Question

Two waves are represented by the equations $y_{1}=a \sin (\omega t+k x+0.57) m$ and $y_{2}=a \cos (\omega t+k x) m$, where $x$ is in meter and $t$ in sec. The phase difference between them is

• A. 1.0 radian
• B. 1.25 radian
• C. 1.57 radian
• D. 0.57 radian

Answer 1

Answer: (A)

$y_{1}=a \sin (\omega t+k x+0.57)$
$\therefore $ Phase, $\phi_{1}=\omega t+k x+0.57$
$y_{2}=a \cos (\omega t+k x)=a \sin \left(\omega t+k x+\frac{\pi}{2}\right)$
$\therefore $ Phase, $\phi_{2}=\omega t+k x+\frac{\pi}{2}$
Phase difference, $\Delta \phi=\phi_{2}-\phi_{1}$
$=\left(\omega t+k x+\frac{\pi}{2}\right)-(\omega t+k x+0.57)=\frac{\pi}{2}-0.57 $
$=(1.57-0.57) $ radian $=1 $ radian