A particle is executing SHM. At a point x =A3, kinetic energy of the particle is K, where A is the amplitude. At a point x = 2A3, kinetic energy of the particle will be:
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a
2 K
b
K2
c
58 K
d
53K
answer is C.
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Detailed Solution
v = ω(A2-x2) ⇒ v ∝ (A2-x2)At x = A3(A2-x2) = (A2-4A29) = A89At x = 2A3 (A2-x2) = (A2-4A29) = A59 K' = 59×98K = 58K