Question

A wave given by equation $y=8\sin 2\pi (0.1x-2t)$ where $x$ and $y$ are in centimeters and $t$ is in seconds. At any instant , the phase difference between two particles, separated by $2.0\text{cm}$ along the $x$-direction, is,

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

Answer 1

The correct option is D ${72}^{\circ }$

Given:
$y=8\sin 2\pi (\frac{x}{10}-2t)$
$\Delta x=2\text{cm}$

writing the equation again we have,
$y=8\sin (\frac{2\pi }{10}x-4\pi t)$
$\Rightarrow k=\frac{2\pi }{10}=\frac{2\pi }{\lambda }$
gives, $\lambda =10\text{cm}$

Now, the phase difference between the points is given by,
$\Delta \varphi =\frac{2\pi }{\lambda }\times \Delta x$

$\Delta \varphi =\frac{2\pi }{10}\times 2=\frac{2\pi }{5}\text{rad}$

$\Rightarrow \Delta \varphi =\frac{2}{5}\times {180}^{\circ }={72}^{\circ }$

Hence, $\left(D\right)$ is the correct answer.