# Sine Wave Addition

1. Mar 7, 2010

### marla11

If two sine waves have the same frequency and amplitude but have different phase shift do they still produce a standing wave?
Thanks for the help.

2. Mar 7, 2010

3. Mar 7, 2010

### marla11

As far as i can see it doesn't say anything about phase shift so does that mean it doesn't affect anything?

4. Mar 7, 2010

### Naty1

wt is the phase shift

see http://en.wikipedia.org/wiki/Phase_(waves [Broken])

Last edited by a moderator: May 4, 2017
5. Mar 7, 2010

### marla11

I'm sorry but I'm still confused. If y1 = Asin(kx-wt+phi) and y2 = Asin(kx+wt)
the addition is y= 2Acos(wt-phi/2)sin(kx+phi/2) where phi is the phase shift between 0 and 2pi. Does this still fit the standing wave equation y=(2Asin(kx))cos(wt) meaning its a standing wave or does the difference in phase shift mean they do not create a standing wave?

6. Mar 7, 2010

### Bob S

You have one forward-traveling wave (wt-kx) and one backward wave (wt+kx) of the same amplitude, which is a standing wave. My CRC Math Tables (10th Ed, 1954) on page 345 shows the sum

sin(x) + sin(y) = 2·sin[(x+y)/2]·cos[(x-y)/2]

Bob S

7. Mar 9, 2010

### meichenl

Marla,

Your equations will be easier to read if you typeset them in LaTeX.

Yes, the equation you give is a standing wave. If you start with

$\Psi(x,t) = A\cos(\omega \left[t-t_0\right]) \sin (k\left[x-x_0\right])$

you can just define a new time coordinate and new space coordinate by

$t' = t - t_0$
$x' = x - x_0$.

Then your original equation is just

$\Psi(x',t') = A \cos(\omega t')\sin(k x')$,

showing that the waveform is exactly the same as the standing wave you're used to.