The speed of sound in an air column of `80 cm` closed at one end is `320 m//s`. Find the natural freqencies of air column between `20 Hz` and `2000 Hz`.

Answer

To find the natural frequencies of an air column closed at one end, we will follow these steps:

### Step 1: Understand the relationship between length, wavelength, and frequency.
For an air column closed at one end, the fundamental frequency (first harmonic) has a wavelength given by:
\[ L = \frac{\lambda}{4} \]
For higher harmonics, the length of the air column can be expressed as:
\[ L = \frac{(2n + 1)\lambda}{4} \]
where \( n = 0, 1, 2, \ldots \)

### Step 2: Relate speed of sound, frequency, and wavelength.
The speed of sound \( V \) is related to frequency \( f \) and wavelength \( \lambda \) by the equation:
\[ V = f \cdot \lambda \]
Given that the speed of sound \( V = 320 \, \text{m/s} \), we can express \( \lambda \) in terms of \( f \):
\[ \lambda = \frac{V}{f} \]

### Step 3: Substitute the expression for wavelength into the length equation.
From the length equation, we have:
\[ L = \frac{(2n + 1)\lambda}{4} \]
Substituting \( \lambda \):
\[ L = \frac{(2n + 1) \cdot \frac{V}{f}}{4} \]
Rearranging gives:
\[ f = \frac{(2n + 1)V}{4L} \]

### Step 4: Substitute known values into the frequency formula.
Using \( V = 320 \, \text{m/s} \) and \( L = 80 \, \text{cm} = 0.8 \, \text{m} \):
\[ f = \frac{(2n + 1) \cdot 320}{4 \cdot 0.8} \]
Calculating the denominator:
\[ 4 \cdot 0.8 = 3.2 \]
Thus:
\[ f = \frac{(2n + 1) \cdot 320}{3.2} \]
Simplifying further:
\[ f = 100(2n + 1) \]

### Step 5: Determine the range of \( n \) for frequencies between 20 Hz and 2000 Hz.
We need to find \( n \) such that:
\[ 20 \leq 100(2n + 1) \leq 2000 \]
Dividing the entire inequality by 100:
\[ 0.2 \leq 2n + 1 \leq 20 \]
Subtracting 1 from all parts:
\[ -0.8 \leq 2n \leq 19 \]
Dividing by 2:
\[ -0.4 \leq n \leq 9.5 \]
Since \( n \) must be a non-negative integer, the possible values for \( n \) are \( 0, 1, 2, \ldots, 9 \).

### Step 6: Calculate the natural frequencies.
Now, substituting the values of \( n \) from 0 to 9 into the frequency formula:
- For \( n = 0 \): \( f = 100(2 \cdot 0 + 1) = 100 \, \text{Hz} \)
- For \( n = 1 \): \( f = 100(2 \cdot 1 + 1) = 300 \, \text{Hz} \)
- For \( n = 2 \): \( f = 100(2 \cdot 2 + 1) = 500 \, \text{Hz} \)
- For \( n = 3 \): \( f = 100(2 \cdot 3 + 1) = 700 \, \text{Hz} \)
- For \( n = 4 \): \( f = 100(2 \cdot 4 + 1) = 900 \, \text{Hz} \)
- For \( n = 5 \): \( f = 100(2 \cdot 5 + 1) = 1100 \, \text{Hz} \)
- For \( n = 6 \): \( f = 100(2 \cdot 6 + 1) = 1300 \, \text{Hz} \)
- For \( n = 7 \): \( f = 100(2 \cdot 7 + 1) = 1500 \, \text{Hz} \)
- For \( n = 8 \): \( f = 100(2 \cdot 8 + 1) = 1700 \, \text{Hz} \)
- For \( n = 9 \): \( f = 100(2 \cdot 9 + 1) = 1900 \, \text{Hz} \)

### Final Frequencies:
The natural frequencies of the air column between 20 Hz and 2000 Hz are:
- \( 100 \, \text{Hz}, 300 \, \text{Hz}, 500 \, \text{Hz}, 700 \, \text{Hz}, 900 \, \text{Hz}, 1100 \, \text{Hz}, 1300 \, \text{Hz}, 1500 \, \text{Hz}, 1700 \, \text{Hz}, 1900 \, \text{Hz} \)