The phase difference between two waves, represented by & y_1=10^ … - Vidyadip

MCQ

The phase difference between two waves, represented by
$
\begin{aligned}
& y_1=10^{-6} \sin \left[100 t+\left(\frac{x}{50}\right)+0.5\right] m \\
& y_2=10^{-6} \cos \left[100 t+\left(\frac{x}{50}\right)\right] m
\end{aligned}
$
where x is expressed in metres and $t$ is expressed in seconds, is approximately

A
1.07 rad
B
1.5 rad
C
0.5 rad
D
2.07 rad

✓

Answer

1.07 rad
$
\begin{aligned}
& \text { Explanation: } y_1=10^{-6} \sin \left[100 t+\left(\frac{x}{50}\right)+0.5\right] m \\
& y_2=10^{-6} \cos \left[100 t+\left(\frac{x}{50}\right)\right] m \\
& =10^{-6} \sin \left[100 t+\left(\frac{x}{50}\right)+\left(\frac{\pi}{2}\right)\right] \\
& \Delta \phi=\frac{\pi}{2}-0.05=\frac{3.14}{2}-0.5=1.57-0.5 \\
& =1.07 rad
\end{aligned}
$

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