# Simple doubt in superposition of waves

1. May 4, 2014

### Tanya Sharma

1. The problem statement, all variables and given/known data

While analyzing superposition of waves from two coherent sources ,the displacement from the two sources are written as y1=A1cos(kx-ωt) and y2=A2cos(kx-ωt+∅) .The resultant displacement at any point is given by y=y1+y2 .

I am having doubt that why does argument of sine in both y1 and y2 contain the same variable ‘x’ . Since ‘x’ represents the distance of the point from the originating source which may be different in the two waves,the displacements should be written as y1=A1cos(kx1-ωt) and y2=A2cos(kx2-ωt+∅) .But this is not the case .

I am pretty sure I am missing something very obvious .I would be grateful if somebody could help me understand this.

Last edited: May 4, 2014
2. May 4, 2014

### TSny

I suspect that x denotes the coordinate of a point on the x axis. You want to add the two waves at the same point, so x is the same for both functions. It might have been clearer if they had said that the resultant displacement at any point x is given by y1 + y2.

The phase angle ø represents the net phase shift between the waves due to any intrinsic phase difference of the sources and the phase shift due to the sources being located at different points.

3. May 5, 2014

### BvU

Hello Tanya,

You are quite right. For analysis purposes, however, usually some coordinate system is chosen (and, not to forget, some t=0 as well, since the sources don't have to be switched on at the same time either -- or they can be otherwise out of phase) and everything is lumped into this one ∅ (nice symbol; I'm more used to $\phi$).

Mind you, it's all math.

Must admit I don't completely understand the change of sign you introduce by "adjusting the phase angle": if y(0,t) = Asinωt and the wave travels to the right (I read myself establishing a +x axis that way) y(x,t) = Asin(ωt-kx ). Period. y=Asin(kx-ωt) describes a wave travelling to the left, which is not at all just a phase change.

Draw a sine on a transparency and play around with it !

4. May 5, 2014

### ian_dsouza

It's been a while since I've worked with waves. So, take this with a pinch of salt.

Assuming the sources point along the same direction, say towards the right:

The $y_{1}$ equation does not contain a phase term. I assume this must mean that if the $y_{1}$ source emitted a 'nice' cosine wave that started at the maximum amplitude at the source at t=0, the origin (of measurement of x) would be placed at the source itself or at an integral multiple of the wavelenth λ, to the left of that source, ie, at a distance nλ to the left. 'n' is an integer.

If the cosine wave did not begin at the maximum amplitude at the source at t=0 and instead had an initial phase $ø_{1}$, the origin would be placed at a distance of nλ+$d_{1}$ to the left of the source, where $d_{1}$ is the distance traversed by the wave corresponding to a phase $ø_{1}$. Hence, the absence of the phase term in the $y_{1}$ equation.

Coming to the $y_{2}$ equation, you still measure from the same origin you utilized for the first source. The ø term in the $y_{2}$ equation is a known constant specific to the given setup of this problem. It takes into account the phase difference between the waves emitted by the sources at any given instant of time (measured at each source) as well as the distance between the two sources. The distance between the two sources is important information because even if you time synchronize the waves from the two sources, you can vary the phase difference of the waves at a fixed arbitrary point to the right of the two waves just by varying the distance between the two sources.

Hope this helps!

5. May 5, 2014

### Tanya Sharma

Hello BvU

Are you sure ?

I think both y=Asin(kx-ωt) and y=Asin(ωt-kx) are travelling towards right with a phase difference of 180° .

OTOH, y=Asin(kx+ωt) would be travelling towards left .

Last edited: May 5, 2014
6. May 5, 2014

### BvU

Yeah, yeah. redface^^2. Noticed my wrongdoing when already on my way, couldn't fix it till now. You got the picture. No need for you to play with an (antique) transparency. Perhaps a good tip for first-timers, though. Much better than gazing at a wikipedia animation.
Must admit that some animations are really, really nice, though.

7. May 5, 2014

### Tanya Sharma

'x' representing the coordinate of a point makes sense .But if this is the case where is the origin ?

Let us take a step back and analyze the equation of a simple sinusoidal wave .The displacement of the particle at the origin y(0,t) = Asinωt .This reaches a distance 'x' at time t-x/v later ,hence y(x,t) = Asin(ωt-kx ) or may be taken as y=Asin(kx-ωt) .

So,don't you think 'x' in the wave equation represents the distance from the origin(source) ? But ,then are we considering two origins for two sources .If that is the case ,the x's in case of two waves should be different ?Well...two origins do not make sense .

If suppose we consider origin at the first source then y1=A1cos(kx-ωt) is fine . But then how do we write y2=A2cos(kx-ωt+ø) considering that the origin is not at source two and the expression for y2 is obtained by considering origin at the second source?

