A block with mass M is connected by a massless spring with stiffness constant k to a rigid wall and

Q: Why is momentum conserved in Case (i) but energy is not?

When mass $m$ is placed on the moving block $M$, they stick together. This is a completely inelastic collision. Since there are no external horizontal forces, horizontal momentum is conserved. However, inelastic collisions always dissipate kinetic energy into heat, which is why mechanical energy is not conserved.

Q: Why does the equilibrium position $x_0$ not shift when the mass is placed?

The spring is horizontal and the motion is on a horizontal frictionless surface. The equilibrium position of a horizontal spring-mass system is determined solely by the natural length of the spring, which does not depend on the mass of the block. Therefore, $x_0$ remains unchanged.

Q: Why is the velocity zero at the extreme position in Case (ii)?

In simple harmonic motion, the extreme positions are the turning points where the velocity of the oscillating mass momentarily becomes zero before reversing direction.

Q: Why does the time period remain the same in both cases?

The time period of a spring-mass system is given by $T = 2\pi\sqrt{m_{total}/k}$. In both cases, after the mass $m$ sticks to $M$, the total oscillating mass is $M+m$ and the spring constant is $k$. Since these parameters are identical, the time period is the same.

Q: How does the maximum speed at the mean position decrease in Case (ii) if the amplitude is unchanged?

The maximum speed is given by $v_{max} = \omega A$. In Case (ii), the amplitude $A$ remains unchanged, but the angular frequency decreases from $\omega_1 = \sqrt{k/M}$ to $\omega_2 = \sqrt{k/(M+m)}$ due to the added mass. Since $\omega_2 < \omega_1$, the maximum speed $v_{max2} = \omega_2 A$ is less than the initial maximum speed $v_{max1} = \omega_1 A$.

Question

A block with mass $M$ is connected by a massless spring with stiffness constant $k$ to a rigid wall and moves without friction on a horizontal surface. The block oscillates with small amplitude $A$ about an equilibrium position $x_0$. Consider two cases : (i) when the block is at $x_0$ and (ii) when the block is at $x = x_0 + A$. In both the cases, a particle with mass $m$ ($m < M$) is softly placed on the block after which they stick to each other. Which of the following statement(s) is (are) true about the motion after the mass $m$ is placed on the mass $M$?

Options:

(A) The amplitude of oscillation in the first case changes by a factor of $\sqrt{\frac{M}{m + M}}$, whereas in the second case it remains unchanged

(B) The final time period of oscillation in both the cases is same

(C) The total energy decreases in both the cases

(D) The instantaneous speed at $x_0$ of the combined masses decreases in both the cases

Sigma Percentile · Accepted Answer

### Introduction to the Problem

Imagine you are watching a playground swing.

If you push someone at the very bottom of their swing, where they are moving fastest, how does it feel compared to gently placing a backpack on their lap when they momentarily stop at the highest point?

This intuitive contrast is exactly what this classic JEE Advanced problem explores.

We are looking at a block of mass $M$ attached to a spring of constant $k$, executing Simple Harmonic Motion (SHM) on a frictionless surface.

Suddenly, a smaller mass $m$ is placed softly on top of it.

Depending on *where* we place this mass—at the mean position or at the extreme position—the resulting motion changes in fascinating ways.

Let's embark on a journey to dissect the mechanics of both scenarios and uncover the elegant truths hidden in the equations.

### Case (i): The High-Speed Collision at the Mean Position

Let's first analyze what happens when the block is at the equilibrium position $x_0$.

At this point, the spring is completely unstretched, meaning the potential energy of the system is zero.

Consequently, all the mechanical energy is stored as kinetic energy, and the block is moving at its maximum speed:

$$v_1 = \omega_1 A$$

Here, the initial angular frequency is:

$$\omega_1 = \sqrt{\frac{k}{M}}$$

Now, we softly place the mass $m$ on top of the moving block.

Because the mass is placed vertically and softly, there are no external horizontal forces acting on the system during this brief interaction.

Therefore, we can confidently apply the **Conservation of Linear Momentum** in the horizontal direction:

$$M v_1 = (M + m) v_2$$

This gives us the velocity of the combined system immediately after the collision:

$$v_2 = \left(\frac{M}{M + m}ight) v_1$$

Notice how the velocity has decreased because the moving mass has increased.

