The bob of a pendulum is released from positi | vedclass

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Question

(A) Let the length of the pendulum be $l$. When the bob is released from position $P$ (horizontal position),it reaches position $Q$ (lowest point). By

Vedclass · Accepted Answer

$2\sqrt{2gl}$

Vedclass · Answer

(A) Let the length of the pendulum be $l$. When the bob is released from position $P$ (horizontal position),it reaches position $Q$ (lowest point). By the law of conservation of energy,the potential energy at $P$ is converted into kinetic energy at $Q$: $mgl = \frac{1}{2}mv^2 \implies v = \sqrt{2gl}$. At position $Q$,the velocity of the bob is $v_b = \sqrt{2gl}$ directed towards the mirror. The velocity of the image of the bob in a plane mirror is equal in magnitude but opposite in direction to the velocity of the object. Thus,the velocity of the image is $v_i = -\sqrt{2gl}$ (directed away from the mirror). The velocity of the image with respect to the bob is given by: $v_{ib} = v_i - v_b = -\sqrt{2gl} - \sqrt{2gl} = -2\sqrt{2gl}$. The magnitude of this relative velocity is $2\sqrt{2gl}$.

The bob of a pendulum is released from position $P$. What is the velocity of the image of the bob on the mirror $M$ at position $Q$ with respect to the bob?

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