Two waves are represented by the equations $y_1=a \sin (\omega t+k x+0.57) \mathrm{m}$ and $y_2=a \cos (\omega t+k x) \mathrm{m}$, where $x$ is in metre and $t$ in second. The phase difference between them is

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Aicharya · Accepted Answer

$y_1=a \sin (\omega t+k x+0.57) \mathrm{m}$ and $y_2=a \cos (\omega t+k x) \mathrm{m}$ or $y_2=a \sin \left(\frac{\pi}{2}+\omega t+k xight) \mathrm{m}$ Phase difference $\begin{aligned} & \Delta \phi=\phi_2-\phi_1 \ & =\frac{\pi}{2}-0.57=1.57-0.57 \ & =1 \mathrm{rad} \end{aligned}$ ~

Two waves are represented by the equations $y_1=a \sin (\omega t+k x+0.57) \mathrm{m}$ and $y_2=a \cos…

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