Two light strings,each of length ell,are fixe | vedclass

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(D) Let the distance between points $A$ and $B$ be $d$. The vertical distance of the bob from the rod is $h = \ell \sin 45^{\circ} = \frac{\ell}{\sqrt

Vedclass · Accepted Answer

$2 \pi \sqrt{\frac{\ell}{\sqrt{2} g}}$

Vedclass · Answer

(D) Let the distance between points $A$ and $B$ be $d$. The vertical distance of the bob from the rod is $h = \ell \sin 45^{\circ} = \frac{\ell}{\sqrt{2}}$.When the bob is displaced by a small angle $	heta$ normal to the plane of the strings,the effective length of the pendulum is the perpendicular distance from the rod to the bob,which is $h = \frac{\ell}{\sqrt{2}}$.The restoring force for small oscillations is provided by the component of gravity,and the motion is equivalent to a simple pendulum of length $L_{eff} = h = \frac{\ell}{\sqrt{2}}$.The time period $T$ is given by $T = 2 \pi \sqrt{\frac{L_{eff}}{g}}$.Substituting $L_{eff} = \frac{\ell}{\sqrt{2}}$,we get $T = 2 \pi \sqrt{\frac{\ell}{\sqrt{2} g}}$.

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