Oscillations

JEE Main level, single correct, medium

91 of 194

Question 91 of 194

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 ƒA. The value of ƒ is:

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Answer: (D) 2\sqrt{2}

Using mometum conservation
m1v1=m2v2mAω=m2A2ω2mAkm=m2A2km/2A2=2A⇒f=2{m}_{1}{v}_{1}={m}_{2}{v}_{2} mA\omega =\dfrac{m}{2}{A}_{2}{\omega }_{2} mA\sqrt{\dfrac{k}{m}}=\dfrac{m}{2}{A}_{2}\sqrt{\dfrac{k}{m/2}} {A}_{2}=\sqrt{2}A\Rightarrow f=\sqrt{2}

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