Question

The phase difference between two waves represented by
$y_1={10}^{-6}\sin \left[100t+\frac{x}{50}+0.5\right]m$
$y_2={10}^{-6}\cos \left[100t+\frac{x}{50}\right]m$
where x is expressed in metres and t is expressed in seconds is approximately

• A. 1.07 rad
• B. 2.07 rad
• C. 0.5 rad
• D. 1.5 rad

Answer 1

The correct option is A 1.07 rad

$y_1={10}^{-6}\sin [100t+\frac{x}{50}+0.5]m$

$y_2={10}^{-6}\cos [100t+\frac{x}{50}]m$

So,$y^1={10}^{-6}\cos [\frac{\pi }{2}-(100t+\frac{x}{50}+0.5)]$

$y^1={10}^{-6}\cos [\frac{-\pi }{2}+100t+\frac{x}{50}+0.5]$

$△\varphi ={\varphi }_2-{\varphi }_1$

${\varphi }_2=0$ where as ${\varphi }_1=0.5-\frac{\pi }{2}$

So,

$△\varphi ={\varphi }_2-{\varphi }_1=\frac{\pi }{2}-0.5=1.07rad$

option $-A$ is correct.