Question No. 14
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 $m$ ($<M$) 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 $M$?

(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

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Solution:

General Notes:

• The spring-mass system has initial mass $M$, spring constant $k$, and amplitude $A$.

• After adding mass $m$ (softly, so no initial impulse), the new mass is $M+m$.

• The new angular frequency is ${\omega }^{\mathrm{\prime }}=\sqrt{\frac{k}{M+m}}$, so the time period increases in both cases. Thus, (B) is true.

Case (i): Block at equilibrium $x_0$ when mass is added

• At $x_0$, the velocity is maximum: $v_{\text{max}}=A\omega =A\sqrt{\frac{k}{M}}$.

• During the soft addition, linear momentum is conserved:

  $$M{v}_{\text{max}}=(M+m)v^{\mathrm{\prime }}$$

  So, the new maximum velocity is:

  $$v_{\text{max}}^{\mathrm{\prime }}=\frac{M}{M+m}{v}_{\text{max}}=\frac{M}{M+m}A\sqrt{\frac{k}{M}}$$

• The new amplitude $A^{\mathrm{\prime }}$ is given by:

  $$v_{\text{max}}^{\mathrm{\prime }}=A^{\mathrm{\prime }}{\omega }^{\mathrm{\prime }}=A^{\mathrm{\prime }}\sqrt{\frac{k}{M+m}}$$

  Equating:

  $$A^{\mathrm{\prime }}\sqrt{\frac{k}{M+m}}=\frac{M}{M+m}A\sqrt{\frac{k}{M}}$$

  Solving:

  $$A^{\mathrm{\prime }}=\frac{M}{M+m}A\sqrt{\frac{M+m}{M}}=A\sqrt{\frac{M}{M+m}}$$

  So, the amplitude changes by a factor of $\sqrt{\frac{M}{M+m}}$.
  Also, the total energy decreases because:

  $$E_{\text{initial}}=\frac{1}{2}k{A}^2,E_{\text{final}}=\frac{1}{2}k(A^{\mathrm{\prime }}{)}^2=\frac{1}{2}k{A}^2\frac{M}{M+m}$$

  So, $E_{\text{final}}<E_{\text{initial}}$.
  The instantaneous speed at $x_0$ (which is the maximum speed) decreases from $v_{\text{max}}$ to $v_{\text{max}}^{\mathrm{\prime }}$.

Case (ii): Block at extreme position $x=x_0+A$ when mass is added

• At the extreme position, velocity is zero.

• Adding mass $m$ softly (with zero velocity) does not change the velocity (still zero).

• The spring extension is $A$ initially. After adding mass, the new equilibrium shifts slightly, but the initial displacement relative to the new equilibrium is still $A$ (since the spring force is balanced by the weight? But horizontal spring, so no gravity effect).

• Actually, for a horizontal spring, the equilibrium position remains at $x_0$ even after adding mass. So the initial displacement is still $A$.

• Since the velocity is zero at the time of addition, the new amplitude remains $A$.

• The total energy remains the same:

  $$E_{\text{initial}}=\frac{1}{2}k{A}^2,E_{\text{final}}=\frac{1}{2}k{A}^2$$

• However, the maximum speed (at $x_0$) decreases because:

  $$v_{\text{max}}^{\mathrm{\prime }}=A{\omega }^{\mathrm{\prime }}=A\sqrt{\frac{k}{M+m}}<A\sqrt{\frac{k}{M}}=v_{\text{max}}$$

  So the instantaneous speed at $x_0$ decreases.

Summary:

• (A) True: In case (i), amplitude changes by $\sqrt{M\mathrm{/}(M+m)}$; in case (ii), amplitude unchanged.

• (B) True: Time period becomes $T^{\mathrm{\prime }}=2\pi \sqrt{(M+m)\mathrm{/}k}$ in both cases.

• (C) False: In case (ii), total energy remains the same.

• (D) True: In both cases, the maximum speed (at $x_0$) decreases.

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Final Answer:

$$\boxed{{\displaystyle A,B,D}}$$