A single pulse, given by$\mathit{h}\mathbf{(}\mathbf{x}\mathbf{-}\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{t}\mathbf{)}$is shown 1in Fig. 16-45 for t=0. The scale of the vertical axis is set by${\mathbf{h}}_{\mathbf{s}}\mathbf{=}\mathbf{2}$. Here xis in centimeters and tis in seconds. What are the (a) speed and (b) direction of travel of the pulse? (c) Plot$\mathit{h}\mathbf{(}\mathbf{x}\mathbf{-}\mathbf{5}\mathbf{t}\mathbf{)}$as a function of xfor t=2 s. (d) Plot$\mathit{h}\mathbf{(}\mathbf{x}\mathbf{-}\mathbf{5}\mathbf{t}\mathbf{)}$as a function of tfor x=10 cm.

A single pulse $h\left(xcm-5.0t\mathrm{s}\right)$for t=0.

We can find the equation of the wave atgiven in the graph. From the slope of the graph, we can predict the approximate value of the phase constant. Then, from the equation of the wave at t = 0, we can easily get the value of the phase constant.

Formula:

The wave equation is given by,$y(x,t)=f(kx-\omega t)$

The equation of the traveling wave in +x direction is given by:

$$\begin{gathered} y\left(x,t\right)=f\left(kx-\omega t\right) \\ y\left(x,t\right)=f\left(\left(\frac{2\mathrm{\pi }}{\lambda }\right)x-\left(\frac{2\mathrm{\pi v}}{\lambda }\right)t\right) \\ y\left(x,t\right)=f\left(\frac{2\mathrm{\pi }}{\lambda }\right)\left(x-vt\right) \\ y\left(x,t\right)=h\left(x-vt\right) \end{gathered}$$ (1)

Here k is the wave number, is the angular frequency,$\lambda$is the wavelength, and v is the velocity of the wave.

The given pulse,

$y(x,t)=h(xcm-5.0t\hspace{0.33em}s)$ (2)

Comparing (1) and (2), we can find the speed of the given pulse as:

$v=5.0\frac{cm}{s}$

Therefore, the speed of the pulse is 5.0$\frac{cm}{s}$

Plotting the graph of $h\left(xcm-5.0ts\right)$vs. x is shown in the figure below. From the plot, we can conclude that the wave travels towards the right.

Thus, the direction of travel of the pulse is to the right (+x axis).

Plotting the graph of$h\left(xcm-5.0ts\right)$vs. x is shown in the figure below as,

The first pulse in the above figure represents the pulse at t=0, the second pulse represents the pulse at t=1 s, and the third pulse represents the pulse at t=2 s.

From the given figure of the pulse, we can say that the leading edge of the pulse is at x=4 cm, and maxima is reached at x=3 cm at t=0. Since the velocity is 5.0 cm/s, the leading edge will reach x=10 cm in time,

$$\begin{gathered} t=\frac{∆x}{∆v} \\ =\frac{10-4}{5} \\ =1.2s \end{gathered}$$

Also, the maxima will be reached at

$$\begin{gathered} t=\frac{10-3}{5} \\ =1.4s \end{gathered}$$

And finally, minima will be reached at

$$\begin{gathered} t=\frac{10-1.0}{5} \\ =1.8s \end{gathered}$$

Thus we can plot the wave at x=10 cm as,

Hence, it is a required plot.