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We have two waves

$\psi_1(x) = a_1\mathrm{cos}(kx-\omega t + \phi_1)$

$\psi_2(x) = a_2\mathrm{cos}(kx-\omega t + \phi_2)$

We know that the superimposed wave can be written as

$\psi(x) = \psi_1(x) + \psi_2(x)$

How can I derive from this that the amplitude of the superimposed wave $\psi(x)$ is $ \sqrt{a_1^2 + a_2^2 + 2a_1a_2 \mathrm{cos}(\phi_1-\phi_2 )} $

I'm pretty sure it has something to do with the relation that when you have

$\psi(x,t)=A\mathrm{cos}(kx+ ωt + \phi) = B\mathrm{cos}(kx+ ωt) - C\mathrm{sin}(kx+ ωt)$

you can derive that

${A}^{2} = {B}^{2} + {C}^{2}$;

and

$\mathrm{cos}(\phi) = \frac{B}{A}, \mathrm{sin}(\phi) = \frac{C}{A}$

But I can't figure out how to do it...

hm8
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1 Answers1

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I'm setting $kx - \omega t = \theta$, for ease of notation.

Let's take your idea: $A \cos ( \theta + \phi ) = A\cos \phi \cos \theta - A\sin \phi \sin \theta$. Adding the two equations gives:

$(\psi_1 + \psi_2)(\theta) = (a_1\cos \phi_1 + a_2 \cos \phi_2)\cos \theta - (a_1\sin \phi_1 + a_2 \sin \phi_2)\sin \theta$ $ \Rightarrow (\psi_1 + \psi_2)(\theta) = A \cos (\theta + \Phi )$

with $A^2 = (a_1\cos \phi_1 + a_2 \cos \phi_2)^2 + (a_1\sin \phi_1 + a_2 \sin \phi_2)^2$

and $\tan \Phi = \frac{-(a_1\sin \phi_1 + a_2 \sin \phi_2)}{(a_1\cos \phi_1 + a_2 \cos \phi_2)}$

Can you solve the above out to get the new amplitude?

Also, in case you're familiar with complex numbers, a lot of the algebra gets easier with the phasor representation .

stochasticboy321
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