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Question

(A) 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

$\frac{mgl}{2}$

Vedclass · Answer

(A) 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 $0$.At point $B$,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 of $A$ below the point of suspension $O$ is $0$. Therefore,the vertical distance of $A$ above $B$ is $h = l - h' = l - l \cos 30^{\circ} = l(1 - \frac{\sqrt{3}}{2})$.By the law of conservation of mechanical energy,the gain in kinetic energy at $B$ is equal to the loss in potential energy as the bob moves from $A$ to $B$.Loss in potential energy $= mg h = mg l(1 - \cos 30^{\circ}) = mgl(1 - \frac{\sqrt{3}}{2})$.Wait,looking at the provided image,$h$ is defined as the vertical distance from the horizontal level of $A$ to the level of $B$. The vertical distance of $B$ from the horizontal level of $A$ is $h = l \sin 30^{\circ} = l/2$.Loss in potential energy $= mgh = mg(l \sin 30^{\circ}) = mg(l/2) = \frac{mgl}{2}$.Therefore,the gain in kinetic energy at $B$ is $\frac{mgl}{2}$.

$A$ simple pendulum is released from point $A$ 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 point $B$ is:

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