Two sinusoidal 120 Hzwaves, of the same frequency and amplitude, are to be sent in the positive direction of an xaxis that is directed along a cord under tension. The waves can be sent in phase, or they can be phase-shifted. Figure 16-47 shows the amplitude yof the resulting wave versus the distance of the shift (how far one wave is shifted from the other wave). The scale of the vertical axis is set by${\mathit{y}}_{\mathbf{s}}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{0}\mathbf{}\mathit{m}\mathit{m}$. If the equations for the two waves are of the form$\mathit{y}\left(x,t\right)\mathbf{=}{\mathit{y}}_{\mathbf{m}}\mathit{s}\mathit{i}\mathit{n}\left(kx-\omega t\right)$, what are (a)${\mathit{y}}_{\mathbf{m}}$, (b) k, (c)$\mathit{\omega }$, and (d) the correct choice of sign in front of$\mathit{\omega }$?

i) Frequency of the wave,f = 120 Hz .

ii) The maximum amplitude of the wave,$y_s=6.0mm$ .

We use the concept of wave motion. Using the equation of amplitude, we find the amplitude. From the equation of wave number related to wavelength, we can find it. Using the relation between angular velocity and frequency, we can find the angular velocity.

Formulae:

The amplitude of the wave,

$y_s\text{'}=\left|2y_m\cos \left(\frac{\varphi }{2}\right)\right|$ (i)

The wavenumber of the wave,

$k=\frac{2\mathrm{\pi }}{\lambda }$ (ii)

The angular frequency of the wave,

$\omega =2\mathrm{\pi f}$ (iii)

Here $\lambda$is the wavelength, f is the frequency,$y_m$ is the amplitude, and $\varphi$is the phase.

At $\varphi =0$we get maximum amplitude and using equation (i) and the given values, we get the value of$y_m$ as:

$$\begin{gathered} 6.0mm=\left|2y_m\cos \left(0\right)\right| \\ 6.0mm=\left|2y_m\right| \\ {y}_m=\frac{6.0}{2} \\ =3.0mm \end{gathered}$$

Hence, the required value of the amplitude is$3.0mm$ .

We can see, in the graph, that the value of shift is 10 cm when$y_s=0$ ,this occurs when,

$$\begin{gathered} \cos \left(\frac{\varphi }{2}\right)=0 \\ \frac{\varphi }{2}={\cos }^{-1}\left(0\right) \\ =\mathrm{\pi } \\ \mathrm{\varphi }=2\mathrm{\pi } \end{gathered}$$

This is a phase of the full cycle, which means 10 cm shift is for half cycle, so for the full cycle is shift is 20 cm .

So the wavelength corresponding to the full cycle will be,

$\lambda =20cm$

Using equation (ii), we can calculate the wavenumberas:

$$\begin{gathered} k=\frac{2\mathrm{\pi }}{0.20} \\ =31.4m^{-1} \end{gathered}$$

Hence, the required value of wavenumber is$31.4m^{-1}$ .

From the given value of frequency f = 120 Hz using equation (iii), we get the angular velocity as:

$$\begin{gathered} \omega =2\mathrm{\pi }\left(120\right) \\ =754\frac{\mathrm{rad}}{\mathrm{s}} \\ =7.5\times {10}^2\mathrm{rad}/\mathrm{s} \end{gathered}$$

Hence, the value of angular frequency is$7.5\times {10}^2\mathrm{rad}/\mathrm{s}$ .

The sign in front of$\mathit{\omega }$is negative since it is traveling along a positive x direction; the wave equation can be written as,

$y\left(x,t\right)=y_m\sin \left(kx-\omega t\right)$

Hence, the sign of angular frequency is negative.