Question

At time $t=0,$ equation of a wave pulse is $y(x,0)=\frac{5}{3+(x-4{)}^2}$ and at $t=2s$, equation of the same wave pulse is $y(x,2)=\frac{5}{3+(x+8{)}^2}$. Here, $y$ is in $\text{mm}$, $x$ is in $\text{m}$ and $t$ is in $\text{sec}$. The wave speed is

• A. $5\text{m/s}$
• B. $6\text{m/s}$
• C. $6\text{mm/s}$
• D. $3\text{mm/s}$

Answer 1

The correct option is B $6\text{m/s}$

Given, at $t=0$,
$y(x,0)=\frac{5}{3+(x-4{)}^2}.....\left(1\right)$
Also, at $t=2s$,
$y(x,2)=\frac{5}{3+(x+8{)}^2}.....\left(2\right)$

Assuming the wave pulse is travelling towards right, we can replace $x$ with $x-vt$, where $v$ is wave speed in $\text{m/s}$ and $t$ is time (in seconds).
Therefore from $\left(1\right)$, we get
$y(x,t)=\frac{5}{3+\left(\right(x-vt)-4{)}^2}.....\left(3\right)$

Comparing $\left(2\right)$ with $\left(3\right)$ we get,
$-vt-4=8\Rightarrow -v=\frac{12}{t}$
At $t=2\text{s}$, $v=-6\text{m/s}$
Thus, wave speed $|v|=6\text{m/s}$

Hence, option (b) is the correct answer.