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, 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 attached to the spring,
- \( k \) is the spring constant.

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

1. For the initial mass \( M \) with a period of 2 seconds:
\[ T_1 = 2 \text{ seconds} \implies 2 = 2\pi \sqrt{\frac{M}{k}} \]

2. For the increased mass \( M + 2 \) with a period of 3 seconds:
\[ T_2 = 3 \text{ seconds} \implies 3 = 2\pi \sqrt{\frac{M + 2}{k}} \]

### Step 2: Rearrange both equations to express \( \frac{M}{k} \) and \( \frac{M + 2}{k} \).

From the first equation:
\[ \frac{M}{k} = \left(\frac{2}{2\pi}\right)^2 = \frac{1}{\pi^2} \]

From the second equation:
\[ \frac{M + 2}{k} = \left(\frac{3}{2\pi}\right)^2 = \frac{9}{4\pi^2} \]

### Step 3: Set up a ratio of the two equations.

Now we can set up the ratio of the two equations:
\[ \frac{M + 2}{M} = \frac{\frac{9}{4\pi^2}}{\frac{1}{\pi^2}} = \frac{9}{4} \]

### Step 4: Solve for \( M \).

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

### Final Answer:
The initial mass \( M \) is \( 1.6 \text{ kg} \).

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