Question:
A block of mass $m$ attached to 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 $f \mathrm{~A}$. The value of $f$ is:
Correct Option: , 4
Solution:
At equilibrium position
$V_{0}=\omega_{0} A=\sqrt{\frac{K}{m}} A$......(i)
$V=\omega A^{\prime}=\sqrt{\frac{K}{\frac{m}{2}} A^{\prime}}$$\ldots . .(\mathrm{ii})$
$\therefore \quad A^{1}=\frac{A}{\sqrt{2}}$