First slide

Travelling wave

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

Easy

A

1.5 rad
B

1.07 rad
C

2.07 rad
D

0.5 rad

Solution

$y_{1} = 10^{- 6} \sin [100 t + (x / 50) + 0 .5]$
$y_{2} = 10^{- 6} \sin [100 t + (\frac{x}{50}) + (\frac{π}{2})]$
Phase difference $ϕ$
$= [100 t + (x / 50) + 1.57] - [100 t + (x / 50) + 0.5] = 1.07 radians .$

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