8. May 5, 2014

### dauto

You can place the origin wherever you want. It doesn't have to be located at the source. Changing the placement of the origin will just add or subtract an arbitrary phase.

9. May 5, 2014

### TSny

As dauto says, you don't need to assume that either source is placed at x = 0.

Note that no matter where you place source #1, the wave traveling in the positive x direction from the source can be written

y1 = A sin(kx-ωt+ø1) for some phase constant ø1.

But you can always redefine your zero of time to make ø1 = 0.

Then, the wave from the second source will be y2 = A sin(kx-ωt+ø) for some phase constant ø. To determine ø you would need to know the value of any intrinsic phase difference between the two sources as well as the distance between the sources.

10. May 5, 2014

### BruceW

Starting with a simpler example: if you're given two functions y1(x) and y2(x') then if someone tells you to add those two functions together, what they almost certainly mean is to use the same input into both functions, and then add the output of those two functions together. In other words, you should let x=x' which gives y(x) = y1(x) + y2(x)

But it is possible that you are not meant to use the same input into each function. i.e. use x and x'. So in this case, y(x,x') = y1(x) + y2(x') You end up with a function of two variables. This could be useful in some cases. But 99% of the time, when someone tells you to add two functions together, they want you to use the same input into both functions.

11. May 5, 2014

### Tanya Sharma

Thank you very much .

12. May 8, 2014

### Tanya Sharma

I tried looking up in the book and on the web but didn't find much on this topic.

In all the intro physics books the wave equation y= Asin(kx-ωt) is derived by keeping origin at the source.

I am unable to understand how the second wave can be written as y2 = A sin(kx-ωt+ø) or in other words how the absence of origin at the source is compensated by a phase constant .

13. May 9, 2014

### ehild

Place the source of the wave in a coordinate system, at distance D to the left from the origin. The distance from the source is x', y=Asin(kx'-ωt).
x' can be expressed by the x coordinate: x'=x+D. Substitute that for x' in the equation of the wave:

y=A sin(k(x+D)-ωt)= Asin(kx-ωt+(Dk)) Dk=θ is the phase constant of the wave in the coordinate system. The wave is advanced by Dk at the origin.

ehild

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Last edited: May 9, 2014
14. May 9, 2014

### Tanya Sharma

Hi ehild !

While studying interference we are looking for superposition of waves at different points in coordinate plane.

Suppose Source one S1 is at origin ,source 2 S2 is at (0,-3) . Both sources are assumed coherent generating waves of equal angular frequency ω and equal amplitudes A.We are looking at interference at point P (4,0) .

In this case wave by S1 is y1 = Asin(kx-ωt) .

What should be the expression for wave by S2 i.e y2 , considering origin is vertically above S2 ?

Last edited: May 9, 2014
15. May 9, 2014

### ehild

What is the direction of travel of the second wave? How can you write it? ( the general form of the plane wave is y=Asin(kΔr-ωt+θ), where k is the wave vector and Δr is the change of the position vector)
ehild

Last edited: May 9, 2014
16. May 9, 2014

### Tanya Sharma

At an angle tan-1(3/4) above horizontal .

Okay .I didn't know that wave equation involved vectors . Thanks for letting me know .
I just have preliminary knowledge about waves where wave equation is given by y = Asin(kx-ωt) if wave is travelling towards right .

In this equation y=Asin(kr-ωt+θ) , r = 4i ,but what is k ? I understand k = ω/v = 2π/λ .

17. May 9, 2014

### ehild

k is the wave vector or propagation vector. Its magnitude is 2pi/λ, and its direction is the direction of propagation of the light.

With respect to the second source, the wave travels from (0,-3) to (4,0). How much does its phase change between the source and point P?

18. May 9, 2014

### Tanya Sharma

2π(path diff)/λ = 10π/λ

19. May 9, 2014

### BvU

Might still be a good idea nevertheless, but now with two transparencies! Note how you have to rotate the x-axes over different angles to let the individual sine waves travel from their starting point to the observation point. You mention S1 and S2 which evokes the double slit interference expressions. But you can also play with only one extended S1 to understand single slit diffraction patterns and gently move in to the really beautiful world of the all-powerful and quasi omnipresent fourier transforms. I get carried away, sorry :shy:.

And the links do have a lot of useful stuff (wiki or the very nice Aussie)

20. May 9, 2014

### ehild

If the second wave travels in the direction of the vector k, and it starts from point r0 it is described by the function y=Asin(k(r-r0)-ωt). The starting point is (0,-3), you want the phase at P(4-0). r-r0=(4,3). k=2pi/λ(0.8, 0.6)

The phase at t =0 is 2pi/λ *5. 5 is the length of path of the second wave. You can also say, that the wave is y2=Asin(ks-ωt) where s it length of the distance travelled and k is 2pi/λ.