# Question about the formula of wave used and plane wave.

1. May 13, 2013

### Outrageous

y= 4sin(ωt) this is only used in SHM
y=4sin(ωt±kx) this is a travelling wave
1)the above statement correct?
2)the x is in the direction of propagation?
3)the ± mean 2 wave?
4) can please give me a simple example of plane wave that we have in our daily life.
Thank you.

2. May 13, 2013

### sophiecentaur

Is this a homework question? What have you done towards getting an answer?
Where does the "4" come from? You would normally expect a letter in its place (say A) which describes the maximum amplitude.
The +- is there to account for the wave travelling in either direction. Try plotting out the function and see how the ω and k affect the waveform in distance and in time.
There are no true plane waves but many waves, once they are a long way from their source, become spherical waves and, as the radius of the sphere increases, the wave front at any point, approximates to a plane wave. I'm sure that some examples will come to mind if you bear that in mind.

3. May 13, 2013

### Outrageous

Not homework. I tried to answer on my questions.
That is an example, eg.A=4.
I can only plot out y against time or y against x. like this
I can understand y=4sin(ωt) and y=4sin(kx) used to describe the two graphs, but how scientist can form y=4sin(ωt±kt)

4. May 13, 2013

### sophiecentaur

The function y=4sin(ωt±kx) just shows how y varies as both x and t vary. An output variable can be dependent on any number of input variables. Look at a film of a water wave and see how the same 'shape' (function of x) varies with t (changes position with time).
If you plot two graphs, one above the other (use the -kx option - for the wave going to the right on the graph). Put t=0 in the upper one and t=pi/8, say. The result will be the waveforms at two different times - showing how the wave progresses in time.

5. May 13, 2013

### Outrageous

do you mean plotting two graphs of y against x with the function of y=4sin(-kx) and the other is y=4sin(wπ/8-kx) .

6. May 13, 2013

### sophiecentaur

Those two graphs should show you the state of the wave at two different times. Sorry but I meant the time to be π/(8ω), giving you a phase value of π/8, or 22.5 degrees. (Bad time to type the wrong thing - just at a crucial time in the flow of your understanding). This shows the effect of movement of the wave, which at any particular point in space, is just oscillating at ω.

7. May 13, 2013

### Outrageous

Yup , but I wonder how scientist know they can form y=4sin(ωt±kx),just simply looking at the graphs or is there any derivation, using trigonometry?
Thank you.

8. May 13, 2013

### Outrageous

y=4sin [(2∏/λ)(x±vt)]
putting x=0, we can get y=4sin(ωt)
putting t=0, we can get y=4sin(kx)
the two formula combine to describe the wave at any time and place.
Correct?

9. May 13, 2013

### ZombieFeynman

Scientists know this because any function of the form f(kx-wt) satisfies by the wave equation for velocities of w/k. Plug it in for yourself and see. A way of thinking about such an equation of two variables is to hold say t fixed snd plot for many instances of x and then hold x fixed and plot many instances of t.

10. May 14, 2013

### Outrageous

What do you mean?
from any function , we will get v=w/k?
plug in what ?

11. May 14, 2013

### ZombieFeynman

Find the wave equation. A good place to start would be the wikipedia article on it. Other places that will have it are books on Optics (Perhaps Hecht) or Oscillations and Waves (say by A.P. French).

Then take any function of the form f(kx+wt) or f(kx-wt) and you will see that they are solutions of the one dimensional wave equation, for arbitrary f.

12. May 14, 2013

### Outrageous

Thank you.
One more to ask, we have wave equation first , then only we derive equation like y=4sin(wt±kx) in order to satisfy the wave equation. correct?

13. May 14, 2013

### ZombieFeynman

No, this is not gereally true. By making a substitution of variables, you can derive that any function of the form I mentioned is a solution.

14. May 15, 2013

Thanks