In a resonance tube the first resonance with a tuning fork occurs at 16 cm and second at 49 cm . If the velocity of sound is 330 m/s , the frequency of tuning fork is

Answer

To solve the problem, we need to find the frequency of the tuning fork using the information given about the resonance tube.

### Step-by-Step Solution:

1. **Identify the lengths of the first and second resonances**:
- The first resonance occurs at \( L_1 = 16 \, \text{cm} \).
- The second resonance occurs at \( L_2 = 49 \, \text{cm} \).

2. **Calculate the difference in lengths**:
- The difference between the two resonance lengths is:
\[ \Delta L = L_2 - L_1 = 49 \, \text{cm} - 16 \, \text{cm} = 33 \, \text{cm} \]

3. **Relate the difference in lengths to the wavelength**:
- The difference in lengths corresponds to half the wavelength (\( \frac{\lambda}{2} \)), since in a resonance tube, the distance between two consecutive resonances corresponds to half a wavelength.
- Therefore, we have:
\[ \frac{\lambda}{2} = 33 \, \text{cm} \]
- To find the wavelength (\( \lambda \)):
\[ \lambda = 2 \times 33 \, \text{cm} = 66 \, \text{cm} \]

4. **Convert the wavelength to meters**:
- Since the velocity of sound is given in meters per second, we convert \( \lambda \) to meters:
\[ \lambda = 66 \, \text{cm} = 0.66 \, \text{m} \]

5. **Use the wave equation to find the frequency**:
- The wave equation relates velocity (\( v \)), frequency (\( f \)), and wavelength (\( \lambda \)):
\[ v = f \cdot \lambda \]
- Rearranging this equation to solve for frequency gives:
\[ f = \frac{v}{\lambda} \]

6. **Substitute the values**:
- Given that the velocity of sound \( v = 330 \, \text{m/s} \):
\[ f = \frac{330 \, \text{m/s}}{0.66 \, \text{m}} = 500 \, \text{Hz} \]

7. **Final answer**:
- The frequency of the tuning fork is \( 500 \, \text{Hz} \).