(iv) A closed organ pipe of length 83.2 cm and 6 cm diameter is vibrated. The velocity of sound is 340 m/s. find the number of overtones in this tube having frequency below 1000 Hz ?

Options

A. 2
B. 4
C. 1
D. 3

Explanation

To solve the problem of finding the number of overtones in a closed organ pipe of length 83.2 cm and diameter 6 cm, with a velocity of sound of 340 m/s, we can follow these steps:

### Step 1: Calculate the Effective Length of the Closed Organ Pipe
The effective length \( L \) of a closed organ pipe is given by:
\[ L = L' + \epsilon \]
where \( L' \) is the actual length of the pipe and \( \epsilon \) is the correction factor due to the diameter of the pipe. The correction factor is calculated as:
\[ \epsilon = 0.3 \times D \]
where \( D \) is the diameter of the pipe.

Given:
- \( L' = 83.2 \, \text{cm} = 0.832 \, \text{m} \)
- \( D = 6 \, \text{cm} = 0.06 \, \text{m} \)

Calculating \( \epsilon \):
\[ \epsilon = 0.3 \times 0.06 = 0.018 \, \text{m} \]

Now, calculate the effective length \( L \):
\[ L = 0.832 + 0.018 = 0.85 \, \text{m} \]

### Step 2: Calculate the Fundamental Frequency
The fundamental frequency \( f_0 \) of a closed organ pipe is given by the formula:
\[ f_0 = \frac{V}{4L} \]
where \( V \) is the velocity of sound.

Given:
- \( V = 340 \, \text{m/s} \)
- \( L = 0.85 \, \text{m} \)

Calculating \( f_0 \):
\[ f_0 = \frac{340}{4 \times 0.85} = \frac{340}{3.4} = 100 \, \text{Hz} \]

### Step 3: Calculate the Frequencies of Overtones
In a closed organ pipe, the frequencies of the overtones are given by:
\[ f_n = f_0 (2n + 1) \]
where \( n \) is the overtone number (starting from \( n = 0 \) for the fundamental frequency).

Calculating the frequencies for \( n = 0, 1, 2, 3, 4, 5 \):
- For \( n = 0 \):
\[ f_0 = 100 \, \text{Hz} \]
- For \( n = 1 \):
\[ f_1 = 100 \times (2 \times 1 + 1) = 100 \times 3 = 300 \, \text{Hz} \]
- For \( n = 2 \):
\[ f_2 = 100 \times (2 \times 2 + 1) = 100 \times 5 = 500 \, \text{Hz} \]
- For \( n = 3 \):
\[ f_3 = 100 \times (2 \times 3 + 1) = 100 \times 7 = 700 \, \text{Hz} \]
- For \( n = 4 \):
\[ f_4 = 100 \times (2 \times 4 + 1) = 100 \times 9 = 900 \, \text{Hz} \]
- For \( n = 5 \):
\[ f_5 = 100 \times (2 \times 5 + 1) = 100 \times 11 = 1100 \, \text{Hz} \]

### Step 4: Count the Overtones Below 1000 Hz
The frequencies we calculated are:
- \( 100 \, \text{Hz} \) (fundamental)
- \( 300 \, \text{Hz} \) (1st overtone)
- \( 500 \, \text{Hz} \) (2nd overtone)
- \( 700 \, \text{Hz} \) (3rd overtone)
- \( 900 \, \text{Hz} \) (4th overtone)
- \( 1100 \, \text{Hz} \) (5th overtone)

The overtones below 1000 Hz are:
- \( 300 \, \text{Hz} \)
- \( 500 \, \text{Hz} \)
- \( 700 \, \text{Hz} \)
- \( 900 \, \text{Hz} \)

Thus, the number of overtones below 1000 Hz is **4**.