The phase difference between two waves repres | vedclass

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(B) Given equations are:$y_1 = 10^{-6} \sin [100t + (x/50) + 0.5]$$y_2 = 10^{-6} \cos [100t + (x/50)]$To compare the phases,convert the cosine functio

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$1.07$

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(B) Given equations are:$y_1 = 10^{-6} \sin [100t + (x/50) + 0.5]$$y_2 = 10^{-6} \cos [100t + (x/50)]$To compare the phases,convert the cosine function into a sine function using the identity $\cos(	heta) = \sin(	heta + \pi/2)$:$y_2 = 10^{-6} \sin [100t + (x/50) + \pi/2]$Since $\pi/2 \approx 1.57$,we have:$y_2 = 10^{-6} \sin [100t + (x/50) + 1.57]$The phase of the first wave is $\phi_1 = 100t + (x/50) + 0.5$.The phase of the second wave is $\phi_2 = 100t + (x/50) + 1.57$.The phase difference $\Delta\phi$ is given by:$\Delta\phi = |\phi_2 - \phi_1|$$\Delta\phi = |(100t + x/50 + 1.57) - (100t + x/50 + 0.5)|$$\Delta\phi = 1.57 - 0.5 = 1.07 \, rad$.

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 expressed in meters and $t$ is expressed in seconds,is approximately .... $rad$.

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