Two elastic waves move along the same direction in the same medium. The pressure amplitudes of both the waves are equal, but the wavelength of the first wave is three times that of the second. If the average power transmitted through unit area by the first wave is `W_(1)` and that by the second is `W_(2)`, then

Options

A. `W_(1)=W_(2)`
B. `W_(1)=3W_(2)`
C. `W_(2)=3W_(1)`
D. `W_(1)=9W_(2)`

Explanation

To solve the problem, we need to analyze the relationship between the average power transmitted through unit area by two elastic waves moving in the same medium.

### Step-by-Step Solution:

1. **Understanding the Problem**: We have two elastic waves traveling in the same medium with equal pressure amplitudes. The wavelength of the first wave (\( \lambda_1 \)) is three times that of the second wave (\( \lambda_2 \)). We need to find the relationship between the average power transmitted through unit area by both waves, denoted as \( W_1 \) and \( W_2 \).

2. **Power Transmission Formula**: The average power transmitted through unit area (intensity) for a wave can be expressed as:
\[ W = \frac{p_0^2}{\rho v} \]
where \( p_0 \) is the pressure amplitude, \( \rho \) is the density of the medium, and \( v \) is the speed of sound in the medium.

3. **Given Information**:
- Pressure amplitudes are equal: \( p_{01} = p_{02} = p_0 \)
- Wavelengths: \( \lambda_1 = 3\lambda_2 \)

4. **Relationship Between Wavelength and Frequency**: The speed of sound \( v \) in a medium is related to the frequency \( f \) and wavelength \( \lambda \) by the equation:
\[ v = f \lambda \]
Therefore, if \( \lambda_1 = 3\lambda_2 \), we can express the frequencies:
\[ f_1 = \frac{v}{\lambda_1} = \frac{v}{3\lambda_2} = \frac{1}{3} f_2 \]
This shows that the frequency of the first wave is one-third that of the second wave.

5. **Calculating the Average Power for Each Wave**:
- For the first wave:
\[ W_1 = \frac{p_0^2}{\rho v} \]
- For the second wave:
\[ W_2 = \frac{p_0^2}{\rho v} \]

6. **Conclusion**: Since both waves have the same pressure amplitude and are traveling in the same medium, their average power transmitted through unit area is equal:
\[ W_1 = W_2 \]

### Final Answer:
\[ W_1 = W_2 \]