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

Answer 1

Answer: (B)

$\frac{- \pi}{4}$

Answer 2

Equations of waves
\hspace30mm y_1 = \cos (4t - 2x) = \sin \big(4t - 2x + \frac{\pi}{2} \big)
and \hspace20mm y_2 = \sin \big(4t - 2x + \frac{\pi}{4} \big)
Therefore, phase difference between the two waves is
\hspace40mm ? \phi = \big(4t - 2x + \frac{\pi}{4}\big) - \big(4t - 2x + \frac{\pi}{2}\big)
\hspace40mm = \frac{\pi}{4} - \frac{\pi}{2} = - \frac{\pi}{4}