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.

Answer

To solve the problem, we need to find the value of the mass \( m \) suspended from an ideal spring, given the conditions of the oscillation periods.

### Step-by-Step Solution:

1. **Understanding the Time Period Formula**:
The time period \( T \) of a mass \( m \) suspended from a spring with spring constant \( K \) is given by the formula:
\[ T = 2\pi \sqrt{\frac{m}{K}} \]
From the problem, we know that the initial time period \( T_1 = 2 \) seconds.

2. **Setting Up the First Equation**:
Using the time period formula for the initial mass \( m \):
\[ 2 = 2\pi \sqrt{\frac{m}{K}} \]
Dividing both sides by \( 2\pi \):
\[ \frac{2}{2\pi} = \sqrt{\frac{m}{K}} \]
Squaring both sides gives:
\[ \left(\frac{1}{\pi}\right)^2 = \frac{m}{K} \]
Thus, we can express \( m \) in terms of \( K \):
\[ m = \frac{K}{\pi^2} \]

3. **Adding the Additional Mass**:
When an additional mass of \( 2 \) kg is added, the new total mass becomes \( m + 2 \). The new time period \( T_2 \) is given as \( 3 \) seconds. Using the time period formula again:
\[ 3 = 2\pi \sqrt{\frac{m + 2}{K}} \]
Dividing both sides by \( 2\pi \):
\[ \frac{3}{2\pi} = \sqrt{\frac{m + 2}{K}} \]
Squaring both sides gives:
\[ \left(\frac{3}{2\pi}\right)^2 = \frac{m + 2}{K} \]
Rearranging gives:
\[ m + 2 = \frac{9}{4\pi^2} K \]

4. **Setting Up the System of Equations**:
Now we have two equations:
1. \( m = \frac{K}{\pi^2} \)
2. \( m + 2 = \frac{9}{4\pi^2} K \)

5. **Substituting the First Equation into the Second**:
Substitute \( m \) from the first equation into the second:
\[ \frac{K}{\pi^2} + 2 = \frac{9}{4\pi^2} K \]
To eliminate \( K \), multiply through by \( 4\pi^2 \):
\[ 4K + 8\pi^2 = 9K \]
Rearranging gives:
\[ 9K - 4K = 8\pi^2 \]
Thus:
\[ 5K = 8\pi^2 \]
Therefore:
\[ K = \frac{8\pi^2}{5} \]

6. **Finding the Value of \( m \)**:
Substitute \( K \) back into the equation for \( m \):
\[ m = \frac{K}{\pi^2} = \frac{\frac{8\pi^2}{5}}{\pi^2} = \frac{8}{5} \]
Therefore, the value of \( m \) is:
\[ m = 1.6 \text{ kg} \]

### Final Answer:
The value of \( m \) is \( 1.6 \) kg.