A mass m attached to a spring oscillates with a period of `3s`. If the mass is increased by `1kg` the period increases by `1s`. The initial mass m is

Options

A. `7/9kg`
B. `9/7kg`
C. `(14)/(7)kg`
D. `(18)/(7)kg`

Explanation

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

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

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

### Step 1: Set up the equations for the periods

1. The initial period \( T_1 \) when the mass is \( m \) is given as \( 3 \) seconds:
\[ T_1 = 2\pi \sqrt{\frac{m}{k}} = 3 \]

2. When the mass is increased by \( 1 \) kg, the new mass becomes \( m + 1 \) kg, and the new period \( T_2 \) is \( 4 \) seconds:
\[ T_2 = 2\pi \sqrt{\frac{m + 1}{k}} = 4 \]

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

1. From the first equation:
\[ (2\pi)^2 \frac{m}{k} = 3^2 \]
\[ 4\pi^2 \frac{m}{k} = 9 \quad \text{(Equation 1)} \]

2. From the second equation:
\[ (2\pi)^2 \frac{m + 1}{k} = 4^2 \]
\[ 4\pi^2 \frac{m + 1}{k} = 16 \quad \text{(Equation 2)} \]

### Step 3: Express \( k \) in terms of \( m \)

From Equation 1, we can express \( k \):
\[ k = \frac{4\pi^2 m}{9} \]

### Step 4: Substitute \( k \) into Equation 2

Substituting \( k \) from Equation 1 into Equation 2:
\[ 4\pi^2 \frac{m + 1}{\frac{4\pi^2 m}{9}} = 16 \]
This simplifies to:
\[ \frac{4\pi^2 (m + 1) \cdot 9}{4\pi^2 m} = 16 \]
\[ \frac{9(m + 1)}{m} = 16 \]

### Step 5: Solve for \( m \)

Cross-multiplying gives:
\[ 9(m + 1) = 16m \]
Expanding and rearranging:
\[ 9m + 9 = 16m \]
\[ 9 = 16m - 9m \]
\[ 9 = 7m \]
\[ m = \frac{9}{7} \approx 1.29 \text{ kg} \]

### Final Answer

The initial mass \( m \) is approximately \( 1.29 \) kg.

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