A mass M, attached to a horizontal spring, executes SHM with amplitude A1. When the mass M passes th

Q: Why can't we just use the conservation of mechanical energy here? The spring is conservative!

While the spring force is conservative, the act of dropping mass $m$ onto mass $M$ constitutes a perfectly inelastic collision in the horizontal direction. The mass $m$ has to be abruptly accelerated to match $M$'s velocity, which causes internal friction and dissipates kinetic energy as heat. Therefore, total mechanical energy is not conserved.

Q: Why do we conserve momentum only at the mean position? What if it was dropped somewhere else?

We can conserve momentum here specifically because at the mean position ($x=0$), the spring is at its natural length, meaning the spring force ($F = -kx$) is zero. If the mass was dropped at any other position, the spring would be exerting an external horizontal force on the system during the collision, and momentum would not be strictly conserved.

Q: Does the time period of the oscillation change after the mass is dropped?

Yes, absolutely. The time period of a spring-mass system is given by $T = 2\pi\sqrt{\frac{\text{mass}}{k}}$. Since the total oscillating mass increases from $M$ to $(M+m)$, the time period will increase. The system will oscillate more slowly.

Q: What would happen to the amplitude if the mass $m$ was dropped at the extreme position instead?

At the extreme position, the velocity of mass $M$ is zero. If you drop mass $m$ there, no horizontal collision occurs (both have zero horizontal velocity). The total energy is purely potential ($\frac{1}{2}kA^2$), which remains unchanged. Therefore, the amplitude would remain exactly the same, $A_1 = A_2$.

Q: Why does the $\sqrt{k}$ term cancel out in the final equation?

The $\sqrt{k}$ term cancels out because the spring itself doesn't change. The stiffness of the spring remains constant before and after the mass is dropped. The change in amplitude is purely a consequence of the change in mass and the resulting loss of velocity due to momentum conservation.

Question

A mass $M$, attached to a horizontal spring, executes SHM with amplitude $A_1$. When the mass $M$ passes through its mean position, then a smaller mass $m$ is placed over it and both of them move together with amplitude $A_2$. The ratio of $\left(\frac{A_1}{A_2}\right)$ is

Options:

(A) $\frac{M+m}{M}$

(B) $\left(\frac{M}{M+m}\right)^{1/2}$

(C) $\left(\frac{M+m}{M}\right)^{1/2}$

(D) $\frac{M}{M+m}$

Sigma Percentile · Accepted Answer

### The Physics of a Sudden Drop: Mass on an Oscillating Spring

Imagine you are watching a block of mass $M$ oscillating back and forth on a frictionless horizontal surface, tethered to a spring. It's a mesmerizing dance of energy, constantly shifting between the kinetic energy of the block and the elastic potential energy of the spring.

But then, we introduce a sudden disruption. Exactly as the block $M$ passes through its mean position (the equilibrium point), a smaller block of mass $m$ is gently dropped onto it. They stick together and continue the oscillation. The question is: how does this sudden addition of mass affect the amplitude of the oscillation?

### The Trap: Why Energy Conservation Fails Here

A very common instinct is to apply the Law of Conservation of Mechanical Energy. You might think, "The total energy before the drop must equal the total energy after the drop."

**This is a trap.**

When the mass $m$ is dropped onto the moving mass $M$, they undergo a perfectly inelastic collision in the horizontal direction. The mass $m$ initially has zero horizontal velocity, and it must be abruptly accelerated to match the velocity of $M$. This sudden "sticking" process generates internal friction and dissipates kinetic energy as heat. Therefore, mechanical energy is **not** conserved during the collision.

### The Master Key: Momentum Conservation at the Mean Position

If energy isn't conserved, what is? Let's look at the forces.

The mass $m$ is dropped vertically, so it exerts no horizontal force on $M$. What about the spring? The problem states that the drop happens exactly at the **mean position**. At the mean position, the spring is at its natural, unstretched length. This means the spring force is exactly zero ($F_{net} = 0$).

Because there are no external horizontal forces acting on the system at that precise instant, we can safely apply the **Conservation of Linear Momentum** in the horizontal direction.

Let the velocity of mass $M$ at the mean position be $v_1$. After mass $m$ is dropped, let their combined velocity be $v_2$.

$$P_{initial} = P_{final}$$
$$M v_1 = (M+m) v_2$$

### Connecting Kinematics to Dynamics

We need to find the ratio of the amplitudes, $A_1$ and $A_2$. How do we connect velocity to amplitude?

In Simple Harmonic Motion (SHM), the velocity is maximum at the mean position, and it is given by the product of the amplitude and the angular frequency:

$$v_{max} = A \omega$$

Substituting this into our momentum equation, we get:

$$M (A_1 \omega_1) = (M+m) (A_2 \omega_2)$$

Now, we need the expressions for the angular frequencies. The angular frequency of a spring-mass system depends only on the spring constant $k$ and the oscillating mass:

$$\omega_1 = \sqrt{\frac{k}{M}}$$
$$\omega_2 = \sqrt{\frac{k}{M+m}}$$

### The Final Mathematical Symphony

Let's substitute these angular frequencies back into our expanded momentum equation:

$$M A_1 \sqrt{\frac{k}{M}} = (M+m) A_2 \sqrt{\frac{k}{M+m}}$$

Now, we perform a bit of elegant algebra. Notice how $M$ divided by $\sqrt{M}$ simplifies to $\sqrt{M}$. The same logic applies to the $(M+m)$ term on the right side:

$$A_1 \sqrt{M} \sqrt{k} = A_2 \sqrt{M+m} \sqrt{k}$$

The spring constant $\sqrt{k}$ is present on both sides, so it beautifully cancels out, leaving us with a pure relationship between the masses and the amplitudes:

$$A_1 \sqrt{M} = A_2 \sqrt{M+m}$$

Finally, rearranging the terms to find the requested ratio $\frac{A_1}{A_2}$:

$$\frac{A_1}{A_2} = \frac{\sqrt{M+m}}{\sqrt{M}} = \sqrt{\frac{M+m}{M}}$$

And there we have it! The amplitude decreases because the system lost kinetic energy during the inelastic collision, but the exact ratio is dictated purely by the conservation of momentum.

### Food for Thought: The Extreme Position Scenario

What if the mass $m$ was dropped when $M$ was at the **extreme position** instead of the mean position?

At the extreme position, the velocity of $M$ is instantaneously zero. Therefore, the initial momentum is zero. When $m$ is dropped, the final momentum is also zero. No kinetic energy is lost to an inelastic collision because nothing is moving horizontally!

In this scenario, the total mechanical energy (which is entirely potential energy $\frac{1}{2}kA^2$ at that instant) remains unchanged. Consequently, the amplitude would remain exactly the same ($A_1 = A_2$). However, because the mass increased, the system would oscillate more slowly (the time period would increase). Physics is truly fascinating when you change just one initial condition!

Select Answer:

Visualized Solution

$v_{ma x}$ \text{ at Mean Position}

\text{The Inelastic Collision}

\text{Conservation of Linear Momentum}

\text{Relating Velocity and Amplitude}

\text{Angular Frequency } (\omega)

\text{Substituting } \omega

\text{Simplifying the Equation}

\text{Final Ratio}

\text{The Way Forward}

The Sigma Insight: Simple Harmonic Motion (SHM)

The Physics of a Sudden Drop

The Trap

The Master Key

Connecting Kinematics to Dynamics

The Final Mathematical Symphony

Food for Thought

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