A block of mass m attached to a massless spring is performing...

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 &fnof;A. The value of &fnof; is:

Options

A $\dfrac{1}{\sqrt{2}}$
B $1$
C $\dfrac{1}{2}$
D $\sqrt{2}$

Correct answer

D. $\sqrt{2}$

Step-by-step solution

Using mometum conservation

${m}_{1}{v}_{1} = {m}_{2}{v}_{2}mA\omega = \dfrac{m}{2}{A}_{2}{\omega}_{2}mA\sqrt{\dfrac{k}{m}} = \dfrac{m}{2}{A}_{2}\sqrt{\dfrac{k}{m / 2}}{A}_{2} = \sqrt{2}A ⇒ f = \sqrt{2}$

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