This is a classic completely inelastic collision, which means some mechanical energy is lost as heat.

But what happens to the amplitude of the subsequent oscillations?

Since the collision occurs at the mean position, the velocity $v_2$ is the new maximum velocity of the system:

$$v_2 = \omega_2 A'$$

The new angular frequency of the combined system is:

$$\omega_2 = \sqrt{\frac{k}{M + m}}$$

By equating these, we can solve for the new amplitude $A'$:

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

Simplifying this expression yields:

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

This is a beautiful result! The amplitude has indeed decreased by a factor of $\sqrt{\frac{M}{M + m}}$.

### Case (ii): The Quiet Addition at the Extreme Position

Now, let's contrast this with the second scenario: placing the mass $m$ at the extreme position $x = x_0 + A$.

At the extreme position, the block momentarily comes to a complete stop to change direction.

Its velocity at this instant is exactly zero:

$$v_1 = 0$$

When we softly place the mass $m$ on top of the stationary block, no horizontal momentum is transferred because nothing was moving in the first place!

The velocity immediately after placement remains zero:

$$v_2 = 0$$

Since the system is at rest at a distance $A$ from the equilibrium position, this starting point must still be the extreme position of the new oscillation.

Because the equilibrium position $x_0$ has not shifted, the maximum displacement from the mean position remains exactly $A$.

Therefore, the new amplitude $A''$ is completely unchanged:

$$A'' = A$$

This is a stark contrast to the first case!

Because the block was at rest, no kinetic energy was present to be dissipated, and the potential energy stored in the spring was preserved perfectly.

### The Time Period: A Universal Constant of the System

Now let's look at the time period of oscillation for both cases.

The time period of a simple harmonic oscillator depends only on the total mass and the spring constant:

$$T = 2\pi \sqrt{\frac{	ext{Total Mass}}{k}}$$

In both cases, once the mass $m$ is placed, the total mass of the oscillating system becomes $M + m$.

Since the spring constant remains $k$, the final time period in both cases is:

$$T = 2\pi \sqrt{\frac{M + m}{k}}$$

This means the time period is identical for both scenarios, regardless of where the mass was added.

This confirms that statement (b) is true.

### Energy and Speed: The Final Verdict

Let's evaluate the total mechanical energy.

In Case (i), the inelastic collision at the mean position dissipates energy, so the total energy decreases.

In Case (ii), the mass is placed at rest, so no energy is dissipated, and the total energy remains constant.

Therefore, statement (c), which claims energy decreases in both cases, is false.

Finally, let's examine the instantaneous speed at the mean position $x_0$, which is the maximum speed of the oscillation.

In Case (i), the speed at $x_0$ immediately after placement is $v_2$, which is clearly less than $v_1$.

In Case (ii), the new maximum speed is:

$$v_{max2} = \omega_2 A = \sqrt{\frac{k}{M + m}} A$$

Since $M + m > M$, the new angular frequency $\omega_2$ is smaller than the initial angular frequency $\omega_1$.

Thus, the new maximum speed is less than the initial maximum speed:

$$v_{max2} < v_{max1}$$

So, in both cases, the instantaneous speed at the mean position decreases, making statement (d) true.

### Summary of Correct Options

By carefully analyzing the physics of both cases, we have shown that:

- **Statement (a)** is true because the amplitude in Case (i) changes by $\sqrt{\frac{M}{M + m}}$ and in Case (ii) remains unchanged.
- **Statement (b)** is true because the final mass and spring constant are identical in both cases.
- **Statement (d)** is true because the maximum speed at the mean position decreases in both cases.

Thus, the correct options are **(a), (b), and (d)**.

Select Answer:

Visualized Solution

Visualizing the Setup & Two Cases

Case (i) - Velocity Just Before Placement

Case (i) - Conservation of Momentum

Case (i) - New Angular Frequency

Case (i) - New Amplitude Calculation

Case (ii) - Placement at Extreme Position

Case (ii) - New Amplitude

Comparing Time Periods

Analyzing Total Energy

Analyzing Instantaneous Speed at $x_{0}$

Conclusion

The Sigma Insight: Simple Harmonic Motion (SHM)

Introduction to the Problem

Case (i)

Case (ii)

The Time Period

Energy and Speed

Summary of Correct Options

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