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 in the same medium, given that their pressure amplitudes are equal and that the wavelength of the first wave is three times that of the second wave.

### Step-by-Step Solution:

1. **Understanding the Given Information**:
- Let the pressure amplitude of both waves be \( P_0 \).
- Let the wavelength of the first wave be \( \lambda_1 \) and that of the second wave be \( \lambda_2 \).
- According to the problem, \( \lambda_1 = 3\lambda_2 \).

2. **Intensity Formula**:
- The intensity \( I \) (or power per unit area) of a wave is given by the formula:
\[ I = \frac{P_0^2}{2 \rho V} \]
where \( P_0 \) is the pressure amplitude, \( \rho \) is the density of the medium, and \( V \) is the speed of the wave in the medium.

3. **Identifying Constants**:
- Since both waves are traveling in the same medium, the density \( \rho \) and the speed \( V \) of the waves are the same for both waves.

4. **Calculating Intensities**:
- For the first wave, the intensity \( W_1 \) is:
\[ W_1 = \frac{P_0^2}{2 \rho V} \]
- For the second wave, the intensity \( W_2 \) is:
\[ W_2 = \frac{P_0^2}{2 \rho V} \]

5. **Comparing the Intensities**:
- Since both \( W_1 \) and \( W_2 \) have the same expression, we can conclude:
\[ W_1 = W_2 \]

6. **Final Conclusion**:
- Therefore, the average power transmitted through unit area by the first wave is equal to that of the second wave:
\[ W_1 = W_2 \]