Q28P Question: (I) An organ pipe is 1... [FREE SOLUTION]

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Chapter 12: Q28P (page 328)

Question: (I) An organ pipe is 116 cm long. Determine the fundamental and first three audible overtones if the pipe is (a) closed at one end, and (b) open at both ends.

Short Answer

Expert verified

The fundamental frequency for the closed pipe is \(74.56{\rm{ Hz}}\). The first three audible overtones are \(223.6{\rm{ Hz}}\), \({\rm{372}}{\rm{.8 Hz}}\) and \({\rm{521}}{\rm{.92 Hz}}\).
The fundamental frequency for the open pipe is\(149.12{\rm{ Hz}}\). The first three audible overtones are \(298.24{\rm{ Hz}}\), \({\rm{447}}{\rm{.36 Hz}}\) and \({\rm{596}}{\rm{.48 Hz}}\).

Step by step solution

Step-1:-Fundamental frequency of an organ pipe

In this problem, the fundamental frequency is calculated by dividing the wave speed (v) by the length of the tube (l).

Step-2:-Given data

The length of the pipe is \(l = 116{\rm{ cm}}\).

The speed of sound at room temperature is \(\nu = 346{\rm{ m/s}}\).

Step-3:-Calculation of fundamental frequency and first three overtones of a closed pipe

The fundamental frequency for the closed pipe is calculated as:

\(\begin{array}{c}{f_1} = \frac{\nu }{{4l}}\\ = \frac{{346{\rm{ m/s}}}}{{4 \times \left( {116\;{\rm{cm}}\left( {\frac{{1\;{\rm{m}}}}{{100\;{\rm{cm}}}}} \right)} \right)}}\\ = 74.56{\rm{ Hz}}\end{array}\)

The first overtone will be,

\(\begin{array}{c}{f_3} = 3{f_1}\\ = 3 \times 74.56{\rm{ Hz}}\\ = 223.6{\rm{ Hz}}\end{array}\)

The second overtone is,

\(\begin{array}{c}{f_5} = 5{f_1}\\ = 5 \times 74.56{\rm{ Hz}}\\{\rm{ = 372}}{\rm{.8 Hz}}\end{array}\)

The third overtone is,

\(\begin{array}{c}{f_7} = 7{f_1}\\ = 7 \times 74.56{\rm{ Hz}}\\{\rm{ = 521}}{\rm{.92 Hz}}\end{array}\)

The fundamental frequency is \(74.56{\rm{ Hz}}\) and the three overtones are \(223.6{\rm{ Hz}}\), \({\rm{372}}{\rm{.8 Hz}}\) and \({\rm{521}}{\rm{.92 Hz}}\).

Step-4:-Calculation of fundamental frequency and first three overtones of an open pipe

The fundamental frequency for an open pipe is calculated as:

\(\begin{array}{c}{f_1} = \frac{\nu }{{2l}}\\ = \frac{{346{\rm{ m/s}}}}{{2 \times \left( {116\;{\rm{cm}}\left( {\frac{{1\;{\rm{m}}}}{{100\;{\rm{cm}}}}} \right)} \right)}}\\ = 149.12{\rm{ Hz}}\end{array}\)

The first overtone is,

\(\begin{array}{c}{f_2} = 2{f_1}\\ = 2 \times 149.12{\rm{ Hz}}\\ = 298.24{\rm{ Hz}}\end{array}\)

The second overtone is,

\(\begin{array}{c}{f_3} = 3{f_1}\\ = 3 \times 149.12{\rm{ Hz}}\\ = 447.36{\rm{ Hz}}\end{array}\)

The third overtone is,

\(\begin{array}{c}{f_4} = 4{f_1}\\ = 4 \times 149.12{\rm{ Hz}}\\ = 596.48{\rm{ Hz}}\end{array}\)

The fundamental frequency is \(149.12{\rm{ Hz}}\) and the three overtones are \(298.24{\rm{ Hz}}\), \({\rm{447}}{\rm{.36 Hz}}\) and \({\rm{596}}{\rm{.48 Hz}}\).