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 rad
• B. 2.07 rad
• C. 0.5 rad
• D. 1.5 rad

Answer 1

Answer: (A)

The given waves are
$y_1=10^{-6}\sin [100t+(x/50)+0.5]m$
and $y_2=10^{-6}\cos [100t+(x/50)]m$
$\Rightarrow y_2=10^{-6}\sin \left[100t+(x/50)+\frac{\pi }{2}\right]m$
$\left[\because \sin \left(\frac{\pi }{2}+\theta \right)=\cos \theta \right]$
Hence, the phase difference between the waves is
$\Delta \varphi =\left(\frac{\pi }{2}-0.5\right)rad$
$=\left(\frac{3.14}{2}-0.5\right)rad$
$=(1.57-0.5)rad$
$=(1.07)rad$
NOTE : The given waves are sine and cosine function, so they are plane progressive harmonic waves.