Open Organ Pipe Physics Explained

This problem delves into the world of sound waves and musical instruments, specifically an open organ pipe. We're given details like the pipe's length, the speed of sound in air, and the range of frequencies a person can hear. Our task is to figure out precisely how many overtones produced by this pipe fall within the audible range for that person.

Organ Pipe Frequencies Explained

An open organ pipe is essentially a tube that is open at both ends. When air is blown into it, the air column inside vibrates, creating sound waves. These vibrations occur at specific frequencies, known as harmonics. The set of frequencies an open organ pipe can produce is determined by its physical properties. The formula for these harmonic frequencies is:

$ f_n = n \times \frac{v}{2L} $

Let's break down this formula:

• $f_n$ represents the frequency of the $n^{th}$ harmonic.
• $n$ is the harmonic number. It must be a positive integer, meaning $n$ can be $1, 2, 3,$ and so on.
• $v$ is the speed of sound in the air inside the pipe.
• $L$ is the length of the organ pipe.

The term 'harmonics' refers to the fundamental frequency and its multiples. The concept of 'overtones' is closely related:

• The fundamental frequency is the lowest possible frequency, which occurs when $n=1$. It's calculated as $f_1 = \frac{v}{2L}$.
• The first overtone occurs at the next possible frequency, corresponding to $n=2$. Its frequency is $f_2 = 2 \times f_1$.
• The second overtone corresponds to $n=3$, with frequency $f_3 = 3 \times f_1$.
• Generally, the frequency $f_n$ represents the $(n-1)^{th}$ overtone.

Pipe Length and Speed Analysis

Here's a summary of the information provided in the question:

• The length of the open organ pipe is $L = 0.1 \text{ m}$.
• The speed of sound in air is $v = 340 \text{ m/s}$.
• The maximum frequency a person can hear distinctly is $20,000 \text{ Hz}$.

Fundamental Frequency Calculation

The first step is to calculate the fundamental frequency ($f_1$) of this specific organ pipe. We use the formula:

$ f_1 = \frac{v}{2L} $

Plugging in the given values:

$ f_1 = \frac{340 \text{ m/s}}{2 \times 0.1 \text{ m}} $

$ f_1 = \frac{340}{0.2} \text{ Hz} $

$ f_1 = 1700 \text{ Hz} $

So, the lowest note this pipe can produce (the fundamental frequency) is $1700 \text{ Hz}$.

Audible Harmonics Determination

All the other frequencies produced by the pipe (the harmonics and overtones) are integer multiples of this fundamental frequency. The general formula for the $n^{th}$ harmonic is:

$ f_n = n \times f_1 = n \times 1700 \text{ Hz} $

We know the person can only hear sounds up to $20,000 \text{ Hz}$. We need to find the highest harmonic number, $n$, whose frequency $f_n$ does not exceed this limit.

We set up the condition:

$ f_n \le 20,000 \text{ Hz} $

Substituting the formula for $f_n$:

$ n \times 1700 \text{ Hz} \le 20,000 \text{ Hz} $

To find the maximum possible value for $n$, we rearrange the inequality:

$ n \le \frac{20,000}{1700} $

$ n \le \frac{200}{17} $

Calculating the value:

$ n \le 11.76\dots $

Since $n$ must be a whole number (an integer), the largest integer value $n$ can take is $11$. This means the $11^{th}$ harmonic is the highest frequency sound wave produced by this pipe that the person can still hear.

Audible Overtones Count

The harmonics produced by the pipe correspond to $n=1, 2, 3, \dots, 11$. These are all potentially audible, as their frequencies are $1700 \text{ Hz}, 3400 \text{ Hz}, \dots, 18700 \text{ Hz}$ (which is $11 \times 1700$).

Now let's focus on the overtones:

• The fundamental frequency ($n=1$) is not an overtone.
• The 1st overtone is the 2nd harmonic ($n=2$).
• The 2nd overtone is the 3rd harmonic ($n=3$).
• ...
• The $k^{th}$ overtone corresponds to the $(k+1)^{th}$ harmonic.

Since the highest harmonic the person can hear is the $11^{th}$ harmonic ($n=11$), this corresponds to the $(11-1)^{th}$ overtone.

Therefore, the highest audible overtone is the $10^{th}$ overtone.

The audible overtones are the 1st, 2nd, 3rd, ..., all the way up to the 10th overtone.

To find the total number of these audible overtones, we count them: there are exactly 10 overtones in this list.

Conclusion on Overtones

The open organ pipe produces a series of harmonics, starting with the fundamental frequency ($n=1$). The overtones begin with the second harmonic ($n=2$). We found that the highest harmonic audible to the person is the $11^{th}$ harmonic ($n=11$). The overtones audible are those corresponding to $n=2, 3, 4, \dots, 11$. Counting these values gives us $11 - 2 + 1 = 10$ audible overtones.

Thus, the person can distinctly hear 10 overtones from this open organ pipe.