Question

The phase difference between two waves, represented by
$y_{1}=10^{-6} \sin [100 t+(x / 50)+0.5] m$
$y_{2}=10^{-6} \cos [100 t+(x / 50)] m$
where $x$ is expressed in metres and $t$ is expressed in seconds, is approximately.

• A. 1.07 radians
• B. 2.07 radians
• C. 0.5 radians
• D. 1.5 radians

Answer 1

Answer: (A)

$y_{1}=10^{-6} \sin [100 t+(x / 50)+0.5]$
$y_{2}=10^{-6} \cos [100 t+(x / 50)]$
$=10^{-6} \sin [100 t+(x / 50)+\pi / 2]$
$=10^{-6} \sin [100 t+(x / 50)+1.57]$
$[$ using $\cos x=\sin (x+\pi / 2)] $
The phase difference $=1.57-0.5=1.07$
[or using $\sin x=\cos (\pi / 2-x) .$ We get the same result.]