Question

A pendulum bob on a 2 m string is displaced 60° from the
vertical and then released. What is the speed of the bob as
it passes through the lowest point in its path (JIPMER 1999)​

Answer 1

Answer:

Use mechanical energy conservation

$mgh = \frac{1}{2} m {v}^{2} \\ \\ {v}^{2} = 2g(l - l\cos(60)) \\ \\ \cos(60) = \frac{1}{2} \\ \\ {v}^{2} = 2g(l - \frac{l}{2} ) = gl \\ \\ v = \sqrt{gl} = \sqrt{ 10 \times 2} = 2 \sqrt{5} \: \frac{m}{s}$