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(D) The given equations are $y_1 = a \sin(\omega t + kx + 0.57)$ and $y_2 = a \cos(\omega t + kx)$.To find the phase difference, we must express both

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$1.0 	ext{ radian}$

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(D) The given equations are $y_1 = a \sin(\omega t + kx + 0.57)$ and $y_2 = a \cos(\omega t + kx)$.To find the phase difference, we must express both waves in terms of the same trigonometric function (sine).We know that $\cos(	heta) = \sin(	heta + \pi/2)$.Therefore, $y_2 = a \sin(\omega t + kx + \pi/2)$.Given $\pi = 3.14$, then $\pi/2 = 3.14 / 2 = 1.57 	ext{ radians}$.So, $y_2 = a \sin(\omega t + kx + 1.57)$.The phase of $y_1$ is $\phi_1 = \omega t + kx + 0.57$ and the phase of $y_2$ is $\phi_2 = \omega t + kx + 1.57$.The phase difference $\Delta\phi = |\phi_2 - \phi_1| = |1.57 - 0.57| = 1.0 	ext{ radian}$.

Two waves are represented by the equations, $y_1 = a \sin(\omega t + kx + 0.57)$ and $y_2 = a \cos(\omega t + kx)$, where $x$ is in metre and $t$ is in second. What is the phase difference between them? $(\pi = 3.14)$

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