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  4. A pendulum is executing simple harmonic motion and its maximum kinetic energy is K1. 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 K2. The ratio K2/K1 is

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

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Solution:

Maximum kinetic energy at lowest point $B =$ Change in potential energy
i.e. $K = mgl (1-cos\, \theta)$
where $\theta =$ angular amp.
image
$K_1 = mg\ell (1 - cos\,\theta) .....(i)$
$K_2 = mg (2\ell)( 1 - cos\,\theta) ....(ii)$
Solving (i) and (ii) we get
$K_2 = 2K_1$.