Question

A wave is represented by $y_1=10cos(5x+25t)$
where, x is measured in metres and t in seconds. A second wave for which $y_2=20cos(5x+25t+\frac{\pi }{3})$
interferes with the first wave. Deduce the amplitude and phase of the resultant wave.

Answer 1

$y_1=10cos(5x+25t)$

$y_2=20cos(5x+25t+\frac{\pi }{3})$

Resultant wave will be $(S=\frac{\pi }{3})$

$y=y_1+y_2=Acos(5x+25t+\alpha )$

Where,

$A=\sqrt{{A_1}^2+{A_2}^2+2A_1{A}_{2}cos\delta }$

$=\sqrt{{\left(10\right)}^2+{\left(20\right)}^2+2\left(200\right)cos\frac{\pi }{3}}$

$A=26.46cm$

$tan\alpha =\frac{A_{2}sin\delta }{A_1+A_{2}cos\delta }\Rightarrow \alpha =0.714rad.$