Question

A simple harmonic wave has the equation$y=6\sin 2\pi (2t–0.1x)$, where$x$ and $y$are in $millimeters$ and $t$is in $seconds$. At any given time, the phase difference between two particles$2mm$ apart is

• A. $18degrees$
• B. $36degrees$
• C. $54degrees$
• D. $72degrees$

Answer 1

The correct option is D

$72degrees$

Step 1. Given data:

$y=6\sin 2\pi (2t–0.1x)$… ….(1)

The phase difference between two particles$2mm$

Step 2. Formula used:

The general equation of progressive wave is given by,

$y=A\sin \left(\omega t-\phi \right)........\left(2\right)$, where and $y$is displacement in $milimeters$ and $t=$time, $A=$amplitude, $\omega =$angular velocity

Step 3. Calculating the required time,

The equation (1) given in the question can also be written as-

$y=6\sin (4\mathrm{\pi t}-0.2\mathrm{\pi x})$

Comparing equation (1) with equation (2)

$\mathrm{phase}\mathrm{\varphi }=-0.2\mathrm{\pi x}$

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

Then the phase difference,

$∆\varphi =\left|{\varphi }_2-{\varphi }_1\right|=\left|-0.2\pi (x_2-x_1)\right|$

Since, the path difference is,

$({\mathrm{x}}_2-{\mathrm{x}}_1)=2\mathrm{mm}$ , by putting this value in the above equation, we get

$∆\varphi =0.2\times {180}^{°}\times 2={72}^{°}$

The phase difference is$72degrees$

Hence, option D is the correct answer.