Question

For the equation, $y=A\sin 2\pi (ax+bt+\pi /4)$, match the following columns.

$\begin{array}{cc}\text{Column I}& \text{Column II}\\ \text{(A) Frequency of wave(Hz)}& \text{(p) a}\\ \text{(B) wavelength of wave(m)}& \text{(q) b}\\ \text{(C) Phase difference between two points}\frac{1}{4a}\text{m}\text{distance apart}& \text{(r)}\pi \\ \text{(D) Phase difference of a particle after a time interval of}\frac{1}{8b}\text{s}& \text{(s)}\frac{\pi }{2}\\ & \text{(t) None of above}\end{array}$

• A. $\begin{array}{cccc}\text{A}& \text{B}& \text{C}& \text{D}\\ \text{q}& \text{t}& \text{s}& \text{t}\end{array}$
• B. $\begin{array}{cccc}\text{A}& \text{B}& \text{C}& \text{D}\\ \text{t}& \text{q}& \text{r}& \text{s}\end{array}$
• C. $\begin{array}{cccc}\text{A}& \text{B}& \text{C}& \text{D}\\ \text{p}& \text{q}& \text{r}& \text{s}\end{array}$
• D. $\begin{array}{cccc}\text{A}& \text{B}& \text{C}& \text{D}\\ \text{t}& \text{s}& \text{p}& \text{r}\end{array}$

Answer 1

The correct option is A $\begin{array}{cccc}\text{A}& \text{B}& \text{C}& \text{D}\\ \text{q}& \text{t}& \text{s}& \text{t}\end{array}$

Given,
Wave function $y=A\sin 2\pi (ax+bt+\frac{1}{4})$
Comparing this with standard equation of a progressive wave along $-ve,x-$direction,
$y=A\sin (kx+\omega t+\varphi )$
We get,
$k=2\pi a,\omega =2\pi b,\varphi =\frac{\pi }{2}$
We know that, $k=\frac{2\pi }{\lambda }$
$\Rightarrow \frac{2\pi }{\lambda }=2\pi a$
$\Rightarrow \lambda =\frac{1}{a}\text{m}$
and $\omega =2\pi f$
$\Rightarrow 2\pi f=2\pi b$
$\Rightarrow f=b\text{Hz}$
Phase difference between two points which are $\frac{1}{4a}$ distance apart i.e $\Delta x=x_2-x_1=\frac{1}{4a}$

Keeping the time constant;
From the wave equation, the phase at position $x_2$ is,
${\varphi }_2=2\pi (a{x}_2+bt+\frac{1}{4})$
Keeping the time constant;
The phase at position $x_1$ is,
${\varphi }_1=2\pi (a{x}_1+bt+\frac{1}{4})$
Hence phase difference is:
$\Rightarrow \Delta \varphi ={\varphi }_2-{\varphi }_1=2\pi a(x_2-x_1)$
$\Rightarrow \Delta \varphi =2\pi a\times \frac{1}{4a}=\frac{\pi }{2}$

Now keeping the distance constant;
The phase at time $t_1$ is,
${\varphi }_1=2\pi (ax+b{t}_1+\frac{1}{4})$
The phase at time $t_2$ is,
${\varphi }_2=2\pi (ax+b{t}_2+\frac{1}{4})$
Hence phase difference is:
$\Rightarrow \Delta \varphi ={\varphi }_2-{\varphi }_1=2\pi b(t_2-t_1)$
Given the time interval, $\Delta t=t_2-t_1=\frac{1}{8b}\text{s}$
$\Rightarrow \Delta \varphi =2\pi b\times \frac{1}{8b}=\frac{\pi }{4}$
Hence, option $\left(a\right)$ is correct.