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 `T_(2)/T_(1)` is

Options

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

Explanation

To solve the problem, we need to find the ratio of the time periods \( T_2 \) and \( T_1 \) after a small mass \( m \) is added to the block of mass \( M \) that is already oscillating due to the spring.

### Step-by-Step Solution:

1. **Identify the Time Period for the Initial System:**
The time period \( T_1 \) of the block of mass \( M \) attached to the spring with spring constant \( k \) is given by the formula:
\[ T_1 = 2\pi \sqrt{\frac{M}{k}} \]

2. **Identify the Time Period for the New System:**
When the small mass \( m \) is added to the block, the total mass that is now oscillating becomes \( M + m \). The new time period \( T_2 \) is given by:
\[ T_2 = 2\pi \sqrt{\frac{M + m}{k}} \]

3. **Calculate the Ratio \( \frac{T_2}{T_1} \):**
To find the ratio of the new time period to the old time period, we divide \( T_2 \) by \( T_1 \):
\[ \frac{T_2}{T_1} = \frac{2\pi \sqrt{\frac{M + m}{k}}}{2\pi \sqrt{\frac{M}{k}}} \]
The \( 2\pi \) and \( \sqrt{k} \) cancel out:
\[ \frac{T_2}{T_1} = \frac{\sqrt{M + m}}{\sqrt{M}} = \sqrt{\frac{M + m}{M}} \]

4. **Final Expression:**
Thus, the ratio of the time periods is:
\[ \frac{T_2}{T_1} = \sqrt{1 + \frac{m}{M}} \]

### Final Answer:
The ratio \( \frac{T_2}{T_1} \) is \( \sqrt{1 + \frac{m}{M}} \).

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