Sound waves of wavelength `alpha` travelling in a medium with a speed of v m `s^(-1)` enter into another medium where its speed is 2v m `s^(-1)`. Wavelength of sound waves in the second medium is

Options

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

Explanation

To solve the problem, we need to determine the wavelength of sound waves in the second medium given the information about the first medium.

### Step-by-Step Solution:

1. **Understand the relationship between speed, frequency, and wavelength**:
The fundamental relationship for waves is given by the equation:
\[ v = f \cdot \lambda \]
where \( v \) is the speed of the wave, \( f \) is the frequency, and \( \lambda \) is the wavelength.

2. **Identify the parameters in the first medium**:
In the first medium, we have:
- Wavelength \( \lambda_1 = \alpha \)
- Speed \( v_1 = v \)

Therefore, we can express the frequency in the first medium as:
\[ f = \frac{v_1}{\lambda_1} = \frac{v}{\alpha} \]

3. **Identify the parameters in the second medium**:
In the second medium, the speed of sound is given as:
- Speed \( v_2 = 2v \)

We need to find the new wavelength \( \lambda_2 \) in this medium.

4. **Use the fact that frequency remains constant**:
Since the frequency of the sound wave does not change when it moves from one medium to another, we have:
\[ f = \frac{v_2}{\lambda_2} \]

5. **Set up the equation using the constant frequency**:
From the first medium, we know:
\[ f = \frac{v}{\alpha} \]
From the second medium, we have:
\[ f = \frac{2v}{\lambda_2} \]

Setting these two expressions for frequency equal to each other gives:
\[ \frac{v}{\alpha} = \frac{2v}{\lambda_2} \]

6. **Solve for \( \lambda_2 \)**:
To eliminate \( v \) from both sides (assuming \( v \neq 0 \)):
\[ \frac{1}{\alpha} = \frac{2}{\lambda_2} \]
Cross-multiplying gives:
\[ \lambda_2 = 2\alpha \]

7. **Conclusion**:
The wavelength of sound waves in the second medium is:
\[ \lambda_2 = 2\alpha \]

### Final Answer:
The wavelength of sound waves in the second medium is \( 2\alpha \).