Question

The equation of a wave travelling on a string stretched along the X-axis is given by

$y=Ae{}^{-}{\left(\frac{x}{a}+\frac{t}{T}\right)}^2.$

(a) Write the dimensions of A, a and T. (b) Find the wave speed. (c) In which direction is the wave travelling? (d) Where is the maximum of the pulse located at t = T? At t = 2 T?

Answer 1

Given,
Equation of the wave travelling on a string stretched along the X-axis:
$y=\mathrm{A}e{}^{\left(\frac{x}{a}+\frac{t}{T}\right){{}^{-}}^2}$
(a) The dimensions of A (amplitude), T (time period) and $a=\frac{\lambda }{2\pi }$, which will have the dimensions of the wavelength, are as follows:
$$\begin{gathered} \left[\mathrm{A}\right]=\left[{\mathrm{M}}^0{\mathrm{L}}^1{\mathrm{T}}^0\right] \\ \left[\mathrm{T}\right]=\left[{\mathrm{M}}^0{\mathrm{L}}^0{\mathrm{T}}^{-1}\right] \\ \left[a\right]=\left[{\mathrm{M}}^0{\mathrm{L}}^1{\mathrm{T}}^0\right] \end{gathered}$$

(b) Wave speed, $\nu =\frac{\lambda }{T}=\frac{a}{T}\left[\lambda =a\right]$

(c) If $y=f\left(t+\frac{x}{\nu }\right)$, then the wave travels in the negative direction; and if $y=f\left(t-\frac{x}{\nu }\right)$, then the wave travels in the positive direction.
Thus, we have:
$$\begin{gathered} y=Ae{}^{{{\left[\left(\frac{x}{a}\right)+\left(\frac{t}{T}\right)\right]}^{-}}^2} \\ =A{e}^{-\frac{1}{T}{\left[t+\left(\frac{xT}{a}\right)\right]}^2} \\ =A{e}^{-\frac{1}{T}\left[t+\frac{x}{V}\right]} \\ =A{e}^{-f\left[t+\frac{x}{V}\right]} \end{gathered}$$
Hence, the wave is travelling is the negative direction.

(d) Wave speed, $v=\frac{a}{t}$
Maximum pulse at t = T = $\left(\frac{a}{T}\right)\times T=a\left(\mathrm{Along}\mathrm{the}\mathrm{negative}x-\mathrm{axis}\right)$
Maximum pulse at t = 2T = $\left(\frac{a}{T}\times 2T\right)=2a\left(\mathrm{Along}\mathrm{the}\mathrm{negative}x-\mathrm{axis}\right)$
Therefore, the wave is travelling in the negative x-direction.