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(B) The swing forms an equilateral triangle with the two ropes of length $L$ and the distance between the suspension points being $L$. The height of t

Vedclass · Accepted Answer

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

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(B) The swing forms an equilateral triangle with the two ropes of length $L$ and the distance between the suspension points being $L$. The height of the center of mass (the man) from the line joining the suspension points is $h = \sqrt{L^2 - (L/2)^2} = \frac{\sqrt{3}}{2} L$. For small oscillations perpendicular to the plane of the swing,the system acts as a physical pendulum. The moment of inertia $I$ about the axis of rotation (the line joining the suspension points) is $I = M h^2 = M (\frac{\sqrt{3}}{2} L)^2 = \frac{3}{4} M L^2$. The time period of a physical pendulum is given by $T = 2\pi \sqrt{\frac{I}{Mgh}}$. Substituting the values: $T = 2\pi \sqrt{\frac{\frac{3}{4} M L^2}{Mg (\frac{\sqrt{3}}{2} L)}} = 2\pi \sqrt{\frac{3 L^2 / 4}{\sqrt{3} g L / 2}} = 2\pi \sqrt{\frac{3 L}{2 \sqrt{3} g}} = 2\pi \sqrt{\frac{\sqrt{3} L}{2g}}$.

$A$ man of mass $M$ is swinging on a swing made of $2$ ropes of equal length $L$. The distance between the points of suspension is also $L$. The time period of the small oscillations about the mean position in a direction perpendicular to the plane of the swing is:

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