Question: How many overtones are present within the audible range for a \({\bf{2}}{\bf{.18 - m}}\) organ pipe at \({\bf{20^\circ C}}\) (a) if it is open, and (b) if it is closed?

The wavelength of an open organ pipe is equal to two times the length of the pipe.

The wavelength of a closed organ pipe is equal to four times the length of the pipe.

The frequency is the function of the wavelength and speed of the sound.

\(f = \frac{v}{\lambda }\).

Given data:

The length of the organ pipe is \(L = 2.18\;{\rm{m}}\).

(a)

The wavelength for an open organ pipe can be calculated as,

\(\begin{array}{c}\lambda = 2L\\ = 2 \times 2.18\;{\rm{m}}\\ = {\rm{4}}{\rm{.36}}\;{\rm{m}}\end{array}\)

The expression for the fundamental frequency is given as,

\(f = \frac{v}{\lambda }\)

Here, v is the speed of sound and its value is 343 m/s and \(\lambda \) is the wavelength.

Substitute the values in the above equation,

The highest audible range can be calculated as,

\(\begin{array}{c}{f_n} = nf\\\left( {20\;{\rm{kHz}}} \right)\left( {\frac{{{\rm{1}}{{\rm{0}}^{\rm{3}}}\;{\rm{Hz}}}}{{{\rm{1}}\;{\rm{kHz}}}}} \right) = \left( {78.67\;{\rm{Hz}}} \right)n\\n = 254\end{array}\)

Here, \({f_n}\) is the audible range frequency and its value is \(20\;{\rm{kHz}}\).

As all the fundamental frequency is not in the audible range so that total overtones will be:

\(254 - 1 = 253\).

Therefore, the total overtones for open end pipe is \(253\;{\rm{overtones}}\).

(b)

The expression for the wavelength for closed organ pipe can be calculated as:

\(\begin{array}{c}\lambda = 4L\\ = 4 \times 2.18\;{\rm{m}}\\ = 8.{\rm{72}}\;{\rm{m}}\end{array}\)

The expression for the fundamental frequency is given as,

\(f = \frac{v}{\lambda }\)

Here, v is the speed of sound and \(\lambda \) is the wavelength.

Substitute the values in the above equation,

The expression for the highest audible range is,

\({f_n} = nf\)

Here, \({f_n}\) is the audible range frequency and its value is \(20\;{\rm{kHz}}\).

Substitute the values in the above equation,

\(\begin{array}{c}\left( {20\;{\rm{kHz}}} \right)\left( {\frac{{{\rm{1}}{{\rm{0}}^{\rm{3}}}\;{\rm{Hz}}}}{{{\rm{1}}\;{\rm{kHz}}}}} \right) = \left( {39.33\;{\rm{Hz}}} \right) \times n\\n = 508\end{array}\)

As only the odd harmonics of the fundamental frequency are present in the case of a closed pipe, so total overtones will be:

\(\begin{array}{c}T = \frac{{n - 1}}{2}\\T = \frac{{508 - 1}}{2}\\T = 254\;{\rm{overtones}}\end{array}\)

Therefore, the total overtones for closed end pipe is \(254\;{\rm{overtones}}\).