Q. A pendulum is executing simple harmonic motion and its maximum kinetic energy is $K_1$. If the length of the pendulum is doubled and it performs simple harmonic motion with the same amplitude as in the first case, its maximum kinetic energy is $K_2$. Then :
Solution:
Maximum kinetic energy at lowest point $B$ is given by
$K = mgl (1 - \cos \theta)$
where $\theta $= angular amp.
$K_1 = mg \ell (1 - \cos \theta)$
$K_2 = mg(2 \ell) (1 - \cos \theta)$
$K_2 = 2K_1. $
