Question

Two waves are represented by the equations $y_1=a\sin (\omega t+kx+0.57)m$ and $y_2=a\cos (\omega t+kx)m$, 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)

$y_1=a\sin (\omega t+kx+0.57)$
$\therefore$ Phase, ${\varphi }_1=\omega t+kx+0.57$
$y_2=a\cos (\omega t+kx)=a\sin \left(\omega t+kx+\frac{\pi }{2}\right)$
$\therefore$ Phase, ${\varphi }_2=\omega t+kx+\frac{\pi }{2}$
Phase difference, $\Delta \varphi ={\varphi }_2-{\varphi }_1$
$=\left(\omega t+kx+\frac{\pi }{2}\right)-(\omega t+kx+0.57)=\frac{\pi }{2}-0.57$
$=(1.57-0.57)$ radian $=1$ radian