Question

Two transverse waves A and B superimpose to produce a node at x = 0. If the equation of wave A is y = a cos ($kx+\omega t$), then the equation of wave B is

• A. $+acos(kx+\omega t)$
• B. $-acos(kx+\omega t)$
• C. $-acos(kx-\omega t)$
• D. $+acos(kx-\omega t)$

Answer 1

The correct option is C $-acos(kx-\omega t)$

Equation of wave A is given by:
$y_1=acos(kx+\omega t)$
$y_1=acos\left(kx\right)cos\left(\omega t\right)-asin\left(kx\right)sin\left(\omega t\right)$
Now there is a node at x = 0;
this implies the cos terms get cancelled, sine terms remain
$y_2=-acos\left(kx\right)cos\left(\omega t\right)-asin\left(kx\right)sin\left(\omega t\right)$
$y_2=-a\left(cos\right(kx-\omega t)$
Option C is correct.