Show that y(x,t)=y1(x,t)+y2(x,t) is a solution to the wave equation

  • Thread starter Thread starter Duderonimous
  • Start date Start date
  • Tags Tags
    Wave Wave equation
Click For Summary

Homework Help Overview

The discussion revolves around demonstrating that the sum of two wave functions, \(y(x,t) = y_1(x,t) + y_2(x,t)\), where \(y_1\) and \(y_2\) are solutions to the wave equation, is also a solution. The subject area is wave equations in physics.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the mathematical implications of the principle of superposition and how to manipulate the wave equation. Some express confusion regarding the use of trigonometric identities, while others suggest focusing on the linearity of differentiation.

Discussion Status

There is an ongoing exploration of how to apply the principle of superposition to show that the sum of the two solutions satisfies the wave equation. Some participants have provided guidance on manipulating the equation without using trigonometric identities, while others are seeking clarification on the steps needed to reach the conclusion.

Contextual Notes

Participants are working under the assumption that both \(y_1\) and \(y_2\) are known solutions to the wave equation, and they are trying to establish the same for their sum. There is a mention of potential complications arising from trigonometric identities.

Duderonimous
Messages
63
Reaction score
1

Homework Statement


Let y[itex]_{1}[/itex](x,t)=Acos(k[itex]_{1}[/itex]x-ω[itex]_{1}[/itex]t) and y[itex]_{2}[/itex](x,t)=Acos(k[itex]_{2}[/itex]-ω[itex]_{2}[/itex]t) be two solutions to the wave equation
[itex]\frac{∂^{2}y}{∂x^{2}}[/itex]=[itex]\frac{1}{v^{2}}[/itex][itex]\frac{∂^{2}y}{∂t^{2}}[/itex]
for the same v. Show that y(x,t)=y[itex]_{1}[/itex](x,t)+y[itex]_{2}[/itex](x,t) is also a solution to the wave equation.

Homework Equations


Equations given in problem statement and

cos(a-b)=cosacosb+sinasinb

The Attempt at a Solution



I am pretty sure I put y[itex]_{1}[/itex](x,t)+y[itex]_{2}[/itex](x,t) in a form that does not involve addition like 2Acosasinb or something close to that. Then I take a second partial derivative with respect to both x and t and show that the ratio of ∂[itex]^{2}[/itex]y/∂t[itex]^{2}[/itex] over ∂[itex]^{2}[/itex]2/∂x[itex]^{2}[/itex] is equal to v[itex]^{2}[/itex].
I am hung up on the trig.

y(x,t)=y[itex]_{1}[/itex](x,t)+y[itex]_{2}[/itex](x,t)

y(x,t)=Acos(k[itex]_{1}[/itex]x-ω[itex]_{1}[/itex]t)+Acos(k[itex]_{2}[/itex]-ω[itex]_{2}[/itex]t)

y(x,t)=A(cosk[itex]_{1}[/itex]xcosω[itex]_{1}[/itex]t+sink[itex]_{1}[/itex]xsinω[itex]_{1}[/itex]t)+A(cosk[itex]_{2}[/itex]xcosω[itex]_{2}[/itex]t+sink[itex]_{2}[/itex]xsinω[itex]_{2}[/itex]t)

Cant figure out where to go from here. Please help! Thanks.
 
Physics news on Phys.org
You're making it too complicated. Don't use any trig identities. You know that y1 and y2 are solutions. What does that mean mathematically?
 
That their sum should also be a solution by the principle of superposition?
 
Well, that's what you're supposed to show. :wink:

When you assume ##y_i## is a solution, you can say that
$$\frac{\partial^2}{\partial x^2}y_i(x,y) = \frac{1}{v^2}\frac{\partial^2}{\partial t^2} y_i(x,t),$$ where ##\frac{\omega_i}{k_i} = v##.

So now plug ##y=y_1+y_2## into the lefthand side of the wave equation. You get
$$\frac{\partial^2}{\partial x^2}y = \frac{\partial^2}{\partial x^2}(y_1+y_2)$$ which you can split into two terms because differentiation is a linear operation. What you want to do is keep manipulating the righthand side until you end up with
$$\frac{1}{v^2} \frac{\partial^2}{\partial t^2}y.$$
 
[itex]\frac{\partial^2}{\partial x^2}y = \frac{\partial^2}{\partial x^2}(y_1+y_2)[/itex]
[itex]\frac{\partial^2}{\partial x^2}y = \frac{\partial^2 y_1}{\partial x^2} + \frac{\partial^2 y_2}{\partial x^2}[/itex]
[itex]\frac{\partial^2}{\partial x^2}y = \frac{1}{v^2}\frac{\partial^2 y_1}{\partial t^2}+\frac{1}{v^2}\frac{\partial^2 y_1}{\partial t^2}[/itex]
[itex]\frac{\partial^2}{\partial x^2}y = \frac{1}{v^2}(\frac{\partial^2 y_1}{\partial t^2}+\frac{\partial^2 y_2}{\partial t^2})[/itex]
[itex]\frac{\partial^2}{\partial x^2}y= \frac{1}{v^2}\frac{\partial^2}{\partial t^2}(y_1+y_2)[/itex]
[itex]\frac{\partial^2}{\partial x^2}y = \frac{1}{v^2}\frac{\partial^2}{\partial t^2}y[/itex]

Is that all I have to show since I already know that y is a solution to the wave equation and that it is assumed that is obeys that principle of superposition? Thanks again for your time.
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
2K
Replies
1
Views
1K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 20 ·
Replies
20
Views
3K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 8 ·
Replies
8
Views
2K
Replies
32
Views
3K
Replies
2
Views
2K
Replies
3
Views
1K
  • · Replies 2 ·
Replies
2
Views
1K