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A pendulum bob has a speed of 3 ${ms}^{- 1}$ at its lowest position. The pendulum is 0.5 m long. The speed of the bob, when the length makes an angle of $60^{0}$ to the vertical, will be:

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3 ${ms}^{- 1}$

$\frac{1}{3}$ ${ms}^{- 1}$

$\frac{1}{2}$ ${ms}^{- 1}$

2 ${ms}^{- 1}$

answer is D.

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Detailed Solution

$\frac{1}{2} {mv}^{2} + mgl (1 - \cos θ) = \frac{1}{2} {mv}^{' 2}$

or $v^{2} + 2 gl (1 - cosθ) = v^{2}$

or $v^{2} + 2 \times 10 \times 0.5 (1 - \frac{1}{2}) = 9$

or $v^{2} = 9 - 5 = 4$

or v = 2 ${ms}^{- 1}$

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