Question

A pulse moving along $x-$axis is represented by the wave function $y(x,t)=\frac{2}{(x-2t{)}^2+1}$ where $x$ and $y$ measured in centimeters and $t$ is measured in seconds. The shape of pulse $y(x,0)=\frac{2}{x^2+1}$ is shown in the figure below. Which of the following correctly represents the shape of pulse at $t=1\text{s}$.

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

Given the displacement of particle as a function of position$\left(x\right)$ and time $\left(t\right)$,
$y(x,t)=\frac{2}{(x-2t{)}^2+1}$
The wave function for a travelling wave (travelling along $+vex-$direction) is in the form of $y=f(x-vt)$.

comparing the equation $y(x,t)$ with the format of $y=f(x-vt)$, it is found that :
$(x-vt)=x-2t$
hence wave speed is $v=2.0\text{cm/s}$.
Furthermore by substituting $x-2.0t=0$ in $y(x,t)$ we find the maximum value of displacement $\left(y_{\text{max}}\right)$
$\Rightarrow y_{\text{max}}=2.0\text{cm}$

Again for $t=0$ seconds,
$y(x,0)=\frac{2}{x^2+1}$
for $t=1$ seconds
$y(x,1)=\frac{2}{(x-2t{)}^2+1}=y(x-2t,0)$
In one second,
$\Rightarrow \Delta x=v\times \Delta t=2\times 1=2\text{cm}$
Therefore the pulse has moved from $x=0\text{cm}$ to $x=2\text{cm}$ towards right.

Hence, the shape of the pulse remains same with maximum displacement occuring at $x=2\text{cm}$.