Stationary waves are produced in 10 m long stretched string. If the string vibrates in 5 segments and wave velocity 20 m/s then the frequency is :-

Options

A. 10 Hz
B. 5 Hz
C. 4 Hz
D. 2 Hz

Explanation

To solve the problem of finding the frequency of stationary waves produced in a 10 m long stretched string that vibrates in 5 segments with a wave velocity of 20 m/s, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Problem**: We have a string of length \( L = 10 \, \text{m} \) vibrating in 5 segments. We need to find the frequency of the waves on this string.

2. **Determine the Wavelength**: When a string vibrates in segments, the number of segments is related to the wavelength. Each segment corresponds to half a wavelength (\( \frac{\lambda}{2} \)). Therefore, if there are 5 segments, the total length of the string can be expressed as:
\[ L = \frac{5\lambda}{2} \]
Rearranging this gives:
\[ \lambda = \frac{2L}{5} \]

3. **Substitute the Length**: Now, substituting the length of the string into the equation:
\[ \lambda = \frac{2 \times 10 \, \text{m}}{5} = \frac{20}{5} = 4 \, \text{m} \]

4. **Use the Wave Velocity Formula**: The relationship between wave velocity (\( V \)), frequency (\( f \)), and wavelength (\( \lambda \)) is given by:
\[ V = f \cdot \lambda \]
We can rearrange this to find the frequency:
\[ f = \frac{V}{\lambda} \]

5. **Substitute the Values**: Now we can substitute the values of wave velocity and wavelength into the frequency formula:
\[ f = \frac{20 \, \text{m/s}}{4 \, \text{m}} = 5 \, \text{Hz} \]

6. **Conclusion**: The frequency of the stationary waves on the string is \( 5 \, \text{Hz} \).

### Final Answer:
The frequency is \( 5 \, \text{Hz} \).

---