Question

Two waves are given by $y_1 = \cos \, (4t - 2x)$ and $y_2 = sin \, \big(4t - 2x + \frac{\pi}{4} \big).$ The phase difference between the two waves is

• A. $\frac{\pi}{4}$
• B. $\frac{- \pi}{4}$
• C. $\frac{3 \pi}{4}$
• D. $\frac{\pi}{2}$
• E. $\frac{3 \pi}{2}$

Answer 1

Answer: (B)

Equations of waves
$ y_1 = \cos (4t - 2x) = \sin \big(4t - 2x + \frac{\pi}{2} \big)$
and $ y_2 = \sin \big(4t - 2x + \frac{\pi}{4} \big)$
Therefore, phase difference between the two waves is
$ \Delta \phi = \big(4t - 2x + \frac{\pi}{4}\big) - \big(4t - 2x + \frac{\pi}{2}\big)$
$ = \frac{\pi}{4} - \frac{\pi}{2} = - \frac{\pi}{4}$