Question

You have learnt that a travelling wave in one dimension is represented by a function y= f(x, t) w ere x must appear in the combination x Vt or x + vt, i.e. y = $F(x\pm vt)$ Is the converse true? Examine if the following function for y can possibly represent a travelling wave :

• A. $(x-vt{)}^2$
• B. $\log \left[\right(x+vt\left)/x_0\right]$
• C. $\exp [-(x+vt\left)/x_0\right]$
• D. none of these

Answer 1

The correct option is D none of these

The is not true, means f any function representing in the form of $y=f(x\pm vt)$ it doesn't necessarily express a travelling wave.

The essential condition is that

if any function $y=f(x\pm vt)$ satisfied

$d^{2}y/d{t}^2=v^2{d}^{2}y/d{x}^2$

(a) It's wave equation,

$y=(x-vt{)}^2$

$dy/dt=-2v(x-vt)$

$dy/dt=2v^2$

Again

$dy/dx=2(x-vt)$

$d^{2}y/d{t}^2=v^2.d^{2}y/d{x}^2$

Hence, it's wave equation

(b)log [(x+vt)/xo]

It's not wave equations because it doesn't follow $d^{2}y/d{t}^2=v^2.d^{2}y/d{x}^2$

Also at x=0 and t=0

$Log\left(0\right)\to \infty$ infinite

Hence, it's not wave equation

(c) 1/(x+vt)

It's not wave equation,

$d^{2}y/d{t}^2=v^2.d^{2}y/d{x}^2$