JEE Main · 2020 · 03 Sep 2020 · Shift 2 · Elasticity & SHM

At the equilibrium point the block moves fastest, $v_{\max }=A\omega =A\sqrt{k/m}$. When half the mass flies off just there, the departing piece carries away none of the restoring force and the remaining block keeps the same velocity at that instant, but its mass is now $m/2$, so its angular frequency becomes ${\omega }^{\prime }=\sqrt{2k/m}$. The new amplitude follows from the same maximum speed, $A^{\prime }{\omega }^{\prime }=v_{\max }$: $A^{\prime }\sqrt{2k/m}=A\sqrt{k/m}$, so $A^{\prime }=\frac{A}{\sqrt{2}}$ and $f=\frac{1}{\sqrt{2}}$.