Question

The displacement equations of two simple harmonic oscillators are given by $x_1=A_1\cos \omega t$; $x_2=A_2\sin (\omega t+\frac{\pi }{6})$. The phase difference between them is:

• A. ${30}^{\circ }$
• B. ${60}^{\circ }$
• C. ${90}^{\circ }$
• D. ${120}^{\circ }$

Answer 1

The correct option is B ${60}^{\circ }$

$X_1=A_{1}cos\omega t=A_{1}sin\left(\omega t{+}^{\pi }{/}_2\right)$
$X_2=A_{2}sin\left(\omega t{+}^{\pi }{/}_6\right)$
so, phase difference $=\frac{\pi }{2}-\frac{\pi }{6}=\frac{\pi }{3}={60}^0$