Question

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

• A. $\frac{\pi}{4}$
• B. π
• C. $\frac{\pi}{8}$
• D. $\frac{\pi}{2}$

Answer 1

Answer: (D)

$\frac{\pi}{2}$

Answer 2

$y_{1} = a\sin(\omega t - kx)$

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 $\frac{\pi}{2}.$