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

Options

A. 500 Hz
B. 300 Hz
C. 330 Hz
D. 165 Hz

Answer

To solve the problem, we need to find the frequency of the tuning fork based on the given resonance tube measurements. Here are the steps to arrive at the solution:

### Step-by-Step Solution:

1. **Identify the Resonance Points**:
- The first resonance occurs at 16 cm and the second resonance occurs at 49 cm.

2. **Calculate the Difference in Resonance Lengths**:
- The difference between the second and first resonance lengths gives us half the wavelength (λ/2).
- \[ \text{Distance} = 49 \, \text{cm} - 16 \, \text{cm} = 33 \, \text{cm} \]
- Therefore, \(\frac{\lambda}{2} = 33 \, \text{cm}\).

3. **Find the Wavelength (λ)**:
- To find the full wavelength, we multiply by 2:
- \[ \lambda = 2 \times 33 \, \text{cm} = 66 \, \text{cm} \]

4. **Convert Wavelength to Meters**:
- Since the speed of sound is given in meters per second, we need to convert the wavelength from centimeters to meters:
- \[ \lambda = 66 \, \text{cm} = 0.66 \, \text{m} \]

5. **Use the Speed of Sound Formula**:
- The formula relating speed (V), frequency (F), and wavelength (λ) is:
- \[ V = F \times \lambda \]
- Rearranging for frequency gives:
- \[ F = \frac{V}{\lambda} \]

6. **Substitute the Known Values**:
- Given the speed of sound \(V = 330 \, \text{m/s}\) and \(\lambda = 0.66 \, \text{m}\):
- \[ F = \frac{330 \, \text{m/s}}{0.66 \, \text{m}} = 500 \, \text{Hz} \]

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

### Final Answer:
The frequency of the tuning fork is **500 Hz**.