Question

Given, $y\left( {x,t} \right) = \dfrac{{0.8}}{{\left[ {{{\left( {4x + 5t} \right)}^2} + 5} \right]}}$ represents a moving pulse, where x and y are in meter and t in second. Then
This question has multiple correct options-
A. pulse is moving in +x direction
B. in 2 s, it will travel a distance of 2.5 m
C. its maximum displacement is 0.16 m
D. it is symmetric pulse

Answer 1

The direction of the wave can be found out from the general form of a wave travelling in positive or negative direction. For distance, we first need to calculate the velocity of the wave which can be obtained from the wave equation. For the maximum displacement, the denominator needs to be minimum. For the symmetry, we need equality of the function for the reversal of x-coordinate.

Complete answer:
1. For a pulse moving in positive x-direction, the general form of the function is given as
$y\left( {x,t} \right) = f\left( {x - vt} \right)$
While for a wave travelling in negative x-direction, the form of the function is given as
$y\left( {x,t} \right) = f\left( {x + vt} \right)$
We notice that for the given pulse, we have the form for the wave travelling in –x direction. So, the pulse is moving in negative x-direction.
2. In order to calculate the distance travelled by the wave, we first need to calculate the velocity of the wave. It can be calculated by using the following wave equation where v is the velocity of the wave.
$\dfrac{{{d^2}y\left( {x,t} \right)}}{{d{t^2}}} = {v^2}\dfrac{{{d^2}y\left( {x,t} \right)}}{{d{x^2}}}$
Now for the given function, we can obtain the following expressions:
\dfrac{{{d^2}y\left( {x,t} \right)}}{{d{t^2}}} = \dfrac{{{d^2}}}{{d{t^2}}}\dfrac{{0.8}}{{\left[ {{{\left( {4x + 5t} \right)}^2} + 5} \right]}} = \dfrac{d}{{dt}}\left[ {\dfrac{{ - 8\left( {4x + 5t} \right)}}{{{{\left\\{ {{{\left( {4x + 5t} \right)}^2} + 5} \right\\}}^2}}}} \right] \\\ = \dfrac{{160{{\left( {4x + 5t} \right)}^2}}}{{{{\left\\{ {{{\left( {4x + 5t} \right)}^2} + 5} \right\\}}^3}}} - \dfrac{{40}}{{{{\left\\{ {{{\left( {4x + 5t} \right)}^2} + 5} \right\\}}^3}}} \\\ = 25\left[ {\dfrac{{6.4{{\left( {4x + 5t} \right)}^2}}}{{{{\left\\{ {{{\left( {4x + 5t} \right)}^2} + 5} \right\\}}^3}}} - \dfrac{{1.6}}{{{{\left\\{ {{{\left( {4x + 5t} \right)}^2} + 5} \right\\}}^3}}}} \right] \\\ \dfrac{{{d^2}y\left( {x,t} \right)}}{{d{x^2}}} = \dfrac{{{d^2}}}{{d{x^2}}}\dfrac{{0.8}}{{\left[ {{{\left( {4x + 5t} \right)}^2} + 5} \right]}} = \dfrac{d}{{dt}}\left[ {\dfrac{{ - 6.4\left( {4x + 5t} \right)}}{{{{\left\\{ {{{\left( {4x + 5t} \right)}^2} + 5} \right\\}}^2}}}} \right] \\\ = 16\left[ {\dfrac{{6.4{{\left( {4x + 5t} \right)}^2}}}{{{{\left\\{ {{{\left( {4x + 5t} \right)}^2} + 5} \right\\}}^3}}} - \dfrac{{1.6}}{{{{\left\\{ {{{\left( {4x + 5t} \right)}^2} + 5} \right\\}}^3}}}} \right] \\\
Comparing the two double derivatives, we get
$\dfrac{{{d^2}y\left( {x,t} \right)}}{{d{t^2}}} = \dfrac{{25}}{{16}} \times \dfrac{{{d^2}y\left( {x,t} \right)}}{{d{x^2}}}$
Comparing this with the wave equation, we get
$v = \sqrt {\dfrac{{25}}{{16}}} = \dfrac{5}{4}m/s$
Now the distance travelled by the wave in two seconds is equal to the product of the velocity of the wave and the time. Therefore, we have
$d = 2 \times v = 2 \times \dfrac{5}{4} = 2.5m$
Hence, in 2 s, it will travel a distance of 2.5 m.
3. For the displacement to be maximum, the value of y needs to be maximum. For that, the value of the denominator needs to minimum and that will happen when
$4x + 5t = 0$
In that case, the value of y is given as
$y = \dfrac{{0.8}}{5} = 0.16m$
Therefore, the maximum displacement is given as 0.16 m.
4. In order to check if the given pulse is symmetric or not, we need to check the behaviour of the pulse at t = 0. For a symmetric wave, $y\left( {x,0} \right) = y\left( { - x,0} \right)$. Let us find these values for the given function. It can be done in the following way.
$y\left( {x,0} \right) = \dfrac{{0.8}}{{\left[ {{{\left( {4x} \right)}^2} + 5} \right]}} \\\ y\left( { - x,0} \right) = \dfrac{{0.8}}{{\left[ {{{\left( { - 4x} \right)}^2} + 5} \right]}} = \dfrac{{0.8}}{{\left[ {{{\left( {4x} \right)}^2} + 5} \right]}} \\\ \Rightarrow y\left( { - x,0} \right) = y\left( {x,0} \right) \\\$
Therefore, the given wave is symmetric in nature.

Note:
It should be noted that in the second part, we have another way of calculating the speed of the wave which is by using the form of the wave mentioned in the first part. For the given wave, we have $y\left( {x,t} \right) = f\left( {x + vt} \right)$ and the corresponding expression is
$y\left( {x,t} \right) = \dfrac{{0.8}}{{\left[ {{{\left( {4x + 5t} \right)}^2} + 5} \right]}} = \dfrac{{0.8}}{{\left[ {4{{\left( {x + \dfrac{5}{4}t} \right)}^2} + 5} \right]}}$
Comparing this with the general form of the equation, we get velocity $v = \dfrac{5}{4}m/s$