Answer
a. 3200 Hz.
b. 1.6 m.
c. 0.11 m.
Work Step by Step
a. The rod is “singing” because standing waves are being excited within it. When the rod “clamped” at its midpoint, it has a node there, and antinodes at its ends. The length of the rod equals a half wavelength. The fundamental frequency for the standing wave in the rod is $f=\frac{v}{\lambda}$.
In Table 12–1 we find the speed of sound in aluminum.
$$f=\frac{5100m/s}{\lambda}=\frac{v}{2(0.8m)}\approx 3200\;Hz$$
b. As just stated, the length of the rod equals a half wavelength, so the wavelength of sound in the rod is twice the rod’s length. The wavelength is 1.6 meters.
c. The wavelength of the sound in air can be calculated from the frequency, and the speed of sound in air, using $f=\frac{v}{\lambda}$.
$$\lambda=\frac{v}{f}=\frac{343m/s}{3187.5Hz}\approx0.11m$$
