Stationary waves of frequency 200 Hz are formed in air. If the velocity of the wave is `360ms^(-1)`, the shortest distance between two antinodes is

Options

A. 1.8 m
B. 3.6 m
C. 0.9 m
D. 0.45 m

Explanation

To solve the problem of finding the shortest distance between two antinodes in stationary waves, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Frequency (f) = 200 Hz
- Velocity (v) = 360 m/s

2. **Use the Wave Equation**:
The relationship between wave velocity (v), frequency (f), and wavelength (λ) is given by the equation:
\[ v = f \cdot \lambda \]
Rearranging this equation to find the wavelength (λ):
\[ \lambda = \frac{v}{f} \]

3. **Calculate the Wavelength (λ)**:
Substitute the values of v and f into the equation:
\[ \lambda = \frac{360 \, \text{m/s}}{200 \, \text{Hz}} = 1.8 \, \text{m} \]

4. **Determine the Distance Between Two Antinodes**:
The distance between two consecutive antinodes in a stationary wave is half the wavelength:
\[ \text{Distance between two antinodes} = \frac{\lambda}{2} \]
Substitute the value of λ:
\[ \text{Distance} = \frac{1.8 \, \text{m}}{2} = 0.9 \, \text{m} \]

5. **Final Answer**:
The shortest distance between two antinodes is:
\[ \text{Distance} = 0.9 \, \text{m} \]