Question

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

$y=A{e}^{-(\frac{x}{a}+\frac{t}{T}{)}^2}$.

(a) Write the dimensions of 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 = 2T ?

Answer 1

Given, $y=A{e}^{-(\frac{x}{a}+\frac{t}{T}{)}^2}$

(a) $\left[A\right]=\left[M^0{L}^1{T}^0\right]$

$\left[T\right]=\left[M^0{L}^0{T}^{-1}\right]$

$\left[a\right]=\left[M^0{L}^1{T}^0\right]$

(b) Wave speed, $v=\frac{\lambda }{T}=\frac{a}{T}$ [Here $\lambda =a$]

(c) If $y=f(t+\frac{x}{v})$

$\Rightarrow$ Wave travelling in negative direction

and if $y=f(t-\frac{x}{v})$

$\Rightarrow$ Wave travelling positive direction

So, $y=A{e}^{-\left[\right(\frac{x}{a})+(\frac{t}{T}){]}^2}$

$=A{e}^{-\frac{1}{T}[t+(\frac{xT}{a}){]}^2}$

$=A{e}^{-\frac{1}{T}[t+\frac{x}{V}]}$

$=A{e}^{-f[t+\frac{x}{V}]}$

Hence wave travelling is negative direction. (d) Wave speed,
$V=\frac{a}{t}$

Maximum pulse at t = T

is $\left(\frac{a}{T}\right)\times T=a$(negative x-axis Maximum pulse at t)

$=2T=(\frac{a}{T}\times 2T)$

$=2a$

(along negative x-axis) So the wave travelling in negative x - direction.