A pendulum bob has a speed of 3, m/s at its lowest position. pendulum is 0.5, m long. speed of bob,

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13.Oscillations

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A pendulum bob has a speed of $3\, m/s$ at its lowest position. The pendulum is $0.5\, m$ long. The speed of the bob, when the length makes an angle of ${60^o}$ to the vertical, will be ..... $m/s$ (If $g = 10\,m/{s^2}$)

A

$3$

B

$0.33$

C

$0.5$

D

$2$

Solution

(d) Let bob velocity be $v$ at point $B$ where it makes an angle of $60^o$ with the vertical, then using conservation of mechanical energy

$K{E_A} + P{E_A} = K{E_B} + P{E_B}$

==> $\frac{1}{2}m \times {3^2} = \frac{1}{2}m{v^2} + mgl(1 – \cos \theta )$

$⇒ 9 = {v^2} + 2 \times 10 \times 0.5 \times \frac{1}{2} $

$⇒ v = 2\,m/s$

Standard 11

Physics

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