Question

The minimum phase difference between two SHM's is:
$y_1=\sin \left(\frac{\pi }{6}\right)\sin \left(\omega t\right)+\sin \left(\frac{\pi }{3}\right)\cos \left(\omega t\right)$
$y_2=\cos \left(\frac{\pi }{6}\right)\sin \left(\omega t\right)+\sin \left(\frac{\pi }{3}\right)\cos \left(\omega t\right)$ is:

• A. $\frac{\pi }{3}$
• B. $\frac{\pi }{6}$
• C. $\frac{\pi }{12}$
• D. $0$

Answer 1

The correct option is B $\frac{\pi }{6}$

$y_1=\sin \left(\omega t\right)\sin \left(\frac{\pi }{6}\right)+\cos$ $\omega t\sin \frac{\pi }{3}$

$=\sin$ $\omega t\cos \left(\frac{\pi }{3}\right)+\cos$ $\omega t\sin \frac{\pi }{3}$

$=\sin (\omega t+\frac{\pi }{3})$

Similarly for the second equation is:
$y_2=\sin (\omega t+\frac{\pi }{6})$

Thus phase difference between the two waves is

$\frac{\pi }{3}-\frac{\pi }{6}=\frac{\pi }{6}$