The period of oscillation of a mass m suspended by an ideal spring is 2s. If an additional mass of 2 kg be suspended, the time period is increased by 1s. Find the value of m.

Options

A. `2 kg`
B. `1 kg`
C. `1.6 kg`
D. `2.6 kg`

Explanation

To solve the problem, we will use the formula for the period of oscillation of a mass-spring system. The period \( T \) is given by:

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

where \( m \) is the mass attached to the spring and \( k \) is the spring constant.

### Step 1: Set up the equations for the two scenarios

1. **Initial scenario** (mass \( m \)):
- Given that the period \( T_1 = 2 \) seconds, we can write:
\[ T_1 = 2\pi \sqrt{\frac{m}{k}} = 2 \]
Dividing both sides by \( 2\pi \):
\[ \sqrt{\frac{m}{k}} = \frac{1}{\pi} \]
Squaring both sides gives:
\[ \frac{m}{k} = \frac{1}{\pi^2} \quad \text{(Equation 1)} \]

2. **After adding 2 kg** (mass \( m + 2 \)):
- The new period \( T_2 = 3 \) seconds, so we have:
\[ T_2 = 2\pi \sqrt{\frac{m + 2}{k}} = 3 \]
Dividing both sides by \( 2\pi \):
\[ \sqrt{\frac{m + 2}{k}} = \frac{3}{2\pi} \]
Squaring both sides gives:
\[ \frac{m + 2}{k} = \frac{9}{4\pi^2} \quad \text{(Equation 2)} \]

### Step 2: Relate the two equations

Now we can relate Equation 1 and Equation 2 by eliminating \( k \).

From Equation 1:
\[ k = \frac{m}{\frac{1}{\pi^2}} = m \pi^2 \]

Substituting \( k \) into Equation 2:
\[ \frac{m + 2}{m \pi^2} = \frac{9}{4\pi^2} \]

### Step 3: Solve for \( m \)

Cross-multiplying gives:
\[ 4(m + 2) = 9m \]
Expanding and rearranging:
\[ 4m + 8 = 9m \]
\[ 8 = 9m - 4m \]
\[ 8 = 5m \]
\[ m = \frac{8}{5} = 1.6 \text{ kg} \]

### Conclusion

Thus, the value of \( m \) is \( 1.6 \) kg.