MCQ
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$\pi$
- C$\frac{\pi }{8}$
- ✓$\frac{\pi }{2}$
✓
Answer
Correct option: D.
$\frac{\pi }{2}$
d
(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 $\frac{\pi }{2}.$
(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 $\frac{\pi }{2}.$
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