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

  1. Aug 28, 2012 #1
    1. The problem statement, all variables and given/known data
    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.


    2. Relevant equations
    Equations given in problem statement and

    cos(a-b)=cosacosb+sinasinb


    3. 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.
     
  2. jcsd
  3. Aug 28, 2012 #2

    vela

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    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?
     
  4. Aug 28, 2012 #3
    That their sum should also be a solution by the principle of superposition?
     
  5. Aug 28, 2012 #4

    vela

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    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
    $$\frac{1}{v^2} \frac{\partial^2}{\partial t^2}y.$$
     
  6. Aug 29, 2012 #5
    [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.
     
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