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JEE Main 2020 (3 Sep, Shift 2) · Spring Systems

Spring Systems question from JEE Main 2020 (3 Sep, Shift 2)

Q80JEE Main 2020 · 3 SEP · SHIFT 2Medium

Spring Systems

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 fA. The value of f is:

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Answer: (d)

When half the mass breaks off, the remaining part keeps the same speed $v$ at equilibrium.
$\omega' = \sqrt{\frac{k}{m/2}} = \sqrt{2}\,\omega$
$A' = \frac{v}{\omega'} = \frac{A}{\sqrt{2}}$ , so $f = \frac{1}{\sqrt{2}}$ .
So the answer is (d).

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