Question

A progressive wave is represented by $y=12sin(5t-4x)cm$. On this wave, how far away are the two points having phase difference of $90^{°}$ ?

• A. $\frac{\pi }{2}cm$
• B. $\frac{\pi }{4}cm$
• C. $\frac{\pi }{8}cm$
• D. $\frac{\pi }{16}cm$

Answer 1

Answer: (C)

Given equation for the wave is $y=12\sin (5t-4x)$
Phase is given by $5t-4x$.
At a given instant, let two points $x_1$ and $x_2$ differ in phase by $90^{\circ }=\pi /2$.
$\Rightarrow \left(5t-4x_1\right)-\left(5t-4x_2\right)=\pi /2$
$\Rightarrow 4\left|x_2-x_1\right|=\pi /2$
Thus, distance between the two points is $\pi /8cm$