Question

The phase difference between two waves, represented by $y_1=10^{-6} \sin[100t +(x/50)+0.5]\,m$ $y_2=10^{-6} \cos[100t +(x/50)]\,m$ 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)

1.07 radians

Answer 2

$y_1=10^{-6} \sin[100t +(x/50)+0.5]$
$y_2=10^{-6} \cos[100t +(x/50)]$
$=10^{-6} \sin[100t +(x/50)+\pi/2]$
$=10^{-6} \sin[100t +(x/50)+1.57]$
[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.]