The water level in a vertical glass tube `1.0 m` long can be adjusted to any position in the tube . `A` tuning fork vibrating at `660 H_(Z)` is held just over the open top end of the tube . At what positions of the water level wil they be in resonance? Speed of sound is `330 m//s` .

Explanation

To solve the problem of finding the positions of the water level in a vertical glass tube where resonance occurs when a tuning fork vibrating at 660 Hz is held over the open top of the tube, we can follow these steps:

### Step 1: Understand the Resonance Condition
In a tube with one end open (the top) and the other end closed (the water surface), the fundamental frequency and its harmonics can be described by the formula:
\[ f_n = \frac{nV}{4L} \]
where:
- \( f_n \) is the frequency of the nth harmonic,
- \( n \) is the harmonic number (1 for fundamental, 2 for first overtone, etc.),
- \( V \) is the speed of sound in air,
- \( L \) is the length of the air column.

### Step 2: Calculate the Fundamental Frequency
Given:
- Speed of sound \( V = 330 \, \text{m/s} \)
- Frequency of tuning fork \( f = 660 \, \text{Hz} \)

For the fundamental frequency (n=1):
\[ L_1 = \frac{V}{4f} = \frac{330}{4 \times 660} = \frac{330}{2640} = \frac{1}{8} \, \text{m} \]

### Step 3: Calculate Higher Harmonics
The lengths of the air column for the first few harmonics can be calculated as follows:
- For the first overtone (n=3):
\[ L_2 = \frac{3V}{4f} = \frac{3 \times 330}{4 \times 660} = \frac{990}{2640} = \frac{3}{8} \, \text{m} \]

- For the second overtone (n=5):
\[ L_3 = \frac{5V}{4f} = \frac{5 \times 330}{4 \times 660} = \frac{1650}{2640} = \frac{5}{8} \, \text{m} \]

### Step 4: Determine Water Level Positions
The total length of the tube is 1 meter. The positions of the water level where resonance occurs can be found by subtracting the lengths of the air column from the total length of the tube.

1. For the fundamental frequency:
\[ \text{Water Level}_1 = 1 - L_1 = 1 - \frac{1}{8} = \frac{7}{8} \, \text{m} \]

2. For the first overtone:
\[ \text{Water Level}_2 = 1 - L_2 = 1 - \frac{3}{8} = \frac{5}{8} \, \text{m} \]

3. For the second overtone:
\[ \text{Water Level}_3 = 1 - L_3 = 1 - \frac{5}{8} = \frac{3}{8} \, \text{m} \]

4. For the third overtone (n=7):
\[ L_4 = \frac{7V}{4f} = \frac{7 \times 330}{4 \times 660} = \frac{2310}{2640} = \frac{7}{8} \, \text{m} \]
\[ \text{Water Level}_4 = 1 - L_4 = 1 - \frac{7}{8} = \frac{1}{8} \, \text{m} \]

### Summary of Water Level Positions
The positions of the water level where resonance occurs are:
- \( \frac{7}{8} \, \text{m} \)
- \( \frac{5}{8} \, \text{m} \)
- \( \frac{3}{8} \, \text{m} \)
- \( \frac{1}{8} \, \text{m} \)