm{l}}_1{\mathrm{l}}_2}{\mathrm{ag}}}$