0
$\begingroup$

Te most general form of a (one dimension) standing wave is $$(A \mathrm{cos}(\omega t)+B \mathrm{sin}(\omega t))(C \mathrm{cos}(k x)+D \mathrm{sin}(k x))=G\mathrm{cos}(\omega t+\phi_1)\mathrm{cos}(kz+\phi_2)\tag{1}$$

Which can be written with exponential notation as

$$G\mathrm{cos}(kz+\phi_2)e^{i(\omega t +\phi_1)}\tag{2}$$

As any standing wave $(2)$ should be the superposition of a progressive and a regressive wave. Nevertheless on textbooks it is usually shown how to get a standing wave like

$$F\mathrm{sin}(\omega t-k x)+F \mathrm{sin}(\omega t+k x)=2F\mathrm{sin}(\omega t)\mathrm{cos}(kz)\tag{3}$$ Or in exponential notation $$F e^{i(\omega t-k x)}+F e^{i(\omega t+k x)}=2F\mathrm{cos}(kz)e^{i\omega t }\tag{4}$$

Which is quite particular, as here $\phi_1=\phi_2=0$.

So what are the progressive and regressive waves (in exponential form) to combine as in $(4)$ but to get an expression as the one in $(2)$ (i.e. where $\phi_1$ and $\phi_2$ are not necessarily zero)?

|cite|improve this question|||||
$\endgroup$
  • 1
    $\begingroup$ Have you played about with putting phase constants into the 'original' progressive waves to see what happens? $\endgroup$ – Philip Wood Sep 8 '17 at 15:51
  • $\begingroup$ @PhilipWood Thanks for the comment, I tried to combine a wave with no phase and a wave with a phase $\phi$ but I do not get the expression I'm looking for: For example I can't write $F e^{i(\omega t-k x)}+F e^{i(\omega t+k x+\phi)}$ as something like $2F\mathrm{cos}(kz+\phi_1 )e^{i\omega t +\phi_2}$ since imaginary part do not cancel out. Is the procedure I'm following the correct one or am I missing something here? $\endgroup$ – Sørën Sep 9 '17 at 1:13
  • $\begingroup$ You seem to have two issues. The one about phase constants I've addressed in my answer. Another seems to be about the use of complex numbers. On this subject note that your eq (1) $can't$ be written as eq (2), without prefacing the rhs of eq (2) with "the real part of". You probably meant this, but just in case... $\endgroup$ – Philip Wood Sep 10 '17 at 17:13
1
$\begingroup$

First note that all we have to do to get $\text{cos} (\omega t + \phi_1)\ \text{cos} (kx+\phi_2)$ from $\text{cos}\ \omega t\ \text{cos}\ kx$ is to measure x from a different origin and t from a different zero of time.

But if we superimpose progressive waves with different phase constants we get $$\text{cos} (\omega t -kx+ \phi_1)\ + \text{cos} (\omega t+kx+\phi_2)= 2\text{cos}\ (\omega t+\frac{\phi_{2}+\phi_{1}}{2})\ \text{cos}\ (kx+\frac{\phi_{2}-\phi_{1}}{2}).$$ So there are your phase constants: $(\frac{\phi_{2}+\phi_{1}}{2})$ and $(\frac{\phi_{2}-\phi_{1}}{2})$.

|cite|improve this answer|||||
$\endgroup$

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.