In the given figure, a mass $$M$$ is attached to a horizontal spring which is fixed on one side to a rigid support. The spring constant of the spring is $$k$$. The mass oscillates on a frictionless surface with time period $$T$$ and amplitude $$A$$. When the mass is in equilibrium position, as shown in the figure, another mass $$m$$ is gently fixed upon it. The new amplitude of oscillation will be:

$$Mv=(M+m)v'$$Initially block M executes SHM with amplitude A.

When it passes through equilibrium position, its speed is maximum:

$$v=\omega A$$

where

$$\omega=\sqrt{\frac{k}{M}}$$

So

$$v=A\sqrt{\frac{k}{M}}$$

Now mass m is gently placed on it and sticks, so use conservation of momentum at that instant:

$$Mv=(M+m)v'$$

Thus

$$v'=\frac{M}{M+m}v$$

$$\frac{M}{M+m}A\sqrt{\frac{k}{M}}$$

Now new mass is

M+m

so new angular frequency is

$$\omega'=\sqrt{\frac{k}{M+m}}$$

At the instant of sticking, system is at equilibrium position, so displacement is zero and all energy is kinetic.

For new SHM, at equilibrium,

$$v′=ω′A′$$

where A′ is new amplitude.

So

$$A'=\frac{v'}{\omega'}$$

Substitute:

$$A'=\frac{\frac{M}{M+m}A\sqrt{k/M}}{\sqrt{k/(M+m)}}$$

Simplifying,

$$A'=A\sqrt{\frac{M}{M+m}}$$