Question

A man is swinging on a swing made of $2$ ropes of equal length $L$ and in direction perpendicular to the plane of paper. The time period of the small oscillations about the mean position is

• A. $2\mathrm{\pi }\sqrt{\frac{\sqrt{3}\mathrm{L}}{2\mathrm{g}}}$
• B. $\mathrm{\pi }\sqrt{\frac{\mathrm{L}}{\mathrm{g}}}$
• C. $2\mathrm{\pi }\sqrt{\frac{\mathrm{L}}{2\mathrm{g}}}$
• D. $2\mathrm{\pi }\sqrt{\frac{\mathrm{L}}{2\sqrt{3\mathrm{g}}}}$

Answer 1

The correct option is A

$2\mathrm{\pi }\sqrt{\frac{\sqrt{3}\mathrm{L}}{2\mathrm{g}}}$

Step 1: Given

Length of ropes is $L$.

Effective length of the swing is $L_{effective}$, which is $AB$ in the below diagram.

Acceleration due to gravity is $g$.

Time period of an oscillation is $T$.

The angle made by the swing is $60°$, as it makes an equilateral triangle.

Step 2: Formula used

1. $T=2\mathrm{\pi }\sqrt{\frac{{\mathrm{L}}_{\mathrm{effective}}}{\mathrm{g}}}$
2. $\sin \theta =\frac{P}{H}$

Step 3: Find the effective length AB of the swing:

$$\begin{gathered} \sin 60°=\frac{P}{H} \\ \Rightarrow \frac{\sqrt{3}}{2}=\frac{AB}{L} \\ \Rightarrow AB=\frac{\sqrt{3}}{2}L \end{gathered}$$

Step 4: Substitute the values to find the time period

$$\begin{gathered} T=2\mathrm{\pi }\sqrt{\frac{{\mathrm{L}}_{\mathrm{effective}}}{\mathrm{g}}} \\ =2\mathrm{\pi }\sqrt{\frac{{\displaystyle \frac{\sqrt{3}\mathrm{L}}{2}}}{\mathrm{g}}} \\ =2\mathrm{\pi }\sqrt{\frac{{\displaystyle \frac{\sqrt{3}\mathrm{L}}{2}}}{\mathrm{g}}} \\ =2\mathrm{\pi }\sqrt{\frac{\sqrt{3}\mathrm{L}}{2\mathrm{g}}} \end{gathered}$$

Hence, option (A) is correct.