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. The phase difference between two particles $2mm$ apart at any instant is

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

Answer 1

The correct option is D ${72}^o$

Given equation can be written as $y=6\sin (4\pi t-0.2\pi x)$

$\therefore$ Phase $\varphi =-0.2\pi x$

$\Rightarrow {\varphi }_1=-0.2\pi x_1$ and ${\varphi }_2=-0.2\pi x_2$

$\therefore$ Phase difference $\Delta \varphi =|{\varphi }_2-{\varphi }_1|=|-0.2\pi (x_2-x_1)|$

given $(x_2-x_1)=2mm$ substituting in above equation we get $\Delta \varphi =0.2\times {180}^0\times 2={72}^0$