MCQ
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
$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$1.5\ \mathrm{rad}$
- ✓$1.07\ \mathrm{rad}$
- C$2.07\ \mathrm{rad}$
- D$0.5\ \mathrm{rad}$
