A mass (M) attached to a spring, oscillates with a period of (2 sec). If the mass in increased by (2 kg) the period increases by one sec. Find the initial mass (M) assuming that Hook's Law is obeyed.

Explanation

To solve the problem step by step, we will use the formula for the period of a mass-spring system and manipulate it according to the given conditions.

### Step 1: Write the formula for the period of a mass-spring system
The formula for the period \( T \) of a mass-spring system 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 2: Set up the first equation using the initial mass
Given that the initial period \( T_1 = 2 \) seconds, we can write:
\[ 2 = 2\pi \sqrt{\frac{M}{k}} \]

### Step 3: Simplify the first equation
Dividing both sides by \( 2 \) gives:
\[ 1 = \pi \sqrt{\frac{M}{k}} \]
Now, squaring both sides:
\[ 1 = \pi^2 \frac{M}{k} \]
Rearranging this, we get:
\[ M = \frac{k}{\pi^2} \]
This is our first equation (Equation 1).

### Step 4: Set up the second equation with increased mass
When the mass is increased by \( 2 \) kg, the new period \( T_2 = 3 \) seconds. Thus, we can write:
\[ 3 = 2\pi \sqrt{\frac{M + 2}{k}} \]

### Step 5: Simplify the second equation
Dividing both sides by \( 3 \):
\[ 1 = \frac{2\pi}{3} \sqrt{\frac{M + 2}{k}} \]
Squaring both sides gives:
\[ 1 = \left(\frac{2\pi}{3}\right)^2 \frac{M + 2}{k} \]
This simplifies to:
\[ 1 = \frac{4\pi^2}{9} \frac{M + 2}{k} \]
Rearranging gives:
\[ M + 2 = \frac{9k}{4\pi^2} \]
This is our second equation (Equation 2).

### Step 6: Solve the two equations simultaneously
Now we have two equations:
1. \( M = \frac{k}{\pi^2} \)
2. \( M + 2 = \frac{9k}{4\pi^2} \)

Substituting Equation 1 into Equation 2:
\[ \frac{k}{\pi^2} + 2 = \frac{9k}{4\pi^2} \]

### Step 7: Clear the fractions by multiplying through by \( 4\pi^2 \)
Multiplying through by \( 4\pi^2 \):
\[ 4k + 8\pi^2 = 9k \]

### Step 8: Rearranging the equation
Rearranging gives:
\[ 9k - 4k = 8\pi^2 \]
\[ 5k = 8\pi^2 \]
Thus,
\[ k = \frac{8\pi^2}{5} \]

### Step 9: Substitute back to find \( M \)
Now substitute \( k \) back into Equation 1:
\[ M = \frac{\frac{8\pi^2}{5}}{\pi^2} = \frac{8}{5} \]

### Final Answer
The initial mass \( M \) is:
\[ M = \frac{8}{5} \text{ kg} \approx 1.6 \text{ kg} \]