# Standing wave and its formation

1. Aug 15, 2012

### pcsx22

I came across a question about wave from univeristy physics (sears n zemansky) and the question was
'can a standing wave.be produced on a string by by superposing two waves travelling in opposite direction with same frequency but different amplitudes? And same amplitude but different frequencies?

I have no idea about this question but i believe two opposite waves of same frequency but different amplitudes can form standing wave..

Last edited by a moderator: Aug 16, 2012
2. Aug 16, 2012

### Jolb

Well, let me show you some equations and maybe you can answer the question for yourself.

How do we represent a wave with wavenumber k moving to the right at a velocity c? The equation would be:
$$\sin(kx-ct)$$
This means we have a sine wave that is periodic in space with a width of one period equal to 2∏/k. However, it's moving to the right at the velocity c.

What if we superpose two waves of equal amplitude and wavenumber, one moving to the right and one moving to the left, both at a velocity c? The combined wave would be:
$$\sin(kx-ct)+\sin(kx+ct)$$
If we apply the angle addition formula:
$$\sin(kx-ct)+\sin(kx+ct)=[\sin(kx)\cos(ct)-\cos(kx)\sin(ct)]+[\sin(kx)\cos(ct)+\cos(kx)\sin(ct)]=2\sin(kx)\cos(ct)$$
Thus the sum of these two waves is just a sine wave of wavenumber k moving at velocity=0, modulated by the factor cos(ct). It is not moving, the only thing changing is its overall amplitude.

Would the same things hold if the two superposed waves had differing amplitudes or frequencies (wavenumbers)?

Last edited: Aug 16, 2012
3. Aug 16, 2012

### pcsx22

So you're saying if two opposite waves of different amplitudes and frequency superimpose, velocity of the resulting wave will not be zero therefore standing wave is not formed?? Correct me if I'm wrong

Last edited by a moderator: Aug 16, 2012
4. Aug 16, 2012

### sophiecentaur

Any reflection at an interface will produce a standing wave - it's just that the anti-nodes will not have zero amplitude because a small reflected wave will not produce perfect cancellation of the forward wave. This sort of thing is easiest to observe with EM waves on an RF feeder with a mis-matched load. The 'Voltage Standing Wave Ratio' (VSWR) gets nearer to unity as you improve the match of the load. It used to be the most convenient way of measuring the match but nowadays, the reflection coefficient is often easier to measure. (You can calculate one from the other.)

5. Aug 16, 2012

### Jolb

A standing wave, by definition, is a wave with zero velocity. For a wave to have a well-defined velocity (in the usual sense), you have to be able to write it in the form $A(t)\sin(kx-ct)$, where A(t) is a possibly time-dependent amplitude. In this case, the constant c is the velocity.

The only reason I was able to write the superposition of the two waves in the following way
$$\sin(kx-ct)+\sin(kx+ct)=[\sin(kx)\cos(ct)-\cos(kx)\sin(ct)]+[\sin(kx)\cos(ct)+\cos(kx)\sin(ct)]=2\sin(kx)\cos(ct)$$
is a lucky coincidence of the fact that there is exact cancellation between the two terms of the form $\cos(kx)\sin(ct)$. If we had waves with arbitrary amplitudes A and B and arbitrary wavenumbers j and k moving in opposite directions, then their superposition $A\sin(jx-ct)+B\sin(kx+ct)$ doesn't in general have the exact cancellation and so the wave cannot be written in the form $A(t)\sin(kx-ct)$ required to identify the wave's velocity.

Last edited: Aug 16, 2012
6. Aug 22, 2012

### alextx

7. Aug 22, 2012

### Alvydas

For you can be interesting standing wave of this type:
sin(ω*(x/(c+v) +t) + sin(ω*(x/(c-v) -t)
It is interesting because visually (if you animate) it will not look like standing wave.
But if you look at any particular x point you will find it waves with the same amplitude
like classical standing wave
sin(ω*k*(x/(c) +t) + sin(ω*k*(x/(c) -t), where k = 1/(1-v^2/c^2)