0
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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?

enter image description here

$\endgroup$
0
$\begingroup$

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}

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.