# Phase difference between two points in a stationary wave

1. Aug 3, 2015

### Janiceleong26

Q6c)

Why is the phase difference between two points in a stationary wave equals to zero?

I understand that a stationary wave is formed by two progressive waves which have the same amplitude, frequency, wavelength and speed, but travelling in opposite directions.

2. Aug 3, 2015

### Shreyas Samudra

i hope you know that - every point on the string (in case of mechanical waves) is in simple harmonic motion
so for a string in resonance frequency (except fundamental) we have a part of the string moving with phase difference of ∏ (180°) with respect to the one adjacent to it (separated by a node).

3. Aug 3, 2015

### Andrew Mason

Welcome to PF Janiceleong26!

A standing wave with a node at x=0 and wavelength $\lambda$ is described by: $y = A\sin(\omega t)\sin(2\pi x/\lambda)$. So the points on the standing wave between two nodes are all reaching their maximum and minimum at the same time. But what about points that are separated by a node?

AM

4. Aug 4, 2015

### Janiceleong26

I've not learn simple harmonic motion yet, but I understand that there is a part on the string moving antiphase to its adjacent part which are both separated by a node.
But how does that link to the phase difference being 0? The points P and Q in the picture attached are not separated by a node.

5. Aug 4, 2015

### Janiceleong26

Hi, thank you for the warm welcome.

How do the points in between nodes in a stationary wave reach their maximum and minimum at the same time? If you have a visual representation of this it'll be great. Points on a stationary wave is in antiphase with its adjacent part separated by a node. But how does that link to a phase difference of zero?

6. Aug 4, 2015

### Andrew Mason

Start with: $y = A\sin(\omega t)\sin(2\pi x/\lambda)$.

When do two points, not separated by a node and having x coordinates x1 and x2 respectively, reach their respective maxima for y? Answer: when $\sin\omega t$ is maximum ie. when $\sin\omega t$ = 1. Similarly, minimas are reached when $\sin\omega t$ = -1 This depends only on time and not on x. So they reach their respective maxima and minima at the same time. (those maxima and minima, of course, will depend on x - for a point (x1,y) the maximum value for y is $y=A\sin(2\pi x_1/\lambda)$ and for (x2,y) the maximum is $y=A\sin(2\pi x_2/\lambda)$)

If they are separated by a node, what happens?

I don't think it does. How are the signs of $\sin(2\pi x_1/\lambda)$ and $\sin(2\pi x_2/\lambda)$ related when there is a node between x1 and x2?

AM

Last edited: Aug 4, 2015
7. Aug 5, 2015

### Janiceleong26

Last edited: Aug 5, 2015
8. Aug 6, 2015

### Andrew Mason

Are the two points P and Q separated by a node?

AM

9. Aug 6, 2015

### Janiceleong26

No, they are not.

10. Aug 6, 2015

### Andrew Mason

So they reach their maxima and minima at the same time as explained in my post #6 - ie. no phase difference. If they were separated by an odd number of nodes, they would be $\pi$ radians out of phase.

AM

11. Aug 8, 2015

### Janiceleong26

I see, thanks. But the two points in the question paper have different maxima / minima.. Do they still reach their respective maxima / minima at the same time too ?

12. Aug 8, 2015

### Andrew Mason

The maxima and minima will depend on x but they will occur at the same time ie. when $\sin\omega t$ = 1 and -1. See my post #6.

AM

13. Aug 9, 2015

### Janiceleong26

Got it. Thank you very much !

14. Aug 10, 2015

### sophiecentaur

I have been searching everywhere for a simple diagram showing the way the phasors of incident and reflected wave but I have failed. What I want (for someone else!!) to show is how the incident and reflected waves, starting with the antiphase condition at the reflecting wall . They will cancel. Imagine one phasor pointing up and the other pointing down. Then, as you move away from the wall, a graph of the phasors against distance will show them each rotating at the same rate but in opposite directions, producing a resultant that always lies on a horizontal axis on the phasor diagram. So the resultant standing wave will always have the same frequency and phase, all the way along the string.
But I think there is a more fundamental reason. That is, the system is linear and cannot generate any frequencies other than the input frequency and that applies to spatial as well as temporal variations. Changing the relative phase would have to imply frequency variation and so the phase of the resultant would have to be the same all the way along the string.