A body of mass m is attached to the lower end of a spring whose upper end is fixed. The spring has negligible mass. When the mass m is slightly pulled down and released, it oscillates with a time period of 3s. When the mass m is increased by `1 kg`, the time period of oscillations becomes 5s. The value of m in kg is

Options

A. `(3)/(4)`kg
B. `(4)/(3)`kg
C. `(16)/(9)`kg
D. `(9)/(16)`kg

Answer

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

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

where \( T \) is the time period, \( m \) is the mass attached to the spring, and \( k \) is the spring constant.

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

1. **For the first case** (mass \( m \) and time period \( T_1 = 3 \, \text{s} \)):
\[ T_1 = 2\pi \sqrt{\frac{m}{k}} \quad \text{(1)} \]

2. **For the second case** (mass \( m + 1 \, \text{kg} \) and time period \( T_2 = 5 \, \text{s} \)):
\[ T_2 = 2\pi \sqrt{\frac{m + 1}{k}} \quad \text{(2)} \]

### Step 2: Square both equations to eliminate the square root

1. From equation (1):
\[ T_1^2 = (2\pi)^2 \frac{m}{k} \implies 9 = 4\pi^2 \frac{m}{k} \]
Rearranging gives:
\[ \frac{m}{k} = \frac{9}{4\pi^2} \quad \text{(3)} \]

2. From equation (2):
\[ T_2^2 = (2\pi)^2 \frac{m + 1}{k} \implies 25 = 4\pi^2 \frac{m + 1}{k} \]
Rearranging gives:
\[ \frac{m + 1}{k} = \frac{25}{4\pi^2} \quad \text{(4)} \]

### Step 3: Set up a system of equations using (3) and (4)

From equations (3) and (4), we can express \( \frac{m + 1}{k} \) in terms of \( \frac{m}{k} \):
\[ \frac{m + 1}{k} = \frac{m}{k} + \frac{1}{k} \]
Substituting the values from equations (3) and (4):
\[ \frac{25}{4\pi^2} = \frac{9}{4\pi^2} + \frac{1}{k} \]

### Step 4: Solve for \( \frac{1}{k} \)

Rearranging gives:
\[ \frac{1}{k} = \frac{25}{4\pi^2} - \frac{9}{4\pi^2} = \frac{16}{4\pi^2} = \frac{4}{\pi^2} \]

### Step 5: Substitute back to find \( m \)

Now substitute \( \frac{1}{k} \) back into equation (3):
\[ \frac{m}{k} = \frac{9}{4\pi^2} \implies m = \frac{9}{4\pi^2} \cdot k \]
Substituting \( k = \frac{4}{\pi^2} \):
\[ m = \frac{9}{4\pi^2} \cdot \frac{4}{\pi^2} = \frac{9}{\pi^4} \]

### Step 6: Calculate \( m \)

Using \( \pi \approx 3.14 \):
\[ m \approx \frac{9}{(3.14)^2} \approx \frac{9}{9.8596} \approx 0.91 \, \text{kg} \]

### Final Answer

The value of \( m \) is approximately \( 0.91 \, \text{kg} \).