Question

For the harmonic travelling wave
$y=2\cos 2\pi (10t-0.0080x+3.5)$
where $x\text{and}y\text{are in}cm\text{and}t\text{is second}$. What is the phase difference between the oscillatory motion at two points separated by a distance of

$A$: $4m$

$B$: $0.5m$

$C$: $\lambda /2$

$D$: $\frac{3\lambda }{4}$ (At a given instant of time)

$E$: What is the phase difference between the oscillation of a particle located at $x=100cm,\text{at}t=Ts\text{and}t=5s$?

Answer 1

$A$:

Given, travelling wave equation,
$y=2\cos 2\pi (10t-0.0080x+3.5)$

Compare this equation with

$y=a\cos (\omega t-kx+\varphi )$

Propagation constant, $k=0.016\pi$

Given, path difference between two points, $\Delta x=4m=400cm$

So, phase difference, $\Delta \varphi =k\left(\Delta x\right)$

$\Delta \varphi =0.016\pi \left(400\right)=6.4\pi rad$

Final Answer: $6.4\pi rad$


$B$:

Given, travelling wave equation,
$y=2\cos 2\pi (10t-0.0080x+3.5)$

Compare this equation with

$y=a\cos (\omega t-kx+\varphi )$

Propagation constant, $k=0.016\pi$

Given, path difference between two points, $\Delta x=0.5m=50cm$

So, phase difference, $\Delta \varphi =k\left(\Delta x\right)$

$\Delta \varphi =0.016\pi \left(50\right)=0.8\pi rad$

Final Answer: $0.8\pi rad$


$C$:

Given, path difference between two points, $\Delta x=\lambda /2$

So, phase difference,
$\Delta \varphi =\frac{2\pi }{\lambda }\left(\Delta x\right)$
$\Delta \varphi =\frac{2\pi }{\lambda }\left(\frac{\lambda }{2}\right)=\pi rad$


Final Answer:$\pi rad$


$D$:


Given, path difference between two points, $\Delta x=3\lambda /4$

So, phase difference,
$\Delta \varphi =\frac{2\pi }{\lambda }\left(\Delta x\right)$
$\Delta \varphi =\frac{2\pi }{\lambda }\left(\frac{3\lambda }{4}\right)=\frac{3\pi }{2}rad$


Final Answer:$\frac{3\pi }{2}rad$



$E$:

Step 1: Find the period of the wave.

Given, wave equation,

$y=2\cos 2\pi (10t-0.0080x+3.5)$

Compare the wave equation with
$y=a\cos (\omega t-kx+\varphi )$
$\omega =20\pi$
So, time period, $T=\frac{2\pi }{\omega }=\frac{2\pi }{20\pi }$

$T=\frac{1}{10}s$

Step 2 : Find phase at $t=Ts$.

Phase of the wave,
$\varphi =2\pi (10t-0.0080x+3.5)$
So, phase at,
$t=T=\frac{1}{10}s$

${\varphi }_1=2\pi (10\times \frac{1}{10}-0.0080x+3.5)$
${\varphi }_1=2\pi (4.5-0.0080x)$

Step 3: Find the phase at $t=5s$ and phase difference.

Phase at $t=5s$
${\varphi }_2=2\pi (10\times 5-0.0080x+3.5)$

$\varphi =2\pi (53.5-0.0080x)$

So, phase difference,
$\Delta \varphi ={\varphi }_2-{\varphi }_1$
$\Delta \varphi =2\pi (53.5-0.0080x)-2\pi (4.5-0.0080x)$
$\Delta \varphi =98\pi rad$

Final Answer: $98\pi rad$