A resonance tube shows resonance with a tuning fork of frequency 256 Hz at column length of 24 cm and 90 cm. What will be the (a) speed of sound in air and (b) end correction ?

Answer

To solve the problem step by step, we will first find the speed of sound in air and then calculate the end correction.

### Step-by-Step Solution

**Given:**
- Frequency of tuning fork (ν) = 256 Hz
- Length of the first resonance (L1) = 24 cm = 0.24 m
- Length of the second resonance (L2) = 90 cm = 0.90 m

**(a) Speed of Sound in Air:**

1. **Use the formula for speed of sound:**
\[ v = 2 \nu (L2 - L1) \]

2. **Substitute the values into the formula:**
\[ v = 2 \times 256 \, \text{Hz} \times (0.90 \, \text{m} - 0.24 \, \text{m}) \]

3. **Calculate the difference in lengths:**
\[ 0.90 \, \text{m} - 0.24 \, \text{m} = 0.66 \, \text{m} \]

4. **Now substitute this back into the equation:**
\[ v = 2 \times 256 \times 0.66 \]

5. **Calculate the speed:**
\[ v = 2 \times 256 \times 0.66 = 337.92 \, \text{m/s} \]

**(b) End Correction:**

1. **Use the formula for end correction (e):**
\[ e = L2 - \frac{3L1}{2} \]

2. **Substitute the values:**
\[ e = 0.90 \, \text{m} - \frac{3 \times 0.24 \, \text{m}}{2} \]

3. **Calculate the term:**
\[ \frac{3 \times 0.24}{2} = \frac{0.72}{2} = 0.36 \, \text{m} \]

4. **Now substitute this back into the equation:**
\[ e = 0.90 \, \text{m} - 0.36 \, \text{m} \]

5. **Calculate the end correction:**
\[ e = 0.54 \, \text{m} \]

6. **Convert to centimeters:**
\[ e = 0.54 \, \text{m} \times 100 = 54 \, \text{cm} \]

### Final Answers:
- (a) Speed of sound in air = 337.92 m/s
- (b) End correction = 54 cm