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 $T = 2 \pi \sqrt{\frac{\ell}{\sqrt{2} g}}$.

Similar Questions

Frequency $f$ of a simple pendulum depends on its length $\ell$ and acceleration $g$ due to gravity according to the following equation $f = \frac{1}{2 \pi} \sqrt{\frac{g}{\ell}}$. $A$ graph between which of the following quantities is a straight line?

The time period of a simple pendulum is $T$. The angular amplitude is $\beta$. How much time will the bob of the pendulum take to move from the equilibrium position $O$ to position $A$,where the string makes an angle $\alpha$ with the vertical?

The motion of a simple pendulum executing $S.H.M.$ is represented by the following equation:
$Y = A \sin (\pi t + \phi)$,where time is measured in $seconds$.
The length of the pendulum is .............$cm$.

Identify the correct statement.

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