Question: A dramatic demonstration, called “singing rods,” involves a long, slender aluminum rod held in the hand near the rod’s midpoint. The rod is stroked with the other hand. With a little practice, the rod can be made to “sing,” or emit a clear, loud, ringing sound. For an \({\bf{80}}\)-cm-long rod, (a) what is the fundamental frequency of the sound? (b) What is its wavelength in the rod, and (c) what is the traveling wavelength of the sound in air at\({\bf{2}}{{\bf{0}}^{\bf{o}}}{\bf{C}}\)?

For an open tube open tube with length \(l\) , the fundamental frequency is \(f = \frac{v}{{2l}}\). Here, \(v = 5000\;{\rm{m/s}}\)is the speed of sound in the metal.

The length of the room is \(l = 80\;{\rm{cm}} = 80\;{\rm{cm}} \times \frac{{1\;{\rm{m}}}}{{100\;{\rm{cm}}}} = 0.80\;{\rm{m}}\).

Part (a)

The frequency of the fundamental tone for the length is,

\(\begin{array}{c}f = \frac{v}{{2l}}\\f = \frac{{5000\,{\rm{m/s}}}}{{2\left( {0.80\,{\rm{m}}} \right)}}\\f \approx 3187.5\,{\rm{Hz}}\end{array}\)

Hence the fundamental frequency is \(3187.5\,{\rm{Hz}}\).

Part (b)

The wavelength of sound in the rod is two times its length which is,

\(\begin{array}{c}\lambda = 2 \times 0.8\;{\rm{m}}\\ = {\rm{1}}{\rm{.6}}\,{\rm{m}}\end{array}\)

Hence, the wavelength is \({\rm{1}}{\rm{.6}}\,{\rm{m}}\).

Part (c)

The frequency and speed of sound in air determine the wavelength of the sound in air.

\(\begin{array}{c}\lambda ' = \frac{v}{f}\\ = \frac{{343\,{\rm{m/s}}}}{{3187.5\,{\rm{Hz}}}}\\ \approx 0.1076\,{\rm{m}}\end{array}\)

Hence, the traveling wavelength of the sound in air is \(0.1076\,{\rm{m}}\).