Question

$y(x,t)=\frac{0.6}{\left[\right(2x+4t{)}^2+5]}$ represents a moving pulse where $x$ and $y$ are in metres and $t$ is in seconds. Then, choose the correct alternative.

• A. Maximum displacement of particle at $x=0$ is $0.12\text{m}$.
• B. Pulse is symmetric about $y-$ axis.
• C. In $2\text{s}$, pulse will travel a distance of $4\text{m}$.
• D. All of these

Answer 1

The correct option is D All of these

Given,
Equation of a wave pulse
$y(x,t)=\frac{0.6}{\left[\right(2x+4t{)}^2+5]}......\left(1\right)$
Assuming shape of the wave pulse does not change with time,
Displacement of the particle at $x=0$
$y(0,t)=\frac{0.6}{[16t^2+5]}$
Maximum displacement of the particle at $x=0$ will be when $t=0$
$\therefore y_{max}=\frac{0.6}{5}=0.12\text{m}$

From equation $\left(1\right)$, at $t=0$
$y(x,0)=\frac{0.6}{4x^2+5}.....\left(2\right)$
As, $y\left(x\right)=y(-x)$, the pulse is symmetric about $y-$ axis.

Comparing $\left(1\right)$ and $\left(2\right)$, we can say that the pulse is moving towards left with a velocity $v$. So, replacing $x$ with $x+vt$ in $\left(2\right)$.
$y(x,t)=\frac{0.6}{4(x+vt{)}^2+5}.......\left(3\right)$
Rewriting $\left(1\right)$, we get
$y(x,t)=\frac{0.6}{\left[4\right(x+2t{)}^2+5]}$
Comparing the above equation with $\left(3\right)$, $v=2\text{m/s}$
So, in $2s$, the pulse will travel a distance of $x=vt=2\times 2=4\text{m}$.

Thus, option (d) is the correct alternative.