Question

A block of mass $M$ attached to a horizontal spring of spring constant $k$ executes SHM with amplitude $A$. When the block passes through its mean position, a small piece of mass $m$ is dropped vertically onto it and sticks to it. The new amplitude of oscillation is:

• A. \( A\sqrt{\left(\frac{M}{M+m}\right)} \)
• B. \( A\left(\frac{M}{M+m}\right) \)
• C. \( A\sqrt{\left(\frac{M+m}{M}\right)} \)
• D. \( A \)

Hint:

This problem can also be solved using Conservation of Energy. The total mechanical energy of a spring-mass system is completely kinetic at the mean position: \(U_{\text{total}} = \frac{1}{2} m_{\text{total}} v_{\text{max}}^2\). Since \(v' = \frac{M}{M+m}v_0\), the new energy is \(E' = \frac{1}{2}(M+m)(v')^2 = \frac{1}{2}(M+m)\left(\frac{M}{M+m}\right)^2 v_0^2 = \left(\frac{M}{M+m}\right) E\). Because energy is proportional to the square of amplitude (\(E \propto A^2\)), the new amplitude scales matching the square root of the energy fraction!

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
This question concerns Simple Harmonic Motion (SHM) combined with the conservation of linear momentum.
A block of mass $M$ is oscillating on a spring, and a mass $m$ is dropped vertically onto it at the mean position. We need to determine the new amplitude of the system after the collision.
Step 2: Key Formulas and Approach:
Velocity at Mean Position: In SHM, the velocity of the block is maximum at the mean position, given by $v_{\text{max}} = \omega A$.

Angular Frequency: The angular frequency of a spring-mass system is $\omega = \sqrt{\frac{k}{M_{\text{total}}}}$.

Conservation of Momentum: Because the mass $m$ is dropped vertically, there is no external horizontal force. Therefore, linear momentum is conserved in the horizontal direction:
\[ M v_1 = (M + m) v_2 \]
Step 3: Detailed Explanation:

Initial State of the System: Before the mass $m$ is added, the block of mass $M$ passes through the mean position with maximum velocity $v_1$:
\[ v_1 = \omega_1 A = \sqrt{\frac{k}{M}} A \]
Applying Momentum Conservation: When the mass $m$ sticks to the block, the total mass becomes $(M + m)$. Let the new velocity at the mean position be $v_2$:
\[ M v_1 = (M + m) v_2 \quad \implies \quad v_2 = \left(\frac{M}{M+m}\right) v_1 \quad \cdots (1) \]
Relating New Velocity to New Amplitude ($A'$): The collision occurs at the mean position, so $v_2$ is the new maximum velocity of the system:
\[ v_2 = \omega_2 A' = \sqrt{\frac{k}{M+m}} A' \quad \cdots (2) \]
Solving for $A'$: Equating the expressions for $v_2$ from equations (1) and (2):
\[ \sqrt{\frac{k}{M+m}} A' = \left(\frac{M}{M+m}\right) \sqrt{\frac{k}{M}} A \] Squaring both sides of the equation:
\[ \left(\frac{k}{M+m}\right) (A')^2 = \left(\frac{M}{M+m}\right)^2 \left(\frac{k}{M}\right) A^2 \] Simplifying this expression:
\[ (A')^2 = \left(\frac{M}{M+m}\right) A^2 \quad \implies \quad A' = A \sqrt{\frac{M}{M+m}} \]
Step 4: Final Answer:
The new amplitude of oscillation is $A\sqrt{\frac{M}{M+m}}$, which corresponds to Option (A).