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Two waves are represented by the equations $y_1 = a\sin(\omega t + kx + 0.57)$ m and $y_2 = a\cos(\omega t + kx)$ m, where x is in meter and t in s. The phase difference between them is
A
0.57 radian
B
1.0 radian
C
1.25 radian
D
1.57 radian
Detailed Solution
Write both waves as sine functions.
$y_2 = a\cos(\omega t + kx) = a\sin\left(\omega t + kx + \frac{\pi}{2}\right)$
$y_1 = a\sin(\omega t + kx + 0.57)$
Phase difference $= \frac{\pi}{2} - 0.57$
$= 1.57 - 0.57 = 1.0$ radian
$y_2 = a\cos(\omega t + kx) = a\sin\left(\omega t + kx + \frac{\pi}{2}\right)$
$y_1 = a\sin(\omega t + kx + 0.57)$
Phase difference $= \frac{\pi}{2} - 0.57$
$= 1.57 - 0.57 = 1.0$ radian