Question

A wave is travelling along positive $x-$ direction with speed $4\text{m/s}$. Further, equation of the wave pulse at $t=0$ is $y(x,0)=\frac{5}{4+(3x+6{)}^2}$. The equation of the same pulse after $t\text{s}$ will be

• A. $\frac{5}{4+9(x+12t+2{)}^2}$
• B. $\frac{5}{4+9(x-4t+2{)}^2}$
• C. $\frac{5}{4+9(x+4t+2{)}^2}$
• D. $\frac{5}{4+(3x+12t+6{)}^2}$

Answer 1

The correct option is B $\frac{5}{4+9(x-4t+2{)}^2}$

Given, at $t=0$
$y(x,0)=\frac{5}{4+(3x+6{)}^2}$
Re-writing the above equation, we get
$y(x,0)=\frac{5}{4+9(x+2{)}^2}.......\left(1\right)$

If the speed of the pulse is $v$, then the pulse will travel to the right by a distance $vt$ in time $t$.
Assuming the shape of the pulse does not change with time, the transverse position of the particles for all positions and times, measured in a stationary frame with the origin at $\text{O}$ can be written as
$y(x,t)=y(x-vt,0)$

Therefore, replacing $x$ with $x-vt$ in the equation $\left(1\right)$, we get
$y(x,t)=\frac{5}{4+9(x-vt+2{)}^2}$
Given that , speed of the wave $v=4\text{m/s}$
$\therefore y(x,t)=\frac{5}{4+9(x-4t+2{)}^2}$

Thus, option (b) is the correct answer.