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Q.

A simple pendulum of length ‘l’ has a bob of mass ‘m’. If it is released when the string makes an angle $60 °$ with the vertical, the tension in the string at the lowest position is

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a

\frac{5}{2} mg

b

\frac{3}{2} mg

c

2 mg

d

3mg

answer is C.

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Detailed Solution

v^{2} = v_{0}^{2} - 2 g l (1 - \cos θ)

$v = 0$

$v_{0}^{2} = 2 g l (1 - \cos θ)$

$v_{0} = \sqrt{2 g l (1 - \cos 60 °)}$

$θ = 60 °$

$v_{0} = \sqrt{2 g l \times \frac{1}{2}}$

$v_{0} = \sqrt{g l}$

$T = \frac{{mv}^{2}}{l} + mg \cos θ$

$θ = 0 °$

$T = \frac{{(m \sqrt{g l})}^{2}}{l} + mg \cos 0 °$

$T = \frac{mg l}{l} + mg = 2 mg$

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A simple pendulum of length ‘l’ has a bob of mass ‘m’. If it is released when the string makes an angle60° with the vertical, the tension in the string at the lowest position is