A block of mass m attached to a massless spring is performing oscillatory motion of amplitude ‘A’ on a frictionless horizontal plane. If half of the mass of the block breaks off when it is passing through its equilibrium point, the amplitude of oscillation for the remaining system become nA. The value of n is :
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a
12
b
1
c
12
d
2
answer is A.
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Detailed Solution
If the block breaks off when it passes through mean position it will carry its own momentum with it. Which means its velocity still remains sameAω=Afωf ⇒Akm=Af2km ⇒Af=A2