(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 with a given length and diameter, we will 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 the formula:
\[ L' = L + 0.3 \times d \]
where \( L \) is the actual length of the pipe and \( d \) is the diameter of the pipe.

Given:
- Length \( L = 83.2 \, \text{cm} = 0.832 \, \text{m} \)
- Diameter \( d = 6 \, \text{cm} = 0.06 \, \text{m} \)

Calculating the effective length:
\[ L' = 0.832 + 0.3 \times 0.06 = 0.832 + 0.018 = 0.850 \, \text{m} \]

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

Given:
- Velocity of sound \( v = 340 \, \text{m/s} \)

Calculating the fundamental frequency:
\[ f_0 = \frac{340}{4 \times 0.850} = \frac{340}{3.4} = 100 \, \text{Hz} \]

### Step 3: Determine the frequencies of the overtones.
The frequencies of the overtones in a closed organ pipe can be calculated using the formula:
\[ f_n = \frac{v}{4L'}(2n + 1) \]
where \( n \) is the overtone number (starting from \( n = 0 \) for the fundamental frequency).

Calculating the first few overtones:
- For \( n = 0 \) (fundamental frequency):
\[ f_0 = 100 \, \text{Hz} \]
- For \( n = 1 \) (first overtone):
\[ f_1 = \frac{340}{4 \times 0.850}(2 \times 1 + 1) = 100 \times 3 = 300 \, \text{Hz} \]
- For \( n = 2 \) (second overtone):
\[ f_2 = \frac{340}{4 \times 0.850}(2 \times 2 + 1) = 100 \times 5 = 500 \, \text{Hz} \]
- For \( n = 3 \) (third overtone):
\[ f_3 = \frac{340}{4 \times 0.850}(2 \times 3 + 1) = 100 \times 7 = 700 \, \text{Hz} \]
- For \( n = 4 \) (fourth overtone):
\[ f_4 = \frac{340}{4 \times 0.850}(2 \times 4 + 1) = 100 \times 9 = 900 \, \text{Hz} \]

### Step 4: Count the number of overtones below 1000 Hz.
The frequencies calculated are:
- \( f_0 = 100 \, \text{Hz} \)
- \( f_1 = 300 \, \text{Hz} \)
- \( f_2 = 500 \, \text{Hz} \)
- \( f_3 = 700 \, \text{Hz} \)
- \( f_4 = 900 \, \text{Hz} \)

All these frequencies are below 1000 Hz. Therefore, the number of overtones (excluding the fundamental frequency) is:
- Overtones: \( f_1, f_2, f_3, f_4 \) (4 overtones)

### Final Answer:
The number of overtones below 1000 Hz is **4**.

---