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Q.

A simple pendulum of length 1 m is oscillated at a place where $g = 9.8 m / s^{2}$ . Find the maximum velocity of the bob of the simple pendulum if the amplitude of oscillation is 3cm.

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a

$7.261 {cms}^{- 1}$

b

$9.42 {cms}^{- 1}$

c

$5.291 {cms}^{- 1}$

d

$6.181 {cms}^{- 1}$

answer is D.

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Detailed Solution

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$V_{\max} = A ω$

$ω for simple pendulum = \sqrt{\frac{g}{l}} = \sqrt{\frac{9.8}{1}} \approx π = 3.14$

$V_{\max} = Aω = 3 x 3.14 = 9.42 cm / \sec$

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