JEE MAIN - Physics (2019 - 11th January Evening Slot - No. 7)
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. Then :
$${K_2}$$ = $${{{K_1}} \over 2}$$
K2 = 2K1
K2 = K1
K2 = $${{{K_1}} \over 4}$$
Explanation
Maximum kinetic energy at lowest point B is given by
K = mg$$\ell $$ (1 $$-$$ cos $$\theta $$)
where $$\theta $$ = angular amp.
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K1 = mg$$\ell $$ (1 $$-$$ cos $$\theta $$)
K2 = mg(2$$\ell $$) (1 $$-$$ cos $$\theta $$)
K2 = 2K1.
K = mg$$\ell $$ (1 $$-$$ cos $$\theta $$)
where $$\theta $$ = angular amp.
K1 = mg$$\ell $$ (1 $$-$$ cos $$\theta $$)
K2 = mg(2$$\ell $$) (1 $$-$$ cos $$\theta $$)
K2 = 2K1.
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