Two sound waves moves in the same direction in the same medium. The pressure amplutides of the waves are equal but the wavelength of the first wave is double the second. Let the average power transmitted across a cross section by the first wave be `P_1` and that by the second wave be `P_2` then.

Options

A. `P_1=P_2`
B. `P_1=4P_2`
C. `P_2=2P_1`
D. `P_2=4P_1`

Answer

To solve the problem, we need to analyze the relationship between the average power transmitted by sound waves and their properties such as pressure amplitude and wavelength.

### Step-by-Step Solution:

1. **Understand the Given Information**:
- We have two sound waves moving in the same direction in the same medium.
- The pressure amplitudes of both waves are equal.
- The wavelength of the first wave (λ₁) is double that of the second wave (λ₂). Therefore, we can write:
\[ λ₁ = 2λ₂ \]

2. **Recall the Formula for Average Power**:
- The average power (P) transmitted by a wave can be expressed as:
\[ P \propto A^2 \]
where \( A \) is the pressure amplitude of the wave. This means that the power is proportional to the square of the amplitude and does not depend on the wavelength.

3. **Apply the Information to Both Waves**:
- Since the pressure amplitudes of both waves are equal, we can denote the amplitude as \( A \) for both waves.
- Therefore, for both waves, we have:
\[ P_1 \propto A^2 \quad \text{and} \quad P_2 \propto A^2 \]

4. **Conclusion About the Powers**:
- Since both waves have the same amplitude, we can conclude that:
\[ P_1 = P_2 \]

5. **Final Answer**:
- The average power transmitted across a cross-section by the first wave \( P_1 \) is equal to the average power transmitted by the second wave \( P_2 \). Thus, the correct answer is:
\[ P_1 = P_2 \]