Question

The phase difference between two waves represented by
$y_1={10}^{-6}\sin [100t+(x/50)+0.5]\text{m}$
$y_2={10}^{-6}\cos [100t+(x/50\left)\right]\text{m}$
Where $x$ is expressed in metre and $t$ is expressed in second, is approximately
[Take $\pi =3.14$]

• A. $2.07$ radian
• B. $0.5$ radian
• C. $1.5$ radian
• D. $1.07$ radian

Answer 1

The correct option is D $1.07$ radian

Given, $y_2={10}^{-6}\cos [100t+\frac{x}{50}]\text{m}$
or we can write it as
$y_2={10}^{-6}\sin [100t+\frac{x}{50}+\frac{\pi }{2}]\text{m}$
Comparing with $y_1={10}^{-6}\sin [100t+\frac{x}{50}+0.5]\text{m}$
Phase difference $\Delta \varphi =\frac{\pi }{2}-0.5=1.57-0.5=1.07\text{rad}$