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 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 [100 t+(x / 50)+0.5] m $
and $ y_{2}=10^{-6} \cos [100 t+(x / 50)] m $
$\Rightarrow y_{2}=10^{-6} \sin \left[100 t+(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 \phi =\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.