MCQ
If $x = a\sin \left( {\omega t + \frac{\pi }{6}} \right)$ and $x' = a\cos \omega t$, then what is the phase difference between the two waves
- ✓$ \pi/ 3$
- B$ \pi/ 6$
- C$ \pi/ 2$
- D$ \pi$
✓
Answer
Correct option: A.
$ \pi/ 3$
a
(a) $x = a\sin \left( {\omega t + \frac{\pi }{6}} \right)$ and $x' = a\cos \omega t = a\sin \,\left( {\omega \,t + \frac{\pi }{2}} \right)$
$\therefore \Delta \phi = \left( {\omega t + \frac{\pi }{2}} \right) - \left( {\omega t + \frac{\pi }{6}} \right) = \frac{\pi }{3}$
(a) $x = a\sin \left( {\omega t + \frac{\pi }{6}} \right)$ and $x' = a\cos \omega t = a\sin \,\left( {\omega \,t + \frac{\pi }{2}} \right)$
$\therefore \Delta \phi = \left( {\omega t + \frac{\pi }{2}} \right) - \left( {\omega t + \frac{\pi }{6}} \right) = \frac{\pi }{3}$
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