# 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 : Option: 1   Option: 2 Option: 3 Option: 4

Let the amplitude of oscillation for the remaining system is $A_1$

then

\begin{aligned} &\text { At equilibrium position }\\ &\mathrm{V}_{0}=\omega_{0} \mathrm{~A}=\sqrt{\frac{\mathrm{K}}{\mathrm{m}}} \mathrm{A}......(1)\\ &\mathrm{V}_0=\omega \mathrm{A}_1=\sqrt{\frac{\mathrm{K}}{\frac{\mathrm{m}}{2}}} \mathrm{~A}_1.......(2)\\ &\text{from equation (1) and (2)}\\ &\text{we get} \ \ \quad \mathrm{A}_1=\frac{\mathrm{A}}{\sqrt{2}} \end{aligned}

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