Find the phase difference between two waves which are 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 meters and $t$ is expressed in seconds]

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TestHub · Accepted Answer

y 1 = 10 –6 sin [100t + (x/50) + 0.5] y 2 = 10 –6 cos[100t + (x/50)] = 1 0 - 6 sin 1 0 0 t + x / 5 0 + π 2 = 10 –6 sin[100t + (x/50) + 1.57] using cos x = sin x + π 2 The phase difference = 1.57 – 0.5 = 1.07 or using sin = cos π 2 - x · We get the same result.

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