Question

Two sinusoidal waves are defined by the functions :
$y_1=\left(2.00\text{cm}\right)\sin (20x-32t)$ and $y_2=\left(2.00\text{cm}\right)\sin (25x-40t)$ where $y_1,y_2$ and $x$ are in centimeters and $t$ is in seconds. What is the phase difference between these two waves at the point $x=5.00\text{cm}$ at $t=2\text{s}$?

• A. ${516}^{\circ }$
• B. ${258}^{\circ }$
• C. ${333}^{\circ }$
• D. ${412}^{\circ }$

Answer 1

The correct option is A ${516}^{\circ }$

Given:
$y_1=\left(2.00\text{cm}\right)\sin (20x-32t)$
$y_2=\left(2.00\text{cm}\right)\sin (25x-40t)$

Phase difference between the waves is given by,
$\Delta \varphi ={\varphi }_1-{\varphi }_2=(20x-32t)-(25x-40t)$
$\therefore \Delta \varphi =-5x+8t$

At $x=5.00\text{cm}$ and $t=2\text{s}$, the phase difference is,
$\Delta \varphi =-5\times 5+8\times 2$

$\Rightarrow |\Delta \varphi |=9\text{rad}\approx {516}^{\circ }$

Hence, $\left(A\right)$ is the correct answer.