Question

Write down the one-dimensional differential equation of a wave and examine which of the following are possible equations of the one-dimensional wave:
\[\begin{align}
  & 1.y=2\sin x\cos ut \\
 & 2.y=5\sin 2x\cos ut? \\
\end{align}\]

Answer 1

Hint : This above question is related to the topic circular wave. A circular wave is a disruption in which energy travels from one place to another and causes some disturbance. In this question first, we will find out the one-dimensional differential equation of a wave, then we will examine the given equation whether the possible equations are of one-dimensional waves.

Complete step-by-step solution:
French scientist D’Alembert first discovered the wave equation in one dimension. Wave is the disruption in which the energy is transferred from one place to another place and causes disturbance
There are 4 types of waves such as microwaves, x-rays, radio waves, and ultraviolet waves.
Waves move in a circular motion and its energy is created by passing through water. Waves which are produced by wind usually create friction between wind and surface water. Wind when blows on the surface of water a continuous disruption takes place there which then result in wave tides. When a light wave strikes an object, they are either reflected or absorbed, that depends on the composition of the object and the speed of the light.
The one-dimensional differential equation of a wave is:
\[\dfrac{\partial {{p}^{2}}}{\partial {{x}^{2}}}-\dfrac{1}{{{c}^{2}}}\dfrac{{{\partial }^{2}}p}{\partial {{t}^{2}}}=0.......Equation\left( 1 \right)\]
Let's solve next part of the question, In this we are given with two equations and we have to check possible equation of one dimensional wave:
1. Given that
\[y=2\sin x\cos ut\]
On differentiating above equation with respect to time t we get,
\[\dfrac{dy}{dt}=-2\sin x\sin ut.....Equation\]
Now on differentiating with respect to x we get:
\[\dfrac{dy}{dx}=2\cos x\cos ut\]
Again differentiating with respect to time x we get:
\[\dfrac{{{\partial }^{2}}y}{\partial {{x}^{2}}}=-2\sin x\cos ut....Equation3\]
Again differentiating equation2 with respect to time t we get:
\[\dfrac{{{\partial }^{2}}y}{\partial {{t}^{2}}}=-2{{u}^{2}}\sin x\cos ut.....Equation4\]
From Equation 4 and Equation3 we get:
\[\dfrac{{{\partial }^{2}}y}{\partial {{t}^{2}}}={{u}^{2}}\dfrac{{{\partial }^{2}}y}{\partial {{x}^{2}}}.....Equation5\]
Equation5 is the one dimensional differential equation. Hence, the solution is possible from equation\[y=2\sin x\cos ut\].
\[2.y=5\sin 2x\cos ut\]
On differentiating above equation with respect to time t t we get:
\[\dfrac{dy}{dt}=-5\sin 2x\sin ut\]
Again on differentiating with respect to time t we get:
\[\dfrac{{{\partial }^{2}}y}{\partial {{t}^{2}}}=-5\sin 2x\cos ut......Equation1\]
Now we differentiate with respect to x we get:
\[\dfrac{dy}{dx}=10\cos 2x\cos ut\]
Again differentiating with respect to x we get:
\[\dfrac{{{\partial }^{2}}y}{\partial {{x}^{2}}}=-20\sin 2x\sin ut.....Equation2\]
Using Equation 1 and 2 we get:
\[\dfrac{{{\partial }^{2}}y}{\partial t}=\dfrac{{{u}^{^{2}}}}{4}\dfrac{{{\partial }^{2}}y}{\partial {{x}^{2}}}....Equation3\]
Equation3 is not a dimensional differential equation. Hence, the solution is not possible from equation\[y=5\sin 2x\cos ut\].

Note: The formation of the wave began million years ago, when desert areas are compacted and solidified into sandstone. The speed of sound travels through air. Light always travels through a vacuum at some million meters per second and it can also travel in any medium.