A simple pendulum, while oscillating rises to a maximum vertical height of $5 c m$ from its rest position. Assume its rest position is at ground level. If the mass of bob of the simple pendulum is $500 g$ and acceleration due to gravity g = $10 m / s^{2}$ . Find the below:
i) The total energy of simple pendulum at any instant while oscillating.
ii) The velocity of bob at its resting position.

0.25 J, 1 m/s

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0.25 J, 5 m/s

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0.30 J, 5 m/s

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0.30 J, 1 m/s

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Solution

The correct option is A
0.25 J, 1 m/s

Given,
Mass $m = 500 g = 0.5 k g$
Height $h " = 5 c m$ = $0.05 m$
Assuming there is no loss in energy due to air friction,
(i) Total energy of simple pendulum
= Potential energy at its extreme position
= $m g h$ = $0.5 \times 10 \times 0.05$ = $0.25 J$
(ii) Kinetic energy at a mean position
Change in PE = Change in KE
= $\frac{1}{2} m v^{2} = m g h$
or $v = \sqrt{2 g h} = \sqrt{2 \times 10 \times 0.05} = \sqrt{1}$
= $1 m s^{- 1}$

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