Question

What is the phase difference between two simple harmonic motions represented by $x_{1}=A \sin \left(\omega t+\frac{\pi}{6}\right)$ and $x_{2}=A \cos (\omega t) ?$

• A. $\frac{\pi}{6}$
• B. $\frac{\pi}{3}$
• C. $\frac{\pi}{2}$
• D. $\frac{2\pi}{3}$

Answer 1

Answer: (B)

Given,
$x_{1}=A \sin \left(\omega t+\frac{\pi}{6}\right)$
$x_{2}=A \cos (\omega t)$
$x_{2}=A \cos t\left(\omega t+\frac{\pi}{2}\right)$
Phase difference
$\Delta \phi=\phi_{2}-\phi_{1}$
$\Delta \phi=\frac{\pi}{2}-\frac{\pi}{6}$
$\Delta \phi=\frac{3 \pi-\pi}{6}=\frac{\pi}{3}$