Question

Question: Two open organ pipes, sounding together, produce a beat frequency of\({\bf{6}}{\bf{.0}}\,{\bf{Hz}}\). The shorter one is \({\bf{2}}{\bf{.40}}\,{\bf{m}}\) long. How long is the other?

Short answer

The length of another pipe is\(2.62\,{\rm{m}}\).

Step-by-step solution

Concept

The fundamental frequency of an open pipe is given by,

\(f = \frac{v}{{2l}}\)

Here,\(v\) is the velocity of sound in air,\(l\) is the length of the pipe.

The larger pipe has the lower frequency.

Given Data

The length of the shorter pipe is\({l_s} = 2.40\;{\rm{m}}\).

The beat frequency is\(\Delta f = 6.0\;{\rm{Hz}}\).

Calculation

The speed of sound in air is\(v = 343\,{\rm{m/s}}\).

The frequency of the shorter pipe is,

\(\begin{array}{c}{f_s} = \frac{v}{{2{l_s}}}\\ = \frac{{343\,{\rm{m/s}}}}{{2\left( {2.40\,{\rm{m}}} \right)}}\\ \approx 71.5\;{\rm{Hz}}\end{array}\)

As the other pipe is larger the frequency is less. The frequency of the other pipe is,

\(\begin{array}{c}{f_l} = {f_s} + \Delta f\\ = 71.5\;{\rm{Hz}} - 6.0\;{\rm{Hz}}\\ = 65.5\;{\rm{Hz}}\end{array}\)

The length of the other pipe is,

\(\begin{array}{c}{f_l} = \frac{v}{{2{l_l}}}\\{l_l} = \frac{v}{{2{f_l}}}\end{array}\)

Substitute the values and get,

\(\begin{array}{c}{l_l} = \frac{{343\,{\rm{m/s}}}}{{2\left( {65.5\;{\rm{Hz}}} \right)}}\\ \approx 2.62\;{\rm{m}}\end{array}\)

Hence, the length of the other pipe is\(2.62\;{\rm{m}}\).