Question

The minimum phase difference between two simple harmonic oscillations,
$y_1=\frac{1}{2}\sin \omega t+\frac{\sqrt{3}}{2}\cos \omega t$
$y_2=\sin \omega t+\cos \omega t$

• A. $\frac{7\pi }{12}$
• B. $\frac{\pi }{12}$
• C. $-\frac{\pi }{3}$
• D. $\frac{\pi }{6}$

Answer 1

The correct option is B $\frac{\pi }{12}$

Phase difference between any two particles in a wave determines lack of harmony in the vibrating state of two particles ie, how far one particle leads the other or lags behind the other.
Here, $y_1=\frac{1}{2}\sin \omega t+\frac{\sqrt{3}}{2}\cos \omega t$
$=\cos \frac{\pi }{3}\sin \omega t+\sin \frac{\pi }{2}\cos \omega t$
$\therefore y_1=\sin (\omega t+\frac{\pi }{3})$
Similarly, $y_2=\sqrt{2}\sin (\omega t+\frac{\pi }{4})$
$\therefore$ Phase difference $\Delta \varphi =\frac{\pi }{3}-\frac{\pi }{4}$
$=\frac{\pi }{12}$