# 16.2 Mathematics of waves  (Page 3/11)

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## Characteristics of a traveling wave on a string

A transverse wave on a taut string is modeled with the wave function

$y\left(x,t\right)=A\phantom{\rule{0.2em}{0ex}}\text{sin}\left(kx-wt\right)=0.2\phantom{\rule{0.2em}{0ex}}\text{m}\phantom{\rule{0.2em}{0ex}}\text{sin}\left(6.28\phantom{\rule{0.2em}{0ex}}{\text{m}}^{-1}x-1.57\phantom{\rule{0.2em}{0ex}}{\text{s}}^{-1}t\right).$

Find the amplitude, wavelength, period, and speed of the wave.

## Strategy

All these characteristics of the wave can be found from the constants included in the equation or from simple combinations of these constants.

## Solution

1. The amplitude, wave number, and angular frequency can be read directly from the wave equation:
$y\left(x,t\right)=A\phantom{\rule{0.2em}{0ex}}\text{sin}\left(kx-wt\right)=0.2\phantom{\rule{0.2em}{0ex}}\text{m}\phantom{\rule{0.2em}{0ex}}\text{sin}\left(6.28\phantom{\rule{0.2em}{0ex}}{\text{m}}^{-1}x-1.57\phantom{\rule{0.2em}{0ex}}{\text{s}}^{-1}t\right).$
$\left(A=0.2\phantom{\rule{0.2em}{0ex}}\text{m;}\phantom{\rule{0.2em}{0ex}}k=6.28\phantom{\rule{0.2em}{0ex}}{\text{m}}^{-1};\phantom{\rule{0.2em}{0ex}}\omega =1.57\phantom{\rule{0.2em}{0ex}}{\text{s}}^{-1}\right)$
2. The wave number can be used to find the wavelength:
$\begin{array}{}\\ k=\frac{2\pi }{\lambda }.\hfill \\ \lambda =\frac{2\pi }{k}=\frac{2\pi }{6.28\phantom{\rule{0.2em}{0ex}}{\text{m}}^{-1}}=1.0\phantom{\rule{0.2em}{0ex}}\text{m}.\hfill \end{array}$
3. The period of the wave can be found using the angular frequency:
$\begin{array}{}\\ \\ \omega =\frac{2\pi }{T}.\hfill \\ T=\frac{2\pi }{\omega }=\frac{2\pi }{1.57\phantom{\rule{0.2em}{0ex}}{\text{s}}^{-1}}=4\phantom{\rule{0.2em}{0ex}}\text{s}.\hfill \end{array}$
4. The speed of the wave can be found using the wave number and the angular frequency. The direction of the wave can be determined by considering the sign of $kx\mp \omega t$ : A negative sign suggests that the wave is moving in the positive x -direction:
$|v|=\frac{\omega }{k}=\frac{1.57\phantom{\rule{0.2em}{0ex}}{\text{s}}^{-1}}{6.28\phantom{\rule{0.2em}{0ex}}{\text{m}}^{-1}}=0.25\phantom{\rule{0.2em}{0ex}}\text{m/s}.$

## Significance

All of the characteristics of the wave are contained in the wave function. Note that the wave speed is the speed of the wave in the direction parallel to the motion of the wave. Plotting the height of the medium y versus the position x for two times $t=0.00\phantom{\rule{0.2em}{0ex}}\text{s}$ and $t=0.80\phantom{\rule{0.2em}{0ex}}\text{s}$ can provide a graphical visualization of the wave ( [link] ).

There is a second velocity to the motion. In this example, the wave is transverse, moving horizontally as the medium oscillates up and down perpendicular to the direction of motion. The graph in [link] shows the motion of the medium at point $x=0.60\phantom{\rule{0.2em}{0ex}}\text{m}$ as a function of time. Notice that the medium of the wave oscillates up and down between $y=+0.20\phantom{\rule{0.2em}{0ex}}\text{m}$ and $y=-0.20\phantom{\rule{0.2em}{0ex}}\text{m}$ every period of 4.0 seconds.

Check Your Understanding The wave function above is derived using a sine function. Can a cosine function be used instead?

Yes, a cosine function is equal to a sine function with a phase shift, and either function can be used in a wave function. Which function is more convenient to use depends on the initial conditions. In [link] , the wave has an initial height of $\text{y}\left(0.00,0.00\right)=0$ and then the wave height increases to the maximum height at the crest. If the initial height at the initial time was equal to the amplitude of the wave $\text{y}\left(0.00,0.00\right)=\text{+}A,$ then it might be more convenient to model the wave with a cosine function.

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