A light spring of force constant k is attached to a block of mass M placed on a horizontal firctionless surface from its one end and other end is fixed to a right support. The system is executing S.H.M. of amplitude `A_(1)` and time period `T_(1)`. At some instant, the block passed through the equilibrium position and a small object of mass m is placed on it. The new amplitude and time period of the system now becomes `A_(2)` and `T_(2)` respectively.
The ratio `A_(2)//A_(1)` is

Options

A. `((M+m)/(M))^(1//2)`
B. `((M)/(M+m))^(1//2)`
C. `((M-m)/(M^(1//2)))`
D. `((M+m)/(M-m)^(1//2))`

Explanation

To solve the problem, we need to find the ratio of the new amplitude \( A_2 \) to the original amplitude \( A_1 \) after a small mass \( m \) is placed on a block of mass \( M \) that is already executing simple harmonic motion (SHM) with amplitude \( A_1 \) and time period \( T_1 \).

### Step-by-Step Solution:

1. **Understanding the Initial Conditions:**
- The system consists of a block of mass \( M \) attached to a spring with force constant \( k \).
- The block is executing SHM with amplitude \( A_1 \) and time period \( T_1 \).

2. **At the Equilibrium Position:**
- When the block passes through the equilibrium position, it has a maximum velocity \( V_1 \).
- The kinetic energy at this position is given by:
\[ KE_1 = \frac{1}{2} M V_1^2 \]
- The potential energy at the extreme position (maximum amplitude) is:
\[ PE_1 = \frac{1}{2} k A_1^2 \]
- At maximum displacement, all kinetic energy converts to potential energy:
\[ \frac{1}{2} M V_1^2 = \frac{1}{2} k A_1^2 \]

3. **Introducing the Small Mass \( m \):**
- When the small mass \( m \) is placed on the block of mass \( M \) at the equilibrium position, the total mass of the system becomes \( M + m \).
- The new velocity \( V_2 \) of the combined mass system can be found using the conservation of momentum:
\[ M V_1 = (M + m) V_2 \]
- Rearranging gives:
\[ V_2 = \frac{M}{M + m} V_1 \]

4. **Calculating the New Kinetic Energy:**
- The new kinetic energy when both masses are moving together at the equilibrium position is:
\[ KE_2 = \frac{1}{2} (M + m) V_2^2 \]
- Substituting \( V_2 \):
\[ KE_2 = \frac{1}{2} (M + m) \left( \frac{M}{M + m} V_1 \right)^2 = \frac{1}{2} \frac{M^2}{M + m} V_1^2 \]

5. **Relating Kinetic Energy to Potential Energy for the New System:**
- The potential energy at the new maximum amplitude \( A_2 \) is:
\[ PE_2 = \frac{1}{2} k A_2^2 \]
- Setting the kinetic energy equal to the potential energy at maximum displacement:
\[ \frac{1}{2} \frac{M^2}{M + m} V_1^2 = \frac{1}{2} k A_2^2 \]

6. **Finding the Ratio of Amplitudes:**
- From the original condition, we have:
\[ \frac{1}{2} M V_1^2 = \frac{1}{2} k A_1^2 \implies V_1^2 = \frac{k A_1^2}{M} \]
- Substituting this into the equation for \( PE_2 \):
\[ \frac{M^2}{M + m} \cdot \frac{k A_1^2}{M} = k A_2^2 \]
- Simplifying gives:
\[ \frac{M}{M + m} k A_1^2 = k A_2^2 \implies A_2^2 = \frac{M}{M + m} A_1^2 \]
- Taking the square root:
\[ \frac{A_2}{A_1} = \sqrt{\frac{M}{M + m}} \]

### Final Result:
The ratio of the new amplitude to the original amplitude is:
\[ \frac{A_2}{A_1} = \sqrt{\frac{M}{M + m}} \]