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

Concept: This problem involves concepts of Simple Harmonic Motion (SHM) and linear momentum conservation:
• At the mean position, the velocity of a mass undergoing SHM is at its maximum, given by \(v_{\text{max}} = \omega A\), where \(\omega = \sqrt{\frac{k}{m_{\text{total}}}}\).
• Since the mass \(m\) is dropped vertically, there is no external horizontal force acting on the system at the instant of collision. Therefore, the linear momentum in the horizontal direction is conserved.
• After the collision, the total combined mass changes, which alters both the angular frequency of the system and the maximum velocity, consequently establishing a new amplitude \(A'\).

Step 1: Applying conservation of linear momentum at the mean position.
Before the collision, the block of mass \(M\) is passing through its mean position with its initial maximum velocity \(v_0\): \[ v_0 = \omega_1 A = \sqrt{\frac{k}{M}} A \] The initial horizontal momentum of the system is: \[ P_{\text{initial}} = M v_0 \] When the mass \(m\) lands vertically and sticks to it, let the new common horizontal velocity of the combined mass \((M + m)\) be \(v'\). The final horizontal momentum is: \[ P_{\text{final}} = (M + m)v' \] Conserving horizontal linear momentum (\(P_{\text{initial}} = P_{\text{final}}\)): \[ M v_0 = (M + m)v' \quad \Rightarrow \quad v' = \left(\frac{M}{M + m}\right) v_0 \quad \cdots (1) \]

Step 2: Relating the new velocity to the new amplitude $A'$.
The collision occurs precisely at the mean position, meaning the potential energy stored in the spring is zero at this instant, and the new velocity \(v'\) represents the new maximum velocity of the altered SHM system: \[ v' = \omega_2 A' \] Where \(\omega_2\) is the new angular frequency for the combined mass: \[ \omega_2 = \sqrt{\frac{k}{M + m}} \] Substitute \(\omega_2\) and the value of \(v_0\) from Step 1 into equation (1): \[ \sqrt{\frac{k}{M + m}} A' = \left(\frac{M}{M + m}\right) \sqrt{\frac{k}{M}} A \] Squaring both sides to eliminate the radical constraints: \[ \left(\frac{k}{M + m}\right) (A')^2 = \left(\frac{M}{M + m}\right)^2 \left(\frac{k}{M}\right) A^2 \] Cancel out the common spring constant \(k\) and one factor of \((M + m)\) from both denominators: \[ (A')^2 = \frac{M^2}{(M + m)^2} \cdot \frac{M + m}{M} \cdot A^2 \] \[ (A')^2 = \left(\frac{M}{M + m}\right) A^2 \] Taking the square root yields the new amplitude: \[ A' = A\sqrt{\frac{M}{M + m}} \]