A particular organ pipe can resonate at \({\bf{264}}\,{\bf{Hz}}\), \({\bf{440}}\,{\bf{Hz}}\), and \({\bf{616}}\,{\bf{Hz}}\), but not at any other frequencies in between. (a) Show why this is an open or a closed pipe. (b) What is the fundamental frequency of this pipe?

In an open pipe, the difference between two consecutive frequencies is an integral multiple of a fundamental frequency.

As for the open pipe, each overtone is an integral multiple of the fundamental frequency.

(a)

Given data:

The frequencies are\(264\;{\rm{Hz,}}\;{\rm{440}}\;{\rm{Hz,}}\;{\rm{and 616 Hz}}\).

The expression to calculate the difference in frequencies is given as,

\(\begin{array}{c}\Delta f = {f_1} - {f_2}\\ = 440\;{\rm{Hz}} - 264\;{\rm{Hz}}\\{\rm{ = 176}}\;{\rm{Hz}}\end{array}\)

The expression for each overtone is an integral multiple of the fundamental frequency is given as,

\(k = \frac{{{f_1}}}{{\Delta f}}\)

Substitute the values in the above equation,

\(\begin{array}{c}k = \frac{{264\;{\rm{Hz}}}}{{176\;{\rm{Hz}}}}\\ = 1.5\;{\rm{Hz}}\end{array}\)

Since this is not an integral, so this is not an open pipe. Thus, this is a closed pipe.

(b)

As the difference between the fundamental overtones is twice of the fundamental frequency,

The expression to calculate frequency is given as,

\({f_0} = \frac{{\Delta f}}{2}\)

Substitute the values in the above equation,

\(\begin{array}{c}{f_0} = \frac{{176\;{\rm{Hz}}}}{2}\\ = 88\;{\rm{Hz}}\end{array}\)

Thus, the fundamental frequency of the pipe is \(88\;{\rm{Hz}}\).