Question

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 $K_2$. The ratio $K_2/K_1$ is

Answer 1

Answer: 2

Maximum kinetic energy at lowest point $B=$ Change in potential energy
i.e. $K=mgl(1-cos\theta )$
where $\theta =$ angular amp.

$K_1=mg\ell (1-cos\theta ).....(i)$
$K_2=mg(2\ell )(1-cos\theta )....(ii)$
Solving (i) and (ii) we get
$K_2=2K_1$.