Question

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

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

Answer 1

Answer: (B)

$\frac{\pi}{12}$

Answer 2

Here $\, \, \, \, \, \, \, y_1 =\frac{1}{2}sin \, \omega t +\frac{\sqrt 3}{2} cos \omega t$
$\, \, \, \, \, \, \, \, =cos\frac{\pi}{3}sin \omega t +sin \frac{\pi}{3}cos \omega t$
$\therefore \, \, \, \, \, \, \, \, y_1 =sin \bigg(\omega t +\frac{\pi}{3}\bigg)$
Similarly $\, \, \, \, \, \, y_2 =\sqrt 2 sin \bigg(\omega t +\frac{\pi}{4}\bigg)$
$\therefore \, \, \, \, \, Phase \, difference \Delta \phi =\frac{\pi}{3}-\frac{\pi}{4}$
$\, \, \, \, \, \, \, \, \, \, \, \, =\frac{\pi}{12}$