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DeIdeal

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This thing is driving me mad, I thought I figured it out already, but it seems I was wrong. Any help would be appreciated.

"Under what conditions does the sum of two sinusoidal waves also satisfy the wave equation?"

The sum wave is

[itex]D(x,t) = A_{1}sin(k_{1} x-\omega_{1} t)+A_{2}sin(k_{2} x-\omega_{2} t)[/itex]

The (1D) wave equation

[itex]\frac{\partial^{2}D}{\partial x^{2}}=\frac{1}{c²} \frac {\partial^{2}D}{\partial t^{2}}[/itex]

Not much, but derivating both sides:

[itex]\frac{\partial^{2}}{\partial x^{2}}(A_{1}sin(k_{1} x-\omega_{1} t)+A_{2}sin(k_{2} x-\omega_{2} t))[/itex]

[itex]=-A_{1}k_{1}^{2} sin(k_{1} x-\omega_{1} t)-A_{2}k_{2}^{2}sin(k_{2} x-\omega_{2} t)[/itex]

[itex]\frac{1}{c^{2}} \frac {\partial^{2}}{\partial t^{2}}( A_{1}sin(k_{1} x-\omega_{1} t)+A_{2}sin(k_{2} x-\omega_{2} t))[/itex]

[itex]=-A_{1}\frac{\omega_{1}^{2}}{c^{2}} sin(k_{1} x-\omega_{1} t)-A_{2}\frac{\omega_{2}^{2}}{c^{2}}sin(k_{2} x-\omega_{2} t)[/itex]

And rearranging gives:

[itex]A_{1}(k_{1}-\frac{\omega_{1}^{2}}{c^{2}}) sin(k_{1} x-\omega_{1} t)=A_{2}(\frac{\omega_{2}^{2}}{c^{2}}-k_{2}^{2})sin(k_{2} x-\omega_{2} t)[/itex]

But to be honest I've got little else I'm able to do after this. I don't think I've ever had to solve something like that. Is there some sort of an obvious trigonometric identity or something I'm missing here?

The answer should be that the velocities of the waves, ie [itex]\frac{\omega}{k}[/itex] are the same (which seems to lead to the velocity of the sum wave being that same velocity).

## Homework Statement

"Under what conditions does the sum of two sinusoidal waves also satisfy the wave equation?"

The sum wave is

[itex]D(x,t) = A_{1}sin(k_{1} x-\omega_{1} t)+A_{2}sin(k_{2} x-\omega_{2} t)[/itex]

## Homework Equations

The (1D) wave equation

[itex]\frac{\partial^{2}D}{\partial x^{2}}=\frac{1}{c²} \frac {\partial^{2}D}{\partial t^{2}}[/itex]

## The Attempt at a Solution

Not much, but derivating both sides:

[itex]\frac{\partial^{2}}{\partial x^{2}}(A_{1}sin(k_{1} x-\omega_{1} t)+A_{2}sin(k_{2} x-\omega_{2} t))[/itex]

[itex]=-A_{1}k_{1}^{2} sin(k_{1} x-\omega_{1} t)-A_{2}k_{2}^{2}sin(k_{2} x-\omega_{2} t)[/itex]

[itex]\frac{1}{c^{2}} \frac {\partial^{2}}{\partial t^{2}}( A_{1}sin(k_{1} x-\omega_{1} t)+A_{2}sin(k_{2} x-\omega_{2} t))[/itex]

[itex]=-A_{1}\frac{\omega_{1}^{2}}{c^{2}} sin(k_{1} x-\omega_{1} t)-A_{2}\frac{\omega_{2}^{2}}{c^{2}}sin(k_{2} x-\omega_{2} t)[/itex]

And rearranging gives:

[itex]A_{1}(k_{1}-\frac{\omega_{1}^{2}}{c^{2}}) sin(k_{1} x-\omega_{1} t)=A_{2}(\frac{\omega_{2}^{2}}{c^{2}}-k_{2}^{2})sin(k_{2} x-\omega_{2} t)[/itex]

But to be honest I've got little else I'm able to do after this. I don't think I've ever had to solve something like that. Is there some sort of an obvious trigonometric identity or something I'm missing here?

The answer should be that the velocities of the waves, ie [itex]\frac{\omega}{k}[/itex] are the same (which seems to lead to the velocity of the sum wave being that same velocity).

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