The phase difference between two waves,repres | vedclass

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(D) The first wave is given by $y_1 = 10^{-6} \sin \{100t + (x/50) + 0.5\}$.The second wave is given by $y_2 = 10^{-6} \cos \{100t + (x/50)\}$.We know

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$1.07$

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(D) The first wave is given by $y_1 = 10^{-6} \sin \{100t + (x/50) + 0.5\}$.The second wave is given by $y_2 = 10^{-6} \cos \{100t + (x/50)\}$.We know that $\cos(	heta) = \sin(	heta + \pi/2)$.Therefore,$y_2 = 10^{-6} \sin \{100t + (x/50) + \pi/2\}$.The phase of the first wave is $\phi_1 = 100t + (x/50) + 0.5$.The phase of the second wave is $\phi_2 = 100t + (x/50) + \pi/2$.The phase difference $\Delta \phi = \phi_2 - \phi_1 = (100t + x/50 + \pi/2) - (100t + x/50 + 0.5)$.$\Delta \phi = \pi/2 - 0.5$.Using $\pi \approx 3.14$,$\Delta \phi \approx 1.57 - 0.5 = 1.07 \ radians$.

The phase difference between two waves,represented by $y_1 = 10^{-6} \sin \{100t + (x/50) + 0.5\} \ m$ and $y_2 = 10^{-6} \cos \{100t + (x/50)\} \ m$,where $x$ is in meters and $t$ is in seconds,is approximately .... $radians$.

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