Question

When a wave traverses a medium, periodic motion executed by the individual particle at distance $x$ at time $t$ is given by the equation $y=8\sin \frac{\pi }{3}(3t+\frac{x}{6})$ where $t$ is in seconds and $x$ and $y$ are in $\text{cm}$. Then, choose the correct option(s):

• A. Frequency of the given wave is $0.5\text{Hz}$.
• B. Wave speed of the given wave in the given medium is $18\text{cm/s}$.
• C. Phase difference for two positions of the same particle at time internal $0.4\text{s}$ is ${72}^{\circ }$.
• D. Phase difference of two particles $18\text{cm}$ apart at a given instant is $\pi \text{rad}$.

Answer 1

Answer: (A), (B), (C), (D)

Phase difference of two particles $18\text{cm}$ apart at a given instant is $\pi \text{rad}$.

Given, equation of the wave is

$y=8\sin \frac{\pi }{3}(3t+\frac{x}{6})$

Rewriting the above equation, we get

$y(x,t)=8\sin \frac{2\pi }{36}(18t+x)$

Comparing with the general equation of wave motion

$y=A\sin \frac{2\pi }{\lambda }(vt+x)$

Amplitude $\left(A\right)=8\text{cm}$
Wavelength $\left(\lambda \right)=36\text{cm}$
Wave velocity $\left(v\right)=18\text{cm/s}$.

$\therefore$ frequency $\left(f\right)=\frac{v}{\lambda }=\frac{18}{36}=0.5\text{Hz}$.

Now, phase of a particle at instant $t_1$ is given by

${\varphi }_1=\frac{\pi }{3}(3t_1+\frac{x}{6})$

and at instant $t_2$,

${\varphi }_2=\frac{\pi }{3}(3t_2+\frac{x}{6})$

The phase difference is given as

$\Delta \varphi ={\varphi }_1-{\varphi }_2=\frac{\pi }{3}\left[3\right(t_1-t_2\left)\right]$

For time difference $\Delta t=0.4\text{s}$,
$\Delta \varphi =\frac{\pi }{3}\times 3\times 0.4=180\times 0.4={72}^{\circ }$

Also, we know that $\Delta \varphi =\frac{2\pi }{\lambda }|\Delta x|$

Phase difference at an instant between two particles with path difference $18\text{cm}$ is given by

$\Delta \varphi =\frac{2\pi }{36}\times 18=\pi \text{rad}$

[substituting the given data]

Hence, options (a) , (b) , (c) and (d) are the correct answers.