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 \frac{1}{\sqrt{2}}  
Option: 2 1
Option: 3 \frac{1}{2}
Option: 4 \sqrt{2}

Answers (1)

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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