Question

The equation of a simple harmonic wave is given by $y=6\sin 2\pi (2t-0.1x)$, where '$x$' and '$y$' are in $mm$ and '$t$' is in seconds. Find the phase difference between two particles 2mm apart at any instant.

Answer 1

Phase difference$=\frac{2\pi }{\lambda }\times$path difference.

$y=6sin2\pi (2t-0.1x)mmy=Asin2\pi (\frac{l}{T}-\frac{x}{\lambda })$

Comparing both we will get,

$\lambda =\frac{1}{0.1}=10mm$

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