# I Satisfying the wave equation with 2 different speeds

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1. Apr 8, 2016

### zazzblam

Hello,
I have a wave of the form

y = Asin(x-vt) + Asin(x+2vt)

which I substituted into the wave equation to find out if it satisfies it. It didn't because of the speed of the left travelling wave being equal to 2v. What I got was:

A[-sin(x-vt)-sin(x-2vt)] = 1/v2 * A[-v2sin(x-vt) - 4v2sin(z-2vt)]

So the v2 terms cancel on the RHS but we're still left with factor of 4 which mean they aren't equal.
According to the superposition principle, if two individual waves satisfy the wave equation then their sum should also satisfy it. Individually they do but together they don't. I'm trying to think of why, physically, this wave wouldn't satisfy the wave equation as putting the equation into Desmos creates what looks like a perfectly normal wave travelling in the negative x direction as t>0.
Any help would be awesome.

2. Apr 8, 2016

### drvrm

i am just reacting in applying common sense-
your waves are having different speeds -after superposition their effect on the medium will be time dilated so the net effect will be a disturbance but it has to be checked whether it obeys the usual wave equation,
you are reporting that the sum of the two solutions do not obey the wave equation.

if one considers the difference of speeds to be effected by change in frequency then we have examples of a variation of intensity of the resultant wave as observed in 'phenomenon of beats' being observed . i do not know whether beats travel with average speed or not?
so one can draw the waves superpose it and draw the resultant nature of disturbance.

3. Apr 8, 2016

### nasu

What wave equation do you think that they both satisfy (individually)?

4. Apr 8, 2016

### Khashishi

It looks like you are trying to add together two solutions to two different wave equations (one with speed v and one with speed 2v). Try adding two solutions to the same wave equation.

5. Apr 8, 2016

### zazzblam

d2y/dx2 = 1/v2 d2y / dt2

6. Apr 8, 2016

### zazzblam

The solution that I posted is substituting the actual form I was given (y = Asin(x-vt) + Asin(x+2vt)) into the wave equation, which doesn't satisfy it. But individually, y = Asin(x-vt) and y = Asin(x+2vt) satisfy the wave equation, so I'm not sure why their sum doesn't.

7. Apr 8, 2016

### zazzblam

So the wave of the form y = Asin(x-vt) + Asin(x+2vt) produces a beat? Does anyone know if beats satisfy the wave equation? I can't seem to find a definite answer online.

8. Apr 8, 2016

### nasu

Are you sure that the second wave satisfies the equation you posted above? Did you actually tried it?

9. Apr 8, 2016

### drvrm

the waves move with different speeds -so try to imagine the net intensity as observed by a static observer - if he finds intensity to vary with certain frequency then get the speed of the resultant intensity variation -why not draw the waves and superpose them.

10. Apr 8, 2016

### zazzblam

I put the two separate waves into Desmos with a t slide and it seems like I get temporal beats that appear and disappear rather than travel. So is it safe to assume that since the sum of the waves with different speeds v and 2v do not satisfy the wave equation and the physical result is beats, then beats don't satisfy the wave equation?

11. Apr 8, 2016

### nasu

The second wave does not satisfy the wave equation you wrote. So why would you expect the sum to do it?

Beats are between waves with slightly different frequencies and not different velocities.

12. Apr 8, 2016

### drvrm

You are getting the temporal occurrence of beats so it must be satisfying some wave equation with the frequency of beats

13. Apr 8, 2016

### zazzblam

The second wave has velocity 2v, so the wave equation is of the form d2y/dx2 = 1/v2 d2y/dt2 and the wave equation is satisfied only when v = 2v as the factor of 4v2 on the RHS will cancel.

I thought the wave speed was related to the frequency through v = f/lambda, so if the second wave has 2v=f/lambda then wouldn't the frequencies also be different?

14. Apr 9, 2016

### nasu

Just try to take the derivatives. It's not hard. It does not satisfy the equation as written. The second wave satisfies a a different equation.

The equation you wrote has non-dispersive solutions. All the solutions represents waves with velocity v. They may have different frequencies and wavelengths but the ratio ω/k=v is always the same.
The second waves satisfies a different equation, one whose solutions are waves with velocity 2v.

15. Apr 10, 2016

### zazzblam

Yes, I know. I had already taken the derivatives of the two individual waves and found they satisfy two different wave equations with velocities v and 2v. I understand now why mathematically the sum wouldn't satisfy the wave equation. I still don't know why the resulting wave from summing the two individual waves y = Asin(x-vt) + Asin(x+2vt) doesn't physically satisfy the wave equation. What is it about having two different velocities that physically causes it to fail?

16. Apr 10, 2016

### nasu

It looks like you almost answered yourself.
Maybe the sum of the two waves is just a mathematical object, not a physical wave?

For non-dispersive waves model, the speed of the wave can change only if it propagates in a different medium.
So your two equations represent waves propagating in two different media. You cannot physically overlap them and get a physical wave, propagating in some medium.

17. Apr 10, 2016

### zazzblam

Thank you for all your help, that makes a lot of sense

18. Apr 11, 2016

### mfig

Just for clarity, the constant v in the wave equation:

$v^2\frac{d^2y}{dx^2}=\frac{d^2y}{dt^2}$

is the link between the mathematics and the physics. The equation can be solved mathematically for arbitrary v, but any manifestation of wave phenomena in a physical system will yield a numerical value for v. In physical systems described by the wave equation, the numerical value of v will be a function of the properties of the system. Thus, if you have two different values for v you either have two different materials or the same material in two different states.