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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1.5 rad

1.07 rad

2.07 rad

0.5 rad

answer is B.

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Detailed Solution

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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