Two springs of force constants `7 Nm^(-1) and 9 Nm^(-1)` are connected in parallel and is fixed to body of mass 100 g. Find the period of oscillation.

Explanation

To find the period of oscillation for a mass connected to two springs in parallel, we can follow these steps:

### Step 1: Identify the force constants of the springs
The force constants given are:
- \( k_1 = 7 \, \text{N/m} \)
- \( k_2 = 9 \, \text{N/m} \)

### Step 2: Calculate the equivalent spring constant
When springs are connected in parallel, the equivalent spring constant \( k_{\text{equiv}} \) is the sum of the individual spring constants:
\[ k_{\text{equiv}} = k_1 + k_2 \]
Substituting the values:
\[ k_{\text{equiv}} = 7 + 9 = 16 \, \text{N/m} \]

### Step 3: Convert the mass into kilograms
The mass given is \( 100 \, \text{g} \). To use SI units, we convert this to kilograms:
\[ m = 100 \, \text{g} = 0.1 \, \text{kg} \]

### Step 4: Use the formula for the period of oscillation
The formula for the period \( T \) of a mass-spring system is given by:
\[ T = 2\pi \sqrt{\frac{m}{k_{\text{equiv}}}} \]
Substituting the values of \( m \) and \( k_{\text{equiv}} \):
\[ T = 2\pi \sqrt{\frac{0.1}{16}} \]

### Step 5: Calculate the period
Now we will compute the value:
1. Calculate \( \frac{0.1}{16} = 0.00625 \)
2. Take the square root: \( \sqrt{0.00625} = 0.079 \)
3. Multiply by \( 2\pi \):
\[ T = 2 \times 3.14 \times 0.079 \approx 0.496 \, \text{seconds} \]

### Final Answer
The period of oscillation is approximately \( 0.496 \, \text{seconds} \).

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