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:

Let the force constant of the spring be $$k$$. For the original block of mass $$m$$ the total mechanical energy in simple harmonic motion is, by definition,

$$E=\frac12\,kA^{2}.$$

Whenever the block is at the equilibrium (mean) position, the spring is momentarily unstretched, so its potential energy is zero and the entire energy is kinetic. Hence the speed $$v_{0}$$ of the block at that position obeys

$$\frac12\,m\,v_{0}^{2}=E=\frac12\,kA^{2}\;\;\Longrightarrow\;\;v_{0}^{2}=\frac{kA^{2}}{m}.$$

Exactly at this instant half the mass is removed - the piece is lifted vertically and therefore has no horizontal velocity. Because the horizontal plane is friction-less, there is no external horizontal impulse; therefore the horizontal linear momentum is conserved during the sudden separation.

Before separation the horizontal momentum was

$$p_{\text{before}}=m\,v_{0}.$$

After separation the remaining block has mass $$m/2$$ and velocity $$v_{1}$$, while the detached half has zero horizontal velocity. Conservation of momentum thus gives

$$m\,v_{0}=\frac{m}{2}\,v_{1}+0\quad\Longrightarrow\quad v_{1}=2\,v_{0}.$$

The kinetic energy of the remaining block immediately after the loss of mass is therefore

$$K'=\frac12\left(\frac{m}{2}\right)v_{1}^{2} =\frac12\left(\frac{m}{2}\right)(2v_{0})^{2} =\frac12\left(\frac{m}{2}\right)4v_{0}^{2} =m\,v_{0}^{2}.$$

Substituting $$v_{0}^{2}=\dfrac{kA^{2}}{m}$$ from the earlier relation, we obtain

$$K' = m\left(\frac{kA^{2}}{m}\right)=kA^{2}.$$

Immediately after the separation the spring is still at its natural length, so its potential energy is zero and the entire mechanical energy of the new system equals this kinetic energy:

$$E' = K' = kA^{2}.$$

Let the new amplitude be $$A' = fA$$. During subsequent oscillations the total energy of the lighter block-spring system is

$$E'=\frac12\,k\,(A')^{2}=\frac12\,k\,(fA)^{2} =\frac12\,k\,f^{2}A^{2}.$$

Equating this to the energy just computed,

$$\frac12\,k\,f^{2}A^{2}=kA^{2} \;\;\Longrightarrow\;\; f^{2}=\;2 \;\;\Longrightarrow\;\; f=\sqrt{2}.$$

Therefore the new amplitude is $$A'=\sqrt{2}\,A$$, so

$$f=\sqrt{2}.$$

Hence, the correct answer is Option D.