two sound waves moves in the same direction .if the average power transmiitted across a cross - section by them are equal while their wavelengths are in the ratio of 1:2 . Their pressure amplitudes would be in the ratio of

Options

A. 1
B. 2
C. 4
D. `1/2`

Answer

To solve the problem, we need to find the ratio of the pressure amplitudes of two sound waves given that their average power transmitted is equal and their wavelengths are in the ratio of 1:2.

### Step-by-Step Solution:

1. **Understanding the Power of Sound Waves**:
The average power (P) transmitted by a sound wave can be expressed as:
\[ P = \frac{1}{2} \rho v A^2 \omega^2 \]
where:
- \( \rho \) = density of the medium
- \( v \) = velocity of sound in the medium
- \( A \) = amplitude of the wave
- \( \omega \) = angular frequency of the wave

2. **Setting Up the Equations**:
Since the average power transmitted by both waves is equal, we can write:
\[ P_1 = P_2 \]
This gives us:
\[ \frac{1}{2} \rho v A_1^2 \omega_1^2 = \frac{1}{2} \rho v A_2^2 \omega_2^2 \]
We can cancel \( \frac{1}{2} \), \( \rho \), and \( v \) (assuming they are the same for both waves), leading to:
\[ A_1^2 \omega_1^2 = A_2^2 \omega_2^2 \]

3. **Relating Wavelengths and Frequencies**:
Given that the wavelengths are in the ratio \( \frac{\lambda_1}{\lambda_2} = \frac{1}{2} \), we can express this in terms of frequency:
\[ \lambda_1 = \frac{v}{f_1} \quad \text{and} \quad \lambda_2 = \frac{v}{f_2} \]
Therefore, the ratio of the wavelengths gives us:
\[ \frac{f_2}{f_1} = \frac{\lambda_1}{\lambda_2} = \frac{1}{2} \]
This implies:
\[ f_2 = \frac{1}{2} f_1 \]

4. **Relating Angular Frequencies**:
The angular frequency \( \omega \) is related to frequency \( f \) by:
\[ \omega = 2\pi f \]
Thus:
\[ \frac{\omega_2}{\omega_1} = \frac{f_2}{f_1} = \frac{1}{2} \]

5. **Substituting Back into the Power Equation**:
Now substituting \( \omega_2 = \frac{1}{2} \omega_1 \) into the power equation:
\[ A_1^2 \omega_1^2 = A_2^2 \left(\frac{1}{2} \omega_1\right)^2 \]
This simplifies to:
\[ A_1^2 \omega_1^2 = A_2^2 \frac{1}{4} \omega_1^2 \]
Canceling \( \omega_1^2 \) from both sides gives:
\[ A_1^2 = \frac{1}{4} A_2^2 \]

6. **Finding the Ratio of Amplitudes**:
Taking the square root of both sides:
\[ \frac{A_1}{A_2} = \frac{1}{2} \]
Therefore, the ratio of the pressure amplitudes \( P_1 \) and \( P_2 \) is:
\[ \frac{P_1}{P_2} = \frac{A_1}{A_2} = \frac{1}{2} \]

### Final Answer:
The ratio of the pressure amplitudes \( P_1 : P_2 \) is \( 1 : 2 \).