How many frequencies below `1 kH_(Z)` of natural oscillations of air column will be produced if a pipe of length `1 m` is closed at one end? [ velocity of sound in air is `340 m//s`]

Options

A. `3`
B. `6`
C. `4`
D. `8`

Explanation

To solve the problem of how many frequencies below 1 kHz of natural oscillations of an air column will be produced in a pipe of length 1 m that is closed at one end, we can follow these steps:

### Step 1: Understand the fundamental frequency formula for a closed pipe
For a pipe closed at one end, the fundamental frequency (first harmonic) is given by the formula:
\[ f_0 = \frac{v}{4L} \]
where:
- \( f_0 \) is the fundamental frequency,
- \( v \) is the velocity of sound in air (340 m/s),
- \( L \) is the length of the pipe (1 m).

### Step 2: Calculate the fundamental frequency
Substituting the values into the formula:
\[ f_0 = \frac{340 \, \text{m/s}}{4 \times 1 \, \text{m}} = \frac{340}{4} = 85 \, \text{Hz} \]

### Step 3: List the harmonic frequencies
For a closed pipe, the frequencies are given by:
\[ f_n = n \cdot f_0 \]
where \( n \) is an odd integer (1, 3, 5, 7, ...). The first few frequencies are:
- \( f_0 = 85 \, \text{Hz} \)
- \( f_1 = 3f_0 = 3 \times 85 = 255 \, \text{Hz} \)
- \( f_2 = 5f_0 = 5 \times 85 = 425 \, \text{Hz} \)
- \( f_3 = 7f_0 = 7 \times 85 = 595 \, \text{Hz} \)
- \( f_4 = 9f_0 = 9 \times 85 = 765 \, \text{Hz} \)
- \( f_5 = 11f_0 = 11 \times 85 = 935 \, \text{Hz} \)
- \( f_6 = 13f_0 = 13 \times 85 = 1105 \, \text{Hz} \)

### Step 4: Determine which frequencies are below 1 kHz
Now, we need to check which of these frequencies are below 1 kHz (1000 Hz):
- \( f_0 = 85 \, \text{Hz} \) (below 1000 Hz)
- \( f_1 = 255 \, \text{Hz} \) (below 1000 Hz)
- \( f_2 = 425 \, \text{Hz} \) (below 1000 Hz)
- \( f_3 = 595 \, \text{Hz} \) (below 1000 Hz)
- \( f_4 = 765 \, \text{Hz} \) (below 1000 Hz)
- \( f_5 = 935 \, \text{Hz} \) (below 1000 Hz)
- \( f_6 = 1105 \, \text{Hz} \) (above 1000 Hz)

### Step 5: Count the frequencies below 1 kHz
The frequencies below 1 kHz are:
1. \( 85 \, \text{Hz} \)
2. \( 255 \, \text{Hz} \)
3. \( 425 \, \text{Hz} \)
4. \( 595 \, \text{Hz} \)
5. \( 765 \, \text{Hz} \)
6. \( 935 \, \text{Hz} \)

Thus, there are **6 frequencies below 1 kHz**.

### Final Answer
The number of frequencies below 1 kHz of natural oscillations of the air column in a closed pipe of length 1 m is **6**.

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