`A` mass `M`, attached to a spring, oscillates with a period of `2 s`. If the mass is increased by `4kg`, the time period increases by one second. Assuming that Hooke's law is obeyed, find the initial mass `M`.

Explanation

To solve the problem, we will use the formula for the period of oscillation of a mass-spring system, which is given by:

\[ T = 2\pi \sqrt{\frac{m}{k}} \]

where:
- \( T \) is the period of oscillation,
- \( m \) is the mass attached to the spring,
- \( k \) is the spring constant.

### Step 1: Write the equations for the two scenarios

1. For the initial mass \( M \) with a period of \( T_1 = 2 \) seconds:
\[ T_1 = 2\pi \sqrt{\frac{M}{k}} \quad \text{(1)} \]

2. For the new mass \( M + 4 \) kg with a period of \( T_2 = 3 \) seconds:
\[ T_2 = 2\pi \sqrt{\frac{M + 4}{k}} \quad \text{(2)} \]

### Step 2: Set up the equations based on the periods

From equation (1):
\[ 2 = 2\pi \sqrt{\frac{M}{k}} \]
Dividing both sides by \( 2\pi \):
\[ \frac{2}{2\pi} = \sqrt{\frac{M}{k}} \quad \text{(3)} \]

From equation (2):
\[ 3 = 2\pi \sqrt{\frac{M + 4}{k}} \]
Dividing both sides by \( 2\pi \):
\[ \frac{3}{2\pi} = \sqrt{\frac{M + 4}{k}} \quad \text{(4)} \]

### Step 3: Square both equations

Squaring equation (3):
\[ \left(\frac{2}{2\pi}\right)^2 = \frac{M}{k} \]
\[ \frac{4}{4\pi^2} = \frac{M}{k} \]
\[ M = \frac{4k}{4\pi^2} \quad \text{(5)} \]

Squaring equation (4):
\[ \left(\frac{3}{2\pi}\right)^2 = \frac{M + 4}{k} \]
\[ \frac{9}{4\pi^2} = \frac{M + 4}{k} \]
\[ M + 4 = \frac{9k}{4\pi^2} \quad \text{(6)} \]

### Step 4: Substitute equation (5) into equation (6)

Substituting \( M \) from equation (5) into equation (6):
\[ \frac{4k}{4\pi^2} + 4 = \frac{9k}{4\pi^2} \]
Multiply through by \( 4\pi^2 \) to eliminate the denominator:
\[ 4k + 16\pi^2 = 9k \]
Rearranging gives:
\[ 16\pi^2 = 9k - 4k \]
\[ 16\pi^2 = 5k \]
\[ k = \frac{16\pi^2}{5} \quad \text{(7)} \]

### Step 5: Substitute \( k \) back into equation (5)

Now substitute \( k \) from equation (7) back into equation (5) to find \( M \):
\[ M = \frac{4 \cdot \frac{16\pi^2}{5}}{4\pi^2} \]
The \( 4\pi^2 \) cancels:
\[ M = \frac{16}{5} = 3.2 \text{ kg} \]

### Final Answer
The initial mass \( M \) is \( 3.2 \) kg.

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