Question

The phase difference between two waves $x_1=A\sin (\omega t+\frac{\pi }{6})$ and $x_2=A\cos \omega t$ is

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

Answer 1

Answer: (B)

Here, the equation of two waves
$x_1=Asin\left(\omega t+\frac{\pi }{6}\right)$ and
$x_2=Acos\left(\omega t\right)$
$\because sin\left(\theta +\frac{\pi }{2}\right)=cos\theta$
Hence, $x_2=Asin\left(\omega t+\frac{\pi }{2}\right)$
Now, the phase difference
$\varphi =\left(\omega t+\frac{\pi }{2}\right)-\left(\omega t+\frac{\pi }{6}\right)$
$\Rightarrow \varphi =\frac{\pi }{2}-\frac{\pi }{6}=\frac{\pi }{3}rad$
Hence, the phase difference between the waves is $\frac{\pi }{3}$ rad.