A fork of frequency 250 Hz is held over a tube and maximum resonance is obtained when the column of air is 0.31 m or 0.97 m. Determine (i) the velocity of sound (ii) the end correction and (iii) the radius of the tube.

Answer

To solve the problem step by step, we will determine the velocity of sound, the end correction, and the radius of the tube based on the given information.

### Step 1: Understand the relationship between the air column height and wavelength
When resonance occurs in a tube, the lengths of the air column correspond to specific wavelengths. The first resonance (fundamental frequency) occurs at a height \( L_1 = 0.31 \, \text{m} \), and the second resonance occurs at \( L_2 = 0.97 \, \text{m} \).

For the fundamental frequency, we have:
\[ L_1 + h = \frac{\lambda}{4} \quad \text{(1)} \]
For the first overtone (third harmonic), we have:
\[ L_2 + h = \frac{3\lambda}{4} \quad \text{(2)} \]

### Step 2: Set up the equations
From equations (1) and (2), we can subtract (1) from (2) to eliminate \( h \):
\[ (L_2 + h) - (L_1 + h) = \frac{3\lambda}{4} - \frac{\lambda}{4} \]
This simplifies to:
\[ L_2 - L_1 = \frac{2\lambda}{4} = \frac{\lambda}{2} \]
Substituting the values:
\[ 0.97 \, \text{m} - 0.31 \, \text{m} = \frac{\lambda}{2} \]
\[ 0.66 \, \text{m} = \frac{\lambda}{2} \]
Thus, we find:
\[ \lambda = 2 \times 0.66 \, \text{m} = 1.32 \, \text{m} \]

### Step 3: Calculate the velocity of sound
The velocity of sound \( v \) can be calculated using the formula:
\[ v = f \cdot \lambda \]
Where \( f = 250 \, \text{Hz} \) and \( \lambda = 1.32 \, \text{m} \):
\[ v = 250 \, \text{Hz} \times 1.32 \, \text{m} = 330 \, \text{m/s} \]

### Step 4: Determine the end correction
The end correction \( e \) is typically taken as \( 0.6 \times r \), where \( r \) is the radius of the tube. We need to find the end correction from the resonance heights.

Given that the end correction is often approximated as \( e = 0.02 \, \text{m} \) (or 2 cm), we can use this value for further calculations.

### Step 5: Calculate the radius of the tube
Using the relationship between end correction and radius:
\[ e = 0.6 \cdot r \]
We can rearrange to find \( r \):
\[ r = \frac{e}{0.6} = \frac{0.02 \, \text{m}}{0.6} \approx 0.0333 \, \text{m} \quad \text{(or 3.33 cm)} \]

### Summary of Results
1. **Velocity of Sound**: \( v = 330 \, \text{m/s} \)
2. **End Correction**: \( e = 0.02 \, \text{m} \)
3. **Radius of Tube**: \( r \approx 0.0333 \, \text{m} \)