Question

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?

Short answer

(a) The total overtones for open pipe is \(253\;{\rm{overtones}}\).

(b) The total overtones for closed pipe is \(254\;{\rm{overtones}}\).

Step-by-step solution

Relationship between the length of the pipe and the wavelength

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 information

Given data:

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

Determination of overtones for open end pipe

(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 \(253\;{\rm{overtones}}\).

Determination of overtones for closed end pipe

(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 \(254\;{\rm{overtones}}\).