Two waves are given by $y_{1} = a\sin{(\omega t - kx)}$ and $y_{2} = a\cos{(\omega t - kx)}$ The phase difference between the two waves is

Question

Dr. Abubacker · Accepted Answer

(d) $y_{1} = a\sin{(\omega t - kx)}$
and $y_{2} = a\cos{(\omega t - kx)} = a\sin\left( \omega t - kx + \frac{\pi}{2} \right)$
Hence phase difference between these two is $\frac{\pi}{2}$

Q. Two waves are given by $y_{1} = a\sin{(\omega t - kx)}$ and $y_{2} = a\cos{(\omega t - kx)}$ The phase difference between the two waves is

Explanation

Reviewed by Dr. Abubacker

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