91Ó°ÊÓ

Travelling wave on a string is

$y=2.0\sin \left[2\mathrm{Ï€}\left(\frac{\mathrm{t}}{0.40}+\frac{\mathrm{x}}{80}\right)\right]$

We can plot as a function of x for different values of t . We can get the values of angular speed and wave number from the graphs. From them, we can find the wave speed using the corresponding relation. From the graphs, we can interpret the direction in which the wave is traveling.

Formula:

The velocity of a body in oscillation,$v=\frac{\mathrm{Ó\neg }}{k}$ …(¾±)

Where, is the angular speed and k is the wave number

Graph of for t = 0.05 s

Graph of x for t = 0.005s

The angular velocity of the wave in the given equation is:

$\mathrm{Ó\neg }=\frac{2\mathrm{Ï€}}{0.40}\mathrm{rad}/\mathrm{s}$

The wave number in the given equation is:

$k=\frac{2\mathrm{Ï€}}{80}\mathrm{rad}/\mathrm{cm}$

So, using equation (i), the velocity of the wave is given as:

$$\begin{gathered} v=\frac{\left({\displaystyle \frac{2\mathrm{Ï€}}{0.40}}\right)\mathrm{rad}/\mathrm{s}}{\left(\frac{2\mathrm{Ï€}}{80}\right)\mathrm{rad}/\mathrm{cm}} \\ =\frac{80}{0.40}\frac{\mathrm{cm}}{\mathrm{s}} \\ =200\frac{\mathrm{cm}}{\mathrm{s}} \\ =2\mathrm{m}/\mathrm{s} \end{gathered}$$

Therefore, the wave speed is 2m/s

We can conclude from the graphs in part (a) that the wave is travelling in the negative x direction.