Question

The fundamental frequency of an air column in a pipe closed at one end is in unison with the third overtone of an open pipe. Calculate the ratio of lengths of the air columns.

Answer 1

Hint: For calculating the ratio of air columns, we should know the frequency of different overtones produced by open organ pipe and closed organ pipe. We can then equate the respective overtone frequencies produced by both pipes.

Formula used:
For closed organ pipe,
${{f}_{o}}=\dfrac{V}{4L}\text{, }{{f}_{1}}=3{{f}_{o}}\text{, }{{f}_{2}}=5{{f}_{o}}$ ....
For open organ pipe,
${{f}_{o}}=\dfrac{V}{2L}\text{, }{{f}_{1}}=2{{f}_{o}}\text{, }{{f}_{2}}=3{{f}_{o}}$ ....

Complete step-by-step answer:
Organ pipes are the 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, closed at one end and Open organ pipes, open at both 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 odd and even harmonics, but the note produced by a closed organ pipe is composed of odd harmonics. Even harmonics are absent in a closed organ pipe.
Fundamental frequency or the natural frequency 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.
We are given that ${{n}_{3}}={{n}_{o}}$
where ${{n}_{3}}$ = frequency of the third overtone of open pipe and ${{n}_{o}}$ = fundamental frequency of closed pipe
Third overtone for open pipe will be,
${{n}_{3}}=4\left( \dfrac{V}{2{{L}_{3}}} \right)$
Fundamental frequency of closed pipe at one end is,
${{n}_{o}}=\dfrac{4V}{{{L}_{o}}}$
As given,

& \dfrac{4V}{{{L}_{o}}}=4\left( \dfrac{V}{2{{L}_{3}}} \right) \\\ & \dfrac{{{L}_{o}}}{{{L}_{3}}}=\dfrac{1}{8} \\\ \end{aligned}$$ The ratio of lengths of air column of closed and open pipe is $ \dfrac{1}{8} $ Note: Students should remember 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 pipes.