Animated Solution for Physics - Oscillations: 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

Oscillations are beautiful, rhythmic, and predictable. But what happens when we disturb an oscillating system mid-flight? This classic JEE problem explores exactly that scenario, blending the kinematics of Simple Harmonic Motion (SHM) with the fundamental laws of mechanics.

Imagine a block of mass $M$ happily oscillating on a frictionless horizontal surface, attached to a spring of constant $k$. It has an amplitude $A$. As it oscillates, its velocity constantly changes. It is momentarily at rest at the extreme positions ($x=\pm A$) and moves the fastest when it crosses the equilibrium or mean position ($x=0$).

We know from the kinematics of SHM that the maximum velocity occurs at the mean position and is given by:

Since the angular frequency $\omega$ for a spring-mass system is $\sqrt{\frac{k}{M}}$, we can write the initial maximum velocity as:

Now, the problem states that exactly when the mass $M$ is at its equilibrium position, another mass $m$ is gently fixed upon it. The word "gently" is a crucial physics code word. It implies that the mass $m$ is dropped with zero initial horizontal velocity, and no external impulsive force is applied in the horizontal direction during the placement.

Because the net external force in the horizontal direction is zero ($\sum F_x=0$), the linear momentum of the system in the x-direction must be conserved during this perfectly inelastic collision.

Let's apply the principle of conservation of linear momentum just before and just after the mass $m$ is dropped. Let the new velocity of the combined mass system be $v^{\prime }$.

Substituting our expression for $v_{max}$, we get:

This allows us to find the new velocity $v^{\prime }$ of the combined system right after the collision:

After the collision, we have a brand new oscillating system. The total mass is now $(M+m)$, but the spring constant $k$ remains the same. This means our new system has a new angular frequency, ${\omega }^{\prime }$:

Crucially, because the collision happened exactly at the mean position, the velocity $v^{\prime }$ we just calculated is the maximum velocity of this new system! For any SHM, the maximum velocity is the product of its amplitude and its angular frequency. Therefore, if the new amplitude is $A^{\prime }$, we can write:

Now, we simply equate our two expressions for $v^{\prime }$:

Let's do the algebra. We can immediately cancel the $\sqrt{k}$ terms from both sides:

Rearranging to solve for $A^{\prime }$:

Simplifying the square roots, we arrive at our final elegant answer:

Why did the amplitude decrease? Because dropping the mass resulted in a perfectly inelastic collision. Kinetic energy was dissipated as heat or sound, leaving the system with less total mechanical energy, and thus a smaller amplitude.

But what if we dropped the mass $m$ when the block $M$ was at its extreme position? At the extreme position, the velocity of $M$ is zero. The total momentum is zero. Dropping a mass there doesn't change the velocity (it stays zero). The system would simply start its return journey from that exact same spot. Therefore, the amplitude would remain exactly $A$! However, because the mass increased, the system would oscillate more slowly, meaning its time period would increase.