Chapter 15, Problem 59A

To determine

To rank: The given tones according to wavelength, from least to greatest.

Explanation of Solution

Introduction:

Wavelength is inversely proportional to frequency.

As $\lambda =v/f$ , where $\lambda$ is the wavelength, $v$ is the speed and $f$ is the frequency. When speed is taken as a constant, the wavelength increases with decreasing frequency. This implies that wavelength is inversely proportional to frequency.

$\lambda \propto 1/f$

Hence wavelength increases with decreasing frequency and vice versa. When the given frequencies are ranked from greatest to least, their wavelengths will be of the order least to greatest.

The tones in order of least to greatest wavelengths can be ranked as:

$\begin{array}{l}\text{E}\text{. 140 Hz}\\ \text{D}\text{. 100 Hz}\\ \text{C}\text{. 83 Hz}\\ \text{B}\text{. 50 Hz}\\ \text{A}\text{. 30 Hz}\end{array}$

Conclusion:

The tones in order of least to greatest wavelengths can be ranked as:

$\begin{array}{l}\text{E}\text{. 140 Hz}\\ \text{D}\text{. 100 Hz}\\ \text{C}\text{. 83 Hz}\\ \text{B}\text{. 50 Hz}\\ \text{A}\text{. 30 Hz}\end{array}$