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

Explanation

To solve the problem of finding the period of oscillation for a mass connected to two springs in series, we can follow these steps:

### Step 1: Identify the given values
We have two springs with force constants:
- \( k_1 = 7 \, \text{N/m} \)
- \( k_2 = 9 \, \text{N/m} \)

The mass attached to the springs is:
- \( m = 100 \, \text{g} = 0.1 \, \text{kg} \) (since 1 kg = 1000 g)

### Step 2: Calculate the equivalent spring constant
For springs connected in series, the equivalent spring constant \( k_{\text{eq}} \) can be calculated using the formula:
\[ \frac{1}{k_{\text{eq}}} = \frac{1}{k_1} + \frac{1}{k_2} \]

Substituting the values:
\[ \frac{1}{k_{\text{eq}}} = \frac{1}{7} + \frac{1}{9} \]

To add these fractions, we find a common denominator. The least common multiple of 7 and 9 is 63:
\[ \frac{1}{k_{\text{eq}}} = \frac{9}{63} + \frac{7}{63} = \frac{16}{63} \]

Now, taking the reciprocal to find \( k_{\text{eq}} \):
\[ k_{\text{eq}} = \frac{63}{16} \, \text{N/m} \]

### Step 3: Calculate the period of oscillation
The period \( T \) of oscillation for a mass-spring system is given by the formula:
\[ T = 2\pi \sqrt{\frac{m}{k_{\text{eq}}}} \]

Substituting the values:
\[ T = 2\pi \sqrt{\frac{0.1}{\frac{63}{16}}} \]

This simplifies to:
\[ T = 2\pi \sqrt{\frac{0.1 \times 16}{63}} = 2\pi \sqrt{\frac{1.6}{63}} \]

Calculating the square root:
\[ T = 2\pi \sqrt{0.0253968253968254} \approx 2\pi \times 0.159 \]

Now calculating \( T \):
\[ T \approx 0.997 \, \text{s} \approx 1 \, \text{s} \]

### Final Answer:
The period of oscillation is approximately \( 1 \, \text{s} \).

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