A sinusoidal wave moving along a string is shown twice in Figure 16-33, as crest Atravels in the positive direction of an xaxis by distance d = 6.0 cmin 4.0 ms. The tick marks along the axis are separated by 10 cm ; height H = 6.00mmIf the wave equation is of the form, $\mathit{y}\left(x,t\right)\mathbf{=}{\mathit{y}}_{\mathbf{m}}\mathit{s}\mathit{i}\mathit{n}\left(kx+\omega t\right)$

(a) What is ${\mathit{y}}_{\mathbf{m}}$,

(b) What is k,

(c) What is $\mathit{\omega }$, and

(d) What is the correct choice of sign in front of $\mathit{\omega }$?

• The distance travelled by the crest,d = 6.0 cm or 0.06 m .
• The time required for this displacement,$t=4.0msor4.0\times {10}^{-3}s$ .
• The scale on x -axis, the distance between the tick marks is10 cm .
• The peak-to-peak distance,$H=6.00mmor6.00\times {10}^{-3}m$ .

A sinusoidal wave traveling in positive x-direction is described by a standard equation. We use the equation and the information from the graph to calculate the required quantities.

Formula:

The maximum amplitude of the wave,$y_m=\frac{1}{2}\times \left(peaktopeakdis\tan ce\right)$ (i)

The transverse speed of a wave, $v=\lambda f=\frac{\omega }{k}$ (ii)

The wavenumber of the wave, $k=\frac{2\mathrm{\pi }}{\lambda }$ (iii)

The velocity of a body in motion, $v=\frac{d}{t}$ (iv)

It is given that thepeak-to-peak distance is$H=6.00\times {10}^{-3}m$.

The maximum displacement is given using equation (i) and the given values as follows:

$$\begin{gathered} y_m=\frac{1}{2}\times 6.00\times {10}^{-3} \\ =3.00\times {10}^{-3}m \end{gathered}$$

Hence, the value of maximum amplitude is$3.00\times {10}^{-3}m$ .

From the graph, we can observe that the graph repeats itself after travelling a distance of 4 tick marks i.e. distance of (4 X 10 cm) = 40 cm

So we get,$\lambda =40.0cmor40.0\times {10}^{-2}m$

Now using equation (iii) and given values, the wavevector can be given as:

$$\begin{gathered} K=\frac{2\times 3.14}{40.0\times {10}^{-2}} \\ \approx 16.0m^{-1} \end{gathered}$$

Hence, the value of wave vector is$16.0m^{-1}$ .

Crest A moves distance d in time t in the positive direction of x axis. Thus, the wave velocity using equation (iv) is given as:

$$\begin{gathered} v=\frac{6.0\times {10}^{-2}}{4.0\times {10}^{-3}} \\ =15m/s \end{gathered}$$

For a travelling wave, the wave velocity using equation (ii) is given as:

$$\begin{gathered} \omega =vk \\ =15\times 16 \\ =240 \\ =2.4\times {10}^{2}rad/s \end{gathered}$$

Hence, the value of angular velocity is$2.4\times {10}^{2}rad/s$ .

The sign of $\omega$should be negative as the wave is moving along the positive direction of x axis.