Question

A simple harmonic progressive wave is represented by the equation $y=8sin2\pi (0.1x-2t)$ where x and y are in centimeters and t is in seconds. At any instant the phase difference between two particles separated by 2.0 cm along the x-direction is

• A. ${18}^{\circ }$
• B. ${36}^{\circ }$
• C. ${54}^{\circ }$
• D. ${72}^{\circ }$

Answer 1

The correct option is D

${72}^{\circ }$

$y=8sin\pi (\frac{x}{10}-2t)2$ given by comparing with

$y=asin2\pi (\frac{t}{T}-\frac{x}{\lambda })$

$\lambda =10cm$

So phase difference = $\left(\frac{2\pi }{\lambda }\right)\times pathdifference$

$=\frac{2\pi }{10}\times 2=\frac{2}{5}\times {180}^{\circ }={72}^{\circ }$