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Simple Pendulum

The bob of a ...

Question

The bob of a pendulum is drawn aside so that it is at a height of $4.5 m$ above the ground and released. If it reaches the equilibrium position which is at a height of $2 m$ above the ground, the speed of the bob will be:

A

49 m / s

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B

7 m / s

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C

7 \sqrt{2} m / s

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D

7 / \sqrt{2} m / s

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Solution

The correct option is A $7 m / s$
Consider pendulum $+$ ball as system conserving M.E since there is no external force.
$(M . E)_{A}$ $=$ $(M . E)_{B}$
$m g h_{A} = m g h_{B} + \frac{1}{2} m V {_{B}}^{2}$
$g (4.5 - 2) = \frac{1}{2} V {_{B}}^{2}$
$V_{B} = 7 m / s$

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