# B Doubts regarding equation of waves

1. Mar 22, 2016

### Dexter Neutron

The wave equation is
$$y = A\sin (\omega t \pm kx \pm \phi_i)$$
What exactly is kx?
Why it is needed? Why can't we directly represent wave as we represent SHM -
$$y = A\sin (\omega t + \phi)$$
I even read that between the waves there is a phase difference of π -
$$y = A\sin (\omega t - kx)$$ and $$y = A\sin (kx - \omega t)$$
What is the difference between these two waves and how do they calculated the phase difference between these two waves?
Can someone please show me the derivation of the wave equation?

Now regarding frequency what I know is that freqeuncy is the number of oscillations or cycles per second.
If a wave is travelling in a string then does one cycle means motion of wave to and fro the string i.e. if the wave starts from one end reflects from other and again reaches first then could this be called as one cycle?

Last edited: Mar 22, 2016
2. Mar 22, 2016

### Staff: Mentor

If every point in space has the same phase, how can a wave propagate? What would propagate in which direction?
The definition of "phase of 0" is arbitrary. Use one and be consistent.

3. Mar 22, 2016

### Dexter Neutron

I can't understand what you are saying? Sorry but have you read the whole question?

4. Mar 22, 2016

### PeroK

The first equation represents a whole line of points, each oscillating up and down with SHM. $x$ represents the coordinate of a point on the line and $y$ represents the displacement of that point (up or down) at time $t$. The wavelength is $2 \pi /k$.

The second represents the motion of a single point, oscillating up and down with SHM. $y$ represents the displacement of that point at time $t$.

5. Mar 22, 2016

### Dexter Neutron

Thank You. You clarified my concepts about wave and SHM. Can you please answer my other questions?

6. Mar 22, 2016

### PeroK

$sin$ is an odd function, so:

$sin (kx - \omega t) = - sin (\omega t - kx) = sin( \omega t - kx + \pi)$

At each time, one wave is just the other upside down, or shifted by $\pi$, whichever you prefer.

If you choose any point on the line $x = x_0$, say. Then:

$y = Asin(\omega t - kx_0) = Asin( \omega t - \phi)$, where $\phi = kx_0$

The point $x_0$, therefore, is oscillating with frequency $\omega/ 2\pi$. And, in fact, every point is oscillating at the same frequency, but they are all continuously out of phase with each other. You have to move $2\pi/k$ along the line to find the next point in phase. I.e. $x_0 + 2\pi/k$ is completely in phase with $x_0$. Hence the wavelength.

You should try drawing the graph of

$y = Asin(\omega t - kx)$

For $t = 0$, $t =$ something small, $t = \pi/ \omega$ and $t = 2\pi/ \omega$. Not least to convince yourself that:

$y = Asin(\omega t - kx)$ is a wave moving to the right, while:

$y = Asin(\omega t + kx)$ is a wave moving to the left.

Or, is it the other way round?

7. Mar 22, 2016

### Dexter Neutron

Thank You ,it was extremely helpful. This clarified the picture of a single particle and a wave in my mind.

8. Mar 22, 2016

### sophiecentaur

That's just what i had planned to suggest. It's not straightforward to do it on a piece of paper if you want to include the time variation but, is you are familiar with Excel (other spreadsheets are available) it is quite convenient to plot a series of graphs, one under the other, to show how snapshots of the wave, in time, move right or left. If you really want to get posh, you can show a 3D representation with not more more effort.
k is the constant of proportionality of phase against distance. For a given value of t, it shows a snapshot of the profile of the wave. It 'relates' distance to the phase in terms of wavelength.

9. Mar 22, 2016

### pixel

If you take a snapshot of the wave at a specific time t, there is a periodic variation of y as a function of x. That is given by the kx term, where k = 2π/wavelength.

If you take a snapshot of the wave at a specific point x, there is a periodic variation of y as a function of t. That is given by the ωt term.

So you need both terms to fully describe the wave.