16.2 Mathematics of waves  (Page 3/11)

Characteristics of a traveling wave on a string

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

Strategy

Solution

1. The amplitude, wave number, and angular frequency can be read directly from the wave equation:
   
   $y\left(x,t\right)=A\text{sin}\left(kx-wt\right)=0.2\text{m}\text{sin}\left(6.28{\text{m}}^{\mathrm{-1}}x-1.57{\text{s}}^{\mathrm{-1}}t\right).$
   $(A=0.2\text{m;}k=6.28{\text{m}}^{\mathrm{-1}};\omega =1.57{\text{s}}^{\mathrm{-1}})$
2. The wave number can be used to find the wavelength:
   
   $\begin{array}{}\\ k=\frac{2\pi }{\lambda }.\hfill \\ \lambda =\frac{2\pi }{k}=\frac{2\pi }{6.28{\text{m}}^{\mathrm{-1}}}=1.0\text{m}.\hfill \end{array}$
3. The period of the wave can be found using the angular frequency:
   
   $\begin{array}{}\\ \\ \omega =\frac{2\pi }{T}.\hfill \\ T=\frac{2\pi }{\omega }=\frac{2\pi }{1.57{\text{s}}^{\mathrm{-1}}}=4\text{s}.\hfill \end{array}$
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:
   
   $\left|v\right|=\frac{\omega }{k}=\frac{1.57{\text{s}}^{\mathrm{-1}}}{6.28{\text{m}}^{\mathrm{-1}}}=0.25\text{m/s}.$

Significance

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 $x=0.60\text{m}$ as a function of time. Notice that the medium of the wave oscillates up and down between $y=+0.20\text{m}$ and $y=\mathrm{-0.20}\text{m}$ every period of 4.0 seconds.