A horizontal spring -block system of mass `2kg` executes `S.H.M` when the block is passing through its equilibrium position an object of mass `1kg` is put on it the two move together The new amplitude of vibration is `(A` being its initial amplitude)

Options

A. `sqrt((2)/(3))A`
B. `sqrt((3)/(2))A`
C. `sqrt(2)A`
D. `(A)/(sqrt(2))`

Explanation

To solve the problem, we need to find the new amplitude of vibration (A') when a mass of 1 kg is added to a spring-block system that initially has a mass of 2 kg and an amplitude A. The system executes simple harmonic motion (SHM).

### Step-by-Step Solution:

1. **Identify Initial Conditions:**
- Mass of the spring-block system (m1) = 2 kg
- Initial amplitude (A) = A
- Initial velocity (v) = v (unknown)

2. **Conservation of Momentum:**
- Before the mass is added, the momentum of the system is given by:
\[ p_{initial} = m_1 \cdot v = 2 \cdot v \]
- After adding the 1 kg mass, the total mass becomes:
\[ m_{total} = m_1 + m_{added} = 2 + 1 = 3 \text{ kg} \]
- Let the new velocity after adding the mass be \( v' \). The momentum after adding the mass is:
\[ p_{final} = m_{total} \cdot v' = 3 \cdot v' \]
- By conservation of momentum:
\[ p_{initial} = p_{final} \]
\[ 2v = 3v' \]
- From this, we can express \( v' \):
\[ v' = \frac{2}{3}v \]

3. **Calculate Initial and Final Kinetic Energy:**
- Initial kinetic energy (KE_initial) is given by:
\[ KE_{initial} = \frac{1}{2} m_1 v^2 = \frac{1}{2} \cdot 2 \cdot v^2 = v^2 \]
- Final kinetic energy (KE_final) after adding the mass is:
\[ KE_{final} = \frac{1}{2} m_{total} (v')^2 = \frac{1}{2} \cdot 3 \cdot \left(\frac{2}{3}v\right)^2 \]
\[ = \frac{1}{2} \cdot 3 \cdot \frac{4}{9}v^2 = \frac{2}{3}v^2 \]

4. **Relate Kinetic Energy to Spring Potential Energy:**
- The total energy in the spring-block system is given by the potential energy stored in the spring:
\[ KE_{initial} = \frac{1}{2} k A^2 \]
- For the new amplitude \( A' \):
\[ KE_{final} = \frac{1}{2} k (A')^2 \]
- Setting the initial energy equal to the final energy:
\[ \frac{1}{2} k A^2 = \frac{2}{3} v^2 \]
\[ \frac{1}{2} k (A')^2 = \frac{2}{3} v^2 \]

5. **Equate Energies:**
- Since both expressions equal \( \frac{2}{3} v^2 \):
\[ \frac{1}{2} k A^2 = \frac{1}{2} k (A')^2 \]
- Canceling \( \frac{1}{2} k \) from both sides:
\[ A^2 = \frac{2}{3} A'^2 \]
- Rearranging gives:
\[ A'^2 = \frac{3}{2} A^2 \]
- Taking the square root:
\[ A' = \sqrt{\frac{2}{3}} A \]

### Final Answer:
The new amplitude of vibration is:
\[ A' = \sqrt{\frac{2}{3}} A \]