A tuning fork of frequency n is held near the open end of tube, the tube is adjusted until resonance occurs. If the two shortest lengths to produce resonance are `L_1 and L_2` , then the speed of the sound is

Options

A. `n(L_2-L_1)`
B. `(n(L_2-L_1))/2`
C. `4n(L_2-L_1)`
D. `2n(L_2-L_1)`

Answer

To solve the problem, we need to find the speed of sound in the tube when resonance occurs with a tuning fork of frequency \( n \).

### Step-by-Step Solution:

1. **Understanding Resonance in the Tube**:
- When a tuning fork is held near the open end of a tube, resonance occurs at specific lengths of the tube. For an open tube, the resonant lengths correspond to odd multiples of a quarter wavelength.

2. **Identifying the Resonant Lengths**:
- The first resonant length \( L_1 \) corresponds to the fundamental frequency, which is \( \frac{\lambda}{4} \).
- The second resonant length \( L_2 \) corresponds to the first overtone, which is \( \frac{3\lambda}{4} \).

3. **Calculating the Difference in Lengths**:
- The difference between the two lengths is given by:
\[ L_2 - L_1 = \frac{3\lambda}{4} - \frac{\lambda}{4} = \frac{2\lambda}{4} = \frac{\lambda}{2} \]

4. **Using the Wave Equation**:
- The relationship between speed \( v \), frequency \( n \), and wavelength \( \lambda \) is given by:
\[ v = n \cdot \lambda \]
- From the previous step, we have \( \lambda = 2(L_2 - L_1) \).

5. **Substituting for Wavelength**:
- Substitute \( \lambda \) into the wave equation:
\[ v = n \cdot 2(L_2 - L_1) \]

6. **Final Expression for Speed of Sound**:
- Therefore, the speed of sound \( v \) in the tube is given by:
\[ v = 2n(L_2 - L_1) \]

### Conclusion:
The speed of sound in the tube is \( v = 2n(L_2 - L_1) \).