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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

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detailed solution

Correct option is B

y1=10−6sin [100 t+(x/50)+0.5]y2=10−6sin 100 t+x50+π2Phase difference ϕ=100t+(x/50)+1.57-100t+(x/50)+0.5 =1.07radians.

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The phase difference between two waves represented by y1=10−6sin[100 t+(x/50)+0.5]m y2=10−6cos [100 t+(x/50)]mwhere x is expressed in metres and t is expressed in seconds, is approximately

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Ready to Test Your Skills?

free study material

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