Question

A closed organ pipe has a fundamental frequency of$1.5KHz$. The number of overtones that can be distinctly heard by a person with this organ pipe will be: (Assume that the highest frequency a person can hear is$20,000Hz$).
$\text{A}\text{. }7$
$\text{B}\text{. 5}$
$\text{C}\text{. 6}$
$\text{D}\text{. 4}$

Answer 1

Hint : For calculating the number of overtones in a closed organ pipe, we will use the expression for relation between the frequency of an overtone and the fundamental frequency. Closed organ pipes only produce odd harmonics. All even harmonics are absent in a closed organ pipe.

Complete step by step answer:
Organ pipes are the type of musical instruments which are used to produce musical sound by the process of blowing air into the pipe. Organ pipes are of two types: Closed organ pipes, which are closed at one end and Open organ pipes, which are open at both the ends.
For closed organ pipe,
${{f}_{o}}=\dfrac{V}{4L}\text{, }{{f}_{1}}=3{{f}_{o}}\text{, }{{f}_{2}}=5{{f}_{o}}$….and so on.
For open organ pipe,
${{f}_{o}}=\dfrac{V}{2L}\text{, }{{f}_{1}}=2{{f}_{o}}\text{, }{{f}_{2}}=3{{f}_{o}}$….and so on.
The note produced by an open organ pipe is composed of both even and odd harmonics, but the note produced by a closed organ pipe is composed of odd harmonics only. Even harmonics are absent in a closed organ pipe.
Fundamental frequency or the natural frequency of an organ pipe is defined as the lowest frequency of a periodic waveform. Any discrete system with $n$ degrees of freedom can have $n$number of natural frequencies. Similarly, a continuous system can have infinite natural frequencies. The lowest natural frequency of a system is called Fundamental natural frequency.
For a closed organ pipe, resonant frequency is odd multiple of fundamental frequency or we can say that, $\text{Overtone = }\left( 2n+1 \right)\times \text{fundamental frequency}$
We are given, the fundamental frequency of a closed organ pipe is $1.5KHz$and the maximum frequency is given as$20,000Hz$.
Therefore,
$\left( 2n+1 \right){{f}_{o}}\le 20,000$
Where,
${{f}_{o}}$is the fundamental frequency, ${{f}_{o}}=1.5KHz=1500Hz$
Putting value of${{f}_{o}}$in above equation, we get,
$\begin{aligned} & \left( 2n+1 \right)1500\le 20,000 \\\ & 2n+1\le \dfrac{40}{3} \\\ & 2n\le \dfrac{37}{3} \\\ & n\le \dfrac{37}{6} \\\ \end{aligned}$
Since$n$signifies the number of overtones, the value of $n$will be a natural number, therefore,
$n\le 6$
Thus, the maximum number of overtones that can be distinctly heard by a person with this organ pipe will be $6$overtones.
Hence, the correct option is C.

Note :
Students should keep in mind that both odd and even harmonics are present in an open organ pipe while only odd harmonics are present in a closed organ pipe. All even harmonics are absent in closed organ pipe. The value of any overtone frequency produced by an organ pipe will be less than its fundamental frequency.