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A simple pendulum of length 90 cm is oscillating with an angular amplitude of $4^{o}$ ,Then the maximum velocity of its bob is $(Take g = 10 m / s^{2})$

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a

$\frac{π}{15} m / s$

b

$\frac{π}{30} m / s$

c

$\frac{π}{10} m / s$

d

$\frac{π}{5} m / s$

answer is A.

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Detailed Solution

$\begin{array}{l} Angular amplitudeϕ = \frac{4 \times π}{180} rad = \frac{π}{45} rad . \\ ∴ Linear amplitude A = ϕd \\ Angular frequency, w = \sqrt{\frac{g}{l}} \\ ∴ V_{\max} = w . A = \sqrt{\frac{g}{l}} \times ϕl = ϕ \sqrt{gl} = \frac{π}{45} \times \sqrt{10 \times 0 .9} m / s = \frac{π}{15} m / s \end{array}$

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A simple pendulum of length 90 cm is oscillating with an angular amplitude of 4o ,Then the maximum velocity of its bob is (Take g=10m/s2)