A bob of a simple pendulum 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 $(60)^{\circ}$ with the vertical, is
(Take $g = 10$ m/s $^{2}$ , $cos 6 0^{\circ} = 1/2$ )

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Let the lowest position be the reference level for potential energy. At the lowest position, the total mechanical energy is purely kinetic:

$E_{i} = \frac{1}{2} m v_{0}^{2}$

When the pendulum makes an angle $θ$ with the vertical, the height of the bob above the lowest position is:

$h = L - L cos θ = L (1 - cos θ)$

At this position, the total mechanical energy is the sum of kinetic and potential energy:

$E_{f} = \frac{1}{2} m v^{2} + m g h$

By conservation of mechanical energy:

$\frac{1}{2} m v_{0}^{2} = \frac{1}{2} m v^{2} + m g h$

$v^{2} = v_{0}^{2} - 2 g h$

Substituting $h = L (1 - cos θ)$ :

$v^{2} = v_{0}^{2} - 2 gL (1 - cos θ)$

Given $v_{0} = 3$ m/s, $L = 0.5$ m, $θ = 6 0^{\circ}$ , and $g = 10$ m/s $^{2}$ :

$v^{2} = (3)^{2} - 2 \times 10 \times 0.5 \times (1 - cos 6 0^{\circ})$

$v^{2} = 9 - 10 \times (1 - \frac{1}{2})$

$v^{2} = 9 - 10 \times \frac{1}{2}$

$v^{2} = 9 - 5 = 4$

$v = 2$ m/s

Answer: $2$ m/s

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