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

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

(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?

#### Solution

Given,

Equation of the wave travelling on a string stretched along the *X*-axis:

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

(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:

\[\left[ A \right] = \left[ M^0 L^1 T^0 \right]\]

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

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

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

(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:

`"y"="Ae"^[[(x/a)+(t/T)]^(-2)]`

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

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

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

Hence, the wave is travelling is the negative direction.

(d) Wave speed, \[v = \frac{a}{t}\] Maximum pulse at *t *= *T* = \[\left( \frac{a}{T} \right) \times T = a\] (Along the negative x - axis)

Maximum pulse at *t* = 2*T** = *\[\left( \frac{a}{T} \times 2T \right) = 2a \] (Along the negative x - axis)

Therefore, the wave is travelling in the negative *x*-direction.