The phase difference between two waves, represented by
$y_{1} = 10^{- 6} \sin [100 t + (\frac{x}{50}) + 0.5] m$
$y_{2} = 10^{- 6} \cos [100 t + (\frac{x}{50})] m$ ,
where $x$ is expressed in metres and $t$ is expressed in seconds, is approximately

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y 1 = 10 – 6 sin [ 100 t + ( x 50 ) + 0 . 5 ] y 2 = 10 - 6 cos [ 100 t + ( x / 50 ) ] = 1 0 - 6 sin 1 0 0 t + x / 5 0 + π 2 = 10 – 6 sin [ 100 t + ( 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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