The frequency of a sound wave is n and its velocity is v. If the frequency is increased to 4n, the velocity of wave will be

Options

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

Answer

To solve the problem, we need to understand the relationship between the frequency, wavelength, and velocity of a sound wave. The key formula we will use is:

\[ v = f \lambda \]

where:
- \( v \) is the velocity of the wave,
- \( f \) is the frequency of the wave,
- \( \lambda \) is the wavelength of the wave.

### Step-by-Step Solution:

1. **Identify the initial conditions:**
- The initial frequency of the sound wave is \( n \).
- The initial velocity of the wave is \( v \).

2. **Write the initial relationship:**
- Using the formula \( v = f \lambda \), we can express the initial conditions as:
\[ v = n \lambda_1 \]
where \( \lambda_1 \) is the initial wavelength.

3. **Change the frequency:**
- The frequency is increased to \( 4n \).

4. **Express the new relationship:**
- The new relationship with the increased frequency becomes:
\[ v = 4n \lambda_2 \]
where \( \lambda_2 \) is the new wavelength.

5. **Set the two expressions for velocity equal:**
- Since the velocity of sound in a given medium remains constant, we can set the two expressions for velocity equal to each other:
\[ n \lambda_1 = 4n \lambda_2 \]

6. **Simplify the equation:**
- Dividing both sides by \( n \) (assuming \( n \neq 0 \)):
\[ \lambda_1 = 4 \lambda_2 \]

7. **Interpret the results:**
- This means that when the frequency increases to \( 4n \), the wavelength decreases to \( \frac{1}{4} \) of its original value (since \( \lambda_2 = \frac{\lambda_1}{4} \)).
- However, the velocity \( v \) remains unchanged regardless of the changes in frequency and wavelength.

### Conclusion:
The velocity of the wave remains \( v \) even after the frequency is increased to \( 4n \).

### Final Answer:
The velocity of the wave will be \( v \).