Let's say I have two arbitrary mechanical waves $y_1$ and $y_2$ propagating on a string in the same direction.

The waves $y_1$ and $y_2$ differ in phase by an arbitrary angle $\phi$ and the resultant wave is given by the sum of these two waves.

Given this information, how can we find the amplitude of the resultant wave?

Given a problem of this nature this is what I would think of doing :

$$y(x,t) = y_1(x,t) + y_2(x, t)$$ $$A_{\ res}cos(kx-\omega t) = A_1cos(kx-\omega t) + A_2cos(kx-\omega t -\phi)$$ $$\text{Setting x=0. t=0 we get:}$$ $$A_{\ res}\ cos(0) = A_1cos(0) + A_2cos(0- \phi)$$ $$\implies A_{\ res}= A_1 + A_2cos(\phi)$$

But this is wrong.

It seems to me that my error is setting $x=0$ and $t=0$, but I'm not sure why that would be wrong as $A_{res}$ should be constant $\forall\ x, t \in \mathbb{R}^+$ (for all values of $x$ and $t$, where $t \geq 0$).

If $A_{res}$ is constant, then no matter what value of $x$ and $t$ I substitute, I should get the same $A_{res}$, subbing in $x=0$ and $t=0$, helps eliminate the unnecessary arguments of the trigonometric functions from the equation, and allows me to solve for $A_{res}.$

I have two questions here :

Q1 : Why is setting $x=0$ and $t=0$, and solving for $A_{res}$ mathematically wrong?

Q2: How would you solve for the amplitude of the resultant wave?

  • $\begingroup$ How do you know your expression for the resulting amplitude is wrong? $\endgroup$ – M. Enns May 8 '16 at 13:17
  • $\begingroup$ @M. Ennds, I've used this derivation for the resulting Amplitude on a few example problems (from Fundamentals of Physics), and found that my answers using $A_{res} = A_1 + A_2cos(\phi)$ were incorrect. $\endgroup$ – Perturbative May 8 '16 at 13:21

Your error is writing this expression

$$A_{\text{ res}}\cos(kx-\omega t) = A_1\cos(kx-\omega t) + A_2\cos(kx-\omega t -\phi)$$

It should be

$$A_{\text{ res}}\cos(kx-\omega t-\psi) = A_1\cos(kx-\omega t) + A_2\cos(kx-\omega t -\phi)$$ Note the resultant is not in phase with $A_1\cos(kx-\omega t)$

I think that a simple way of doing the addition is to draw a phasor diagram and then use the cosine and sine rule?

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What is wrong is to assume that the sum will read $A_{res}\cos(\omega t - kx)$ while in fact the sum should read in general $A_{res}\cos(\omega t - k x + \phi_{res} )$.

The strategy consists in starting from the sought solution $y_{res}= A_{res}\cos(\omega t - k x + \phi_{res})$ and expand the cosine function $y_{res} = A_{res}[\cos(\omega t - k x)\cos \phi_{res} -\sin(\omega t -k x )\sin \phi_{res}]$.

Next, you need to write the sum of waves in exactly the same form (by using the identity $\cos(a+b) = \cos a \cos b - \sin a \sin b$):

\begin{equation} y_1+y_2 = A_1 \cos(\omega t - k x) + A_2 \cos(\omega t -k x)\cos \phi - A_2 \sin(\omega t - k x)\sin \phi \end{equation}

We deduce from it that

\begin{equation} A_{res}\cos \phi_{res} = A_1+A_2 \cos \phi; \: A_{res}\sin \phi_{res} = A_2 \sin \phi \end{equation}

By squaring and adding each term we get that

\begin{equation} A_{res}^2 = (A_1+A_2\cos \phi)^2+A_2^2 \implies \boxed{A_{res} = \pm \sqrt{(A_1+A_2\cos \phi)^2+A_2^2}} \end{equation} and by taking the ratio of the two equations we find that:

\begin{equation} \boxed{\tan \phi_{res} = \frac{A_2\sin \phi}{A_1+A_2 \cos \phi}} \end{equation}


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