A simple pendulum is released from A as shown | vedclass

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(C) Let the point of suspension be $O$. The pendulum is released from $A$,where the string is horizontal. The vertical distance of $A$ from $O$ is $0$

Vedclass · Accepted Answer

$\frac{\sqrt{3}}{2} mgl$

Vedclass · Answer

(C) Let the point of suspension be $O$. The pendulum is released from $A$,where the string is horizontal. The vertical distance of $A$ from $O$ is $0$.The bob moves to position $B$,where the string makes an angle of $30^{\circ}$ with the vertical.The vertical distance of $B$ below the point of suspension $O$ is $h' = l \cos 30^{\circ} = l \frac{\sqrt{3}}{2}$.The vertical distance fallen by the bob from $A$ to $B$ is $h = l - h' = l - l \cos 30^{\circ} = l(1 - \frac{\sqrt{3}}{2})$.By the law of conservation of energy,the gain in kinetic energy is equal to the loss in potential energy.Gain in kinetic energy $= mgh = mgl(1 - \frac{\sqrt{3}}{2})$.Note: The provided options seem to assume the height $h$ is measured from the lowest point or relative to the vertical projection. Given the options,the intended calculation is likely based on the vertical drop from the horizontal level to the position $B$ relative to the lowest point,or a misinterpretation of the angle. Re-evaluating the standard interpretation for this specific problem type: The potential energy at $A$ (relative to the lowest point) is $mgl$. The potential energy at $B$ (relative to the lowest point) is $mgl(1 - \cos 30^{\circ})$. The gain in kinetic energy is $PE_A - PE_B = mgl - mgl(1 - \cos 30^{\circ}) = mgl \cos 30^{\circ} = mgl \frac{\sqrt{3}}{2}$.

$A$ simple pendulum is released from $A$ as shown in the figure. If $m$ and $l$ represent the mass of the bob and the length of the pendulum respectively,the gain in kinetic energy at $B$ is

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