Question

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

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

Answer 1

Answer: (C)

$y=12 \, sin \, \left(\right.5t-4x\left.\right)$
Comparing in $y=Asin \left(w t - k x\right)$
We have $A=12, \, w=5 \, and \, k=4$
$\left(w t - k x\right)=$ phase difference $=\frac{\pi }{2}$

$\therefore \left|\right. 5 t - 4 x \left|\right. = \frac{\pi }{2}$
When $t=0, \, 4x=\frac{\pi }{2}$
$\therefore x=\frac{\pi }{8} \, cm$