A pendulum is executing simple harmonic motio | vedclass

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Question

(A) The maximum kinetic energy of a simple pendulum at its lowest point is equal to the potential energy at its maximum angular displacement $	heta$.

Vedclass · Accepted Answer

$K_2 = 2K_1$

Vedclass · Answer

(A) The maximum kinetic energy of a simple pendulum at its lowest point is equal to the potential energy at its maximum angular displacement $	heta$.The potential energy $U$ at an angular displacement $	heta$ is given by $U = mg\ell(1 - \cos 	heta)$,where $m$ is the mass,$g$ is the acceleration due to gravity,and $\ell$ is the length of the pendulum.For the first case,the maximum kinetic energy is $K_1 = mg\ell(1 - \cos 	heta)$.In the second case,the length is doubled $(\ell' = 2\ell)$ and the angular amplitude $	heta$ remains the same. The new maximum kinetic energy is $K_2 = mg(2\ell)(1 - \cos 	heta)$.Comparing the two expressions,we get $K_2 = 2 	imes [mg\ell(1 - \cos 	heta)] = 2K_1$.

$A$ 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 angular amplitude as in the first case,its maximum kinetic energy is $K_2$,then:

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