Find the number of possible natural oscillations of air column in a pipe whose frequencies lie below `f_(0) = 1250 Hz`. The length of the pipe is `l = 85 cm`. The velocity of sound is `v = 340 m//s`. Consider two cases
the pipe is closed from one end .

Explanation

To find the number of possible natural oscillations of an air column in a closed pipe whose frequencies lie below \( f_0 = 1250 \, \text{Hz} \), we will follow these steps:

### Step 1: Convert the Length of the Pipe
The length of the pipe is given as \( l = 85 \, \text{cm} \). We need to convert this into meters for consistency with the velocity of sound.
\[ l = 85 \, \text{cm} = 0.85 \, \text{m} \]

### Step 2: Calculate the Fundamental Frequency
For a closed pipe, the fundamental frequency \( f_1 \) is given by the formula:
\[ f_1 = \frac{v}{4l} \]
where \( v \) is the velocity of sound and \( l \) is the length of the pipe. Substituting the values:
\[ v = 340 \, \text{m/s}, \quad l = 0.85 \, \text{m} \]
\[ f_1 = \frac{340}{4 \times 0.85} = \frac{340}{3.4} = 100 \, \text{Hz} \]

### Step 3: Determine the Harmonics
The frequencies of the natural oscillations in a closed pipe are given by:
\[ f_n = (2n - 1) f_1 \]
where \( n \) is a natural number (1, 2, 3, ...). Thus, the frequencies are:
\[ f_1 = 100 \, \text{Hz}, \quad f_2 = 300 \, \text{Hz}, \quad f_3 = 500 \, \text{Hz}, \quad f_4 = 700 \, \text{Hz}, \quad f_5 = 900 \, \text{Hz}, \quad f_6 = 1100 \, \text{Hz}, \quad f_7 = 1300 \, \text{Hz}, \ldots \]

### Step 4: Count the Frequencies Below \( f_0 \)
We need to find how many of these frequencies are below \( f_0 = 1250 \, \text{Hz} \):
- \( 100 \, \text{Hz} \) (1st harmonic)
- \( 300 \, \text{Hz} \) (2nd harmonic)
- \( 500 \, \text{Hz} \) (3rd harmonic)
- \( 700 \, \text{Hz} \) (4th harmonic)
- \( 900 \, \text{Hz} \) (5th harmonic)
- \( 1100 \, \text{Hz} \) (6th harmonic)

The next frequency, \( 1300 \, \text{Hz} \), exceeds \( 1250 \, \text{Hz} \).

Thus, the number of possible natural oscillations below \( 1250 \, \text{Hz} \) is **6**.

### Final Answer
The number of possible natural oscillations of the air column in the closed pipe is **6**.

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