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. `sqrt2A`
D. `(A)/(sqrt2)`

Explanation

To solve the problem of finding the new amplitude of the spring-block system after an additional mass is placed on it, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Initial Conditions:**
- The initial mass of the block, \( m_1 = 2 \, \text{kg} \).
- The initial amplitude of vibration, \( A \).
- The additional mass placed on the block, \( m_2 = 1 \, \text{kg} \).
- The total mass after adding the second block, \( m = m_1 + m_2 = 2 \, \text{kg} + 1 \, \text{kg} = 3 \, \text{kg} \).

2. **Conservation of Momentum:**
- When the additional mass is placed on the block at the equilibrium position, the momentum of the system is conserved.
- Let the initial velocity of the 2 kg block be \( v \). After the 1 kg mass is added, let the new velocity be \( v' \).
- By conservation of momentum:
\[ m_1 v = (m_1 + m_2) v' \]
\[ 2v = 3v' \quad \Rightarrow \quad v' = \frac{2}{3}v \]

3. **Kinetic Energy and Potential Energy Relation:**
- The total mechanical energy in the system is conserved. The kinetic energy at the equilibrium position is equal to the potential energy at the extreme position.
- Initially, the kinetic energy is given by:
\[ KE_{\text{initial}} = \frac{1}{2} m_1 v^2 = \frac{1}{2} \cdot 2 \cdot v^2 = v^2 \]
- After adding the mass, the kinetic energy becomes:
\[ KE_{\text{final}} = \frac{1}{2} (m_1 + m_2) v'^2 = \frac{1}{2} \cdot 3 \cdot \left(\frac{2}{3}v\right)^2 = \frac{1}{2} \cdot 3 \cdot \frac{4}{9} v^2 = \frac{2}{3} v^2 \]

4. **Relate Kinetic Energy to Amplitude:**
- The potential energy at the extreme position is given by:
\[ PE = \frac{1}{2} k A^2 \]
- Since the total energy is conserved:
\[ KE_{\text{initial}} = PE_{\text{initial}} \quad \Rightarrow \quad v^2 = \frac{1}{2} k A^2 \]
- For the new mass:
\[ KE_{\text{final}} = PE_{\text{final}} \quad \Rightarrow \quad \frac{2}{3} v^2 = \frac{1}{2} k A'^2 \]

5. **Setting the Equations Equal:**
- From the equations for kinetic energy:
\[ v^2 = \frac{1}{2} k A^2 \quad \text{and} \quad \frac{2}{3} v^2 = \frac{1}{2} k A'^2 \]
- Substitute \( v^2 \) from the first equation into the second:
\[ \frac{2}{3} \cdot \frac{1}{2} k A^2 = \frac{1}{2} k A'^2 \]
- Cancel \( \frac{1}{2} k \) from both sides:
\[ \frac{2}{3} A^2 = A'^2 \]

6. **Solve for the New Amplitude:**
- Taking the square root of both sides:
\[ A' = A \sqrt{\frac{2}{3}} \]

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