The phase difference between two waves represented by $y_{1} = 10^{- 6}\sin{\lbrack 100t + (x/50) + 0.5\rbrack m}$ $y_{2} = 10^{- 6}\cos{\lbrack 100t + (x/50)\rbrack m}$ Where $x$ is expressed in metres and $t$ is expressed in second, is approximately

Question

Dr. Abubacker · Accepted Answer

(b) $y_{1} = 10^{- 6}\sin{\lbrack 100\ t + (x/50) + 0.5\rbrack}$
$y_{2} = 10^{- 6}\sin\left\lbrack 100t + \left( \frac{x}{50} \right) + \left( \frac{\pi}{2} \right) \right\rbrack$
Phase difference $\phi$
$= \left\lbrack 100t + (x/50) + 1.57 \right\rbrack - \lbrack 100t + (x/50) + 0.5\rbrack$
$= 1.07\ radians$

Q. The phase difference between two waves represented by $y_{1} = 10^{- 6}\sin{\lbrack 100t + (x/50) + 0.5\rbrack m}$ $y_{2} = 10^{- 6}\cos{\lbrack 100t + (x/50)\rbrack m}$ Where $x$ is expressed in metres and $t$ is expressed in second, is approximately

Explanation

Reviewed by Dr. Abubacker

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