Question

A simple pendulum has a bob of mass M. The bob is on a light string of length . The string is fixed at C. At position A, the string is horizontal and the bob is at rest. The bob is released from A and swings to B, where the string is vertical. What is the velocity of the bob when it is at B?

Answer 1

Answer:

The answer is v=\sqrt{2gL}

Answer 2

Answer:

$v=\sqrt{2gL}$

Explanation:

mass of bob = M

string is fixed at C, at position A the string is horizontal and at position B teh string is vertical.

Let the length of the string is L.

At the point C, it has maximum potential energy which is equal to

U = M x g x L   ..... (1)

At the position B, it has maximum kinetic energy and the velocity is v.

K = 1/2 Mv²     ...... (2)

According to the conservation of energy

The potential energy at the position A is equal to the kinetic energy at position B.

M x g x L = 1/2 M x v²

v² = 2 x g x L

$v=\sqrt{2gL}$