A wave has SHM (simple harmonic motion) whose period is 4s while another periods 3 s. If both are combined, then the resultant wave will have the period equal to

Options

A. 4 s
B. 5 s
C. 12 s
D. 3 s

Answer

To solve the problem, we need to find the period of the resultant wave when two waves with different periods are combined. Here are the steps to arrive at the solution:

### Step-by-Step Solution:

1. **Identify the periods of the two waves**:
- The first wave has a period \( T_1 = 4 \, \text{s} \).
- The second wave has a period \( T_2 = 3 \, \text{s} \).

2. **Calculate the frequencies of the two waves**:
- The frequency \( \nu_1 \) of the first wave is given by:
\[ \nu_1 = \frac{1}{T_1} = \frac{1}{4} \, \text{Hz} \]
- The frequency \( \nu_2 \) of the second wave is given by:
\[ \nu_2 = \frac{1}{T_2} = \frac{1}{3} \, \text{Hz} \]

3. **Find the beat frequency**:
- The beat frequency \( \nu \) is the difference between the two frequencies:
\[ \nu = \nu_1 - \nu_2 = \frac{1}{4} - \frac{1}{3} \]
- To perform this subtraction, find a common denominator (which is 12):
\[ \nu = \frac{3}{12} - \frac{4}{12} = \frac{-1}{12} \, \text{Hz} \]
- The absolute value is taken since frequency cannot be negative:
\[ \nu = \frac{1}{12} \, \text{Hz} \]

4. **Calculate the period of the resultant wave**:
- The period \( T \) of the resultant wave is the reciprocal of the beat frequency:
\[ T = \frac{1}{\nu} = \frac{1}{\frac{1}{12}} = 12 \, \text{s} \]

### Final Answer:
The period of the resultant wave when both waves are combined is **12 seconds**.

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