Figure 16-26 shows three waves that are separatelysent along a string that is stretched under a certain tension along an xaxis. Rank the waves according to their (a) wavelengths, (b) speeds, and (c) angular frequencies, greatest first.

The figure shows three separate waves.

Use the concept of wavelength, frequency, and speed of wave.

According to the concept of wave motion, wavelength is the distance between the two maxima or minima.

Formulae are as follow:

$$\begin{gathered} v=f\lambda \\ \omega =2\tau \tau f \end{gathered}$$

Where, v is speed of wave, f is frequency, 𝜔 is angular frequency, $\lambda$is wavelength.

a)

Rank of the waves according to their wavelengths:

From the figure, it can be observed that the wavelength of waveis large. The wavelength of waves 1 and 2 is the same but smaller than 3.

Hence, the rank of the wavelength according to the figure is 3>2=1

b)

Rank of the waves according to their speeds:

The speed of a wave on a stretched string is set by properties of the string. The speed of a string with tension$\tau$and linear density$\mu$

$v=\sqrt{\frac{\tau }{\mu }}$

Thus, the speed will be the same for all the three waves i.e.1=2=3 .

c)

Rank of the waves according to their angular frequency:

The frequency is inversely proportional to the wavelength of the wave.

$f\alpha \frac{1}{\lambda }$

Thus, the frequency of wave 3 is small. The frequency of waves 1 and 2 is large.

The angular frequency also depends on the frequency, that is, cycles per second.

The expression of the angular frequency is,

$\omega =2\tau \tau f$

Therefore, the angular frequency of wave 3 is small and that of waves 1 and 2 is large.

Hence, the rank of the angular frequency according to the figure is 1 =2>3

Therefore, the rank of the given waves according to the wavelength, speeds, and angular frequencies can be found by using the wavelength concept and the relation between them.