A closed organ pipe has fundamental frequency 100Hz. What frequencies will be produced, if its other ends is also opend ?

Options

A. 200, 400, 600….
B. 200, 300, 400, 500….
C. 100, 300, 500, 700……
D. 100, 200, 300, 400….

Explanation

To solve the problem step by step, we need to analyze the situation of a closed organ pipe and then determine the frequencies produced when one end is opened.

### Step 1: Understand the fundamental frequency of a closed organ pipe
The fundamental frequency (first harmonic) of a closed organ pipe is given by the formula:
\[ f = \frac{V}{4L} \]
where:
- \( f \) is the fundamental frequency,
- \( V \) is the speed of sound in air,
- \( L \) is the length of the pipe.

Given that the fundamental frequency is 100 Hz, we can express this as:
\[ f = 100 \text{ Hz} \]

### Step 2: Determine the speed of sound in the closed pipe
From the formula, we can rearrange it to find the speed of sound:
\[ V = 4Lf \]
Substituting the known frequency:
\[ V = 4L \times 100 \text{ Hz} = 400L \]

### Step 3: Analyze the situation when one end is opened
When one end of the closed organ pipe is opened, it becomes an open pipe. The frequencies of an open pipe are given by:
\[ f_n = \frac{nV}{2L} \]
where \( n \) is the harmonic number (1, 2, 3, ...).

### Step 4: Substitute the speed of sound into the open pipe formula
Using the speed of sound we found earlier:
\[ f_n = \frac{n(400L)}{2L} \]
This simplifies to:
\[ f_n = 200n \]

### Step 5: List the frequencies produced
The frequencies produced by the open pipe will be:
- For \( n = 1 \): \( f_1 = 200 \times 1 = 200 \text{ Hz} \)
- For \( n = 2 \): \( f_2 = 200 \times 2 = 400 \text{ Hz} \)
- For \( n = 3 \): \( f_3 = 200 \times 3 = 600 \text{ Hz} \)
- And so on...

Thus, the frequencies produced will be \( 200 \text{ Hz}, 400 \text{ Hz}, 600 \text{ Hz}, \ldots \)

### Final Answer
The frequencies produced when the other end of the closed organ pipe is opened are:
\[ 200 \text{ Hz}, 400 \text{ Hz}, 600 \text{ Hz}, \ldots \]

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