Analysis of Mass Addition on Oscillating Block Spring System

To determine the correct statements, let us analyze the simple harmonic motion (SHM) of the system before and after the particle of mass $m$ is placed on the block of mass $M$.

Initial Parameters of the SHM

Before the particle is placed, the block of mass $M$ oscillates with:

• Angular frequency: $\omega_0 = \sqrt{\frac{k}{M}}$
• Time period: $T_0 = 2\pi \sqrt{\frac{M}{k}}$
• Total mechanical energy: $E_0 = \frac{1}{2} k A^2$
• Speed at the equilibrium position $x_0$: $v_0 = \omega_0 A = \sqrt{\frac{k}{M}} A$

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Analysis of Scenario (i): Mass $m$ is placed at $x = x_0$ (Equilibrium Position)

1. Velocity and Amplitude: At the equilibrium position $x_0$, the block $M$ has maximum velocity $v_0$. Since mass $m$ is placed gently, linear momentum in the horizontal direction is conserved during the inelastic impact: $M v_0 = (M + m) v_1 \implies v_1 = \left(\frac{M}{M + m}\right) v_0$

   The angular frequency of the combined system becomes: $\omega' = \sqrt{\frac{k}{M + m}}$

   The new amplitude $A_1$ is related to $v_1$ by $v_1 = \omega' A_1$: $A_1 = \frac{v_1}{\omega'} = \frac{\left(\frac{M}{M + m}\right) \sqrt{\frac{k}{M}} A}{\sqrt{\frac{k}{M + m}}} = \sqrt{\frac{M}{M + m}} A$ Thus, the amplitude in the first scenario changes by a factor of $\sqrt{\frac{M}{M + m}}$.

2. Energy: The total mechanical energy after placing mass $m$ is: $E_1 = \frac{1}{2} k A_1^2 = \frac{1}{2} k \left(\frac{M}{M + m}\right) A^2 = \left(\frac{M}{M + m}\right) E_0 < E_0$ So, the total energy decreases in scenario (i).

3. Speed at $x_0$: The speed at $x_0$ after the mass is placed is $v_1 = \left(\frac{M}{M + m}\right) v_0 < v_0$. Thus, the speed decreases.

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Analysis of Scenario (ii): Mass $m$ is placed at $x = x_0 + A$ (Extreme Position)

1. Velocity and Amplitude: At the extreme position $x = x_0 + A$, the velocity of the block $M$ is instantaneously zero ($v = 0$). Placing mass $m$ gently at rest at this instant transfers zero momentum, so the velocity of the combined mass remains zero at $x = x_0 + A$. Therefore, $x = x_0 + A$ remains the extreme position of the new oscillation, meaning the new amplitude is: $A_2 = A$ The amplitude remains unchanged in the second scenario.

2. Energy: The total mechanical energy after placing mass $m$ is: $E_2 = \frac{1}{2} k A_2^2 = \frac{1}{2} k A^2 = E_0$ Thus, the total energy remains constant in scenario (ii). Consequently, statement (C) is false.

3. Speed at $x_0$: The maximum speed at $x_0$ for the combined mass system in scenario (ii) is: $v_{2, \text{max}} = \omega' A_2 = \sqrt{\frac{k}{M + m}} A$ Since $M + m > M$, we have: $v_{2, \text{max}} = \sqrt{\frac{k}{M + m}} A < \sqrt{\frac{k}{M}} A = v_0$ Thus, the instantaneous speed at $x_0$ decreases in scenario (ii) as well.

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Evaluation of Options

• Option A: The amplitude in the first scenario changes by a factor of $\sqrt{\frac{M}{M + m}}$, whereas in the second scenario it remains unchanged. (TRUE)
• Option B: The final time period of oscillation for both cases depends only on the total mass $(M + m)$ and spring constant $k$: $T_{\text{final}} = 2\pi \sqrt{\frac{M + m}{k}}$ This value is identical in both cases. (TRUE)
• Option C: Total energy decreases in scenario (i) but remains unchanged in scenario (ii). (FALSE)
• Option D: The instantaneous speed at $x_0$ decreases from $v_0$ in both cases ($v_1 < v_0$ and $v_{2, \text{max}} < v_0$). (TRUE)

Correct Options: A, B, D