Question

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

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

Answer 1

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

$y_1=\frac{1}{2}sim\omega t+\frac{\sqrt{3}}{2}cos\omega t$
Thus we get $y_1=cos\frac{\pi }{3}sim\omega t+sin\frac{\pi }{3}cos\omega t$
$\therefore y_1=sim(\omega t+\pi /3)$
$y_2=\sqrt{2}(\frac{1}{\sqrt{2}}sim\omega t+\frac{1}{\sqrt{2}}cos\omega t)$
Similarly, $y_2=\sqrt{2}sin(\omega t+\pi /4)$
Phase difference $=\Delta \Phi =\frac{\pi }{3}-\frac{\pi }{4}=\frac{\pi }{12}$