First slide

Applications of SHM

Question

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 (If $g = 10 m / s^{2}$ )

Moderate

A

$3 m / s$
B

$\frac{1}{3} m / s$
C

$\frac{1}{2} m / s$
D

$2 m / s$

Solution

Let bob velocity be v at point B where it makes an angle of 60^owith the vertical, then using conservation of mechanical energy
${KE}_{A} + {PE}_{A} = {KE}_{B} + {PE}_{B}$
$\Rightarrow \frac{1}{2} m \times 3^{2} = \frac{1}{2} {mv}^{2} + mgl (1 - cosθ)$
$\Rightarrow 9 = v^{2} + 2 \times 10 \times 0.5 \times \frac{1}{2} \Rightarrow v = 2 m / s$

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