When a mass m attached to a spring it oscillates with period 4s. When an additional mass of 2 kg is attached to a spring, time period increases by 1s. The value of m is :-

Options

A. 3.5 kg
B. 8.2 kg
C. 4.7 kg
D. 2.6 kg

Answer

To solve the problem, we need to find the value of the mass \( m \) that is attached to a spring, given the time periods of oscillation with different masses.

### Step-by-Step Solution:

1. **Understand the formula for the time period of a spring-mass system**:
The time period \( T \) of a mass \( m \) attached to a spring with spring constant \( k \) is given by:
\[ T = 2\pi \sqrt{\frac{m}{k}} \]

2. **Set up the equations for the two scenarios**:
- When the mass is \( m \), the time period \( T_1 \) is 4 seconds:
\[ T_1 = 2\pi \sqrt{\frac{m}{k}} = 4 \]
- When an additional mass of 2 kg is added, the new mass becomes \( m + 2 \) and the time period \( T_2 \) is 5 seconds:
\[ T_2 = 2\pi \sqrt{\frac{m + 2}{k}} = 5 \]

3. **Square both equations to eliminate the square root**:
- From the first equation:
\[ T_1^2 = (2\pi)^2 \frac{m}{k} = 16 \implies \frac{m}{k} = \frac{16}{(2\pi)^2} \]
- From the second equation:
\[ T_2^2 = (2\pi)^2 \frac{m + 2}{k} = 25 \implies \frac{m + 2}{k} = \frac{25}{(2\pi)^2} \]

4. **Set up a ratio of the two equations**:
\[ \frac{m + 2}{m} = \frac{25}{16} \]

5. **Cross-multiply to solve for \( m \)**:
\[ 16(m + 2) = 25m \]
Expanding this gives:
\[ 16m + 32 = 25m \]
Rearranging terms:
\[ 25m - 16m = 32 \implies 9m = 32 \implies m = \frac{32}{9} \]

6. **Calculate the numerical value**:
\[ m \approx 3.56 \text{ kg} \]

### Final Answer:
The value of \( m \) is approximately \( 3.56 \) kg.