Question

$y(x,t)=0.8/\left[\right(4x+5t{)}^2+5]$ represents a moving pulse, where $x$ and $y$ are in meter and $t$ in second. Then

• A. pulse is moving in $+x$ direction
• B. in $2s$ it will travel a distance of $2.5m$
• C. its maximum displacement is $0.16m$
• D. it is symmetric pulse

Answer 1

Answer: (B), (C)

The correct options are
B in $2s$ it will travel a distance of $2.5m$
C its maximum displacement is $0.16m$

The maximum displacement is the maximum value of $y(x,t)$

$y(x,t)=\frac{0.8}{{(4x+5t)}^2+5}$

For $y$ to be maximum, denominator of the expression must be minimum. Which means when $(4x+5t)=0$

then

$y=\frac{0.8}{5}=0.16m$

Option $C$ is correct

Now we know that wave equation is given by

$\frac{d^{2}y(x,t)}{d{t}^2}=c^2\frac{d^{2}y(x,t)}{d{x}^2}$

now using differentiation we get

$\frac{d^{2}y(x,t)}{d{t}^2}=\frac{160{(4x+5t)}^2}{{\{{(4x+5t)}^2+5\}}^3}-\frac{40}{{\{{(4x+5t)}^2+5\}}^3}=25[\frac{6.4{(4x+5t)}^2}{{\{{(4x+5t)}^2+5\}}^3}-\frac{1.6}{{\{{(4x+5t)}^2+5\}}^3}]$

and

$\frac{d^{2}y(x,t)}{d{x}^2}=16[\frac{6.4{(4x+5t)}^2}{{\{{(4x+5t)}^2+5\}}^3}-\frac{1.6}{{\{{(4x+5t)}^2+5\}}^3}]\therefore C=\sqrt{\frac{25}{16}}=\frac{5}{4}m/sec$

So in $2$ seconds it will travel distance $=2C=2.5m$

Since $f(-x,t)\ne f(x,t)$ the pulse is not symmetric