Where are the waves out of phase with each other by π radians?

[hidemi]

Homework Statement

Two point sources emit sound waves of 1.0-m wavelength. The source 1 is at x = 0 and source 2 is at x = 2.0 m along x-axis. The sources, 2.0 m apart, emit waves which are in phase with each other at the instant of emission. Where, along the line between the sources, are the waves out of phase with each other by π radians?

The correct answer is x = 0.75 m, 1.25 m.

Relevant Equations

Δ γ/λ = ϕ/2 𝝿

I know that at the antinodes, the diffrence is pi radians and i got 0.25, 0.75, 1.25, and 1.75. for the position of antinodes. but the answer is x = 0.75 m, 1.25 m.

I can use drawing to obtain the answer, but my calculation gives me 0.25, 0.75, 1.25, and 1.75. Where am I doing wrong? Thanks!

 

[BvU]

my calculation

Post, else we can't see it and comment

Does the picture here help ? (it's 5 $\lambda\$, so you have to shield a certain percentage on left and right, total 3 $\lambda\$)

 

[hidemi]

Post, else we can't see it and comment

Does the picture here help ? (it's 5 $\lambda\$, so you have to shield a certain percentage on left and right, total 3 $\lambda\$)

Δr / λ = Φ / 2π,
Δr = Φ/2π λ
———————-
λ = 1 m
Δr = Φ *1 / 2π

constructive interference:
Φ : 0 , 2π , 4π
Δr: 0, 1, 2

Destructive interference occurs by
Φ = π /2, 3π / 2, 5π /2 , 7π /2
Δr= 0.25m, 0,75m, 1.25m, 1.75m

 

[BvU]

And do you distinguish $\phi_1 - \phi_2$ from $\phi_2 - \phi_1$ ?

 

[BvU]

I have to correct myself: I figured the exercise distinguished between $\pi$ and $-\pi$ but it doesn't look like that.

I can use drawing to obtain the answer

Always good to make a drawing. Does it look like mine below ?

Blue: $\phi_1 = \omega t - kx\$ running from the left to the right and

Red: $\phi_2 = \omega t - k(2-x)\$ going he other way.

Green: $\phi_1 - \phi_2$ going from +4$\pi$ to -4$\pi$.

If $\omega t$ increases, the blue and red lines move up, the green line stays in place.

One can argue that the phase difference $|\Delta\phi|\$ is $\pi$ at 0.75 and 1.25 m and that the phase difference is 3$\pi$ at 0.25 and 0.75 m. Hence the book answer.
$\$

 

[hidemi]

I have to correct myself: I figured the exercise distinguished between $\pi$ and $-\pi$ but it doesn't look like that.
Always good to make a drawing. Does it look like mine below ?

View attachment 277616
Blue: $\phi_1 = \omega_t - kx\$ running from the left to the right and

Red: $\phi_2 = \omega_t - k(2-x)\$ going he other way.

Green: $\phi_1 - \phi_2$ going from +4$\pi$ to -4$\pi$.

If $\omega_t$ increases, the blue and red lines move up, the green line stays in place.

One can argue that the phase difference $|\Delta\phi|\$ is $\pi$ at 0.75 and 1.25 m and that the phase difference is 3$\pi$ at 0.25 and 0.75 m. Hence the book answer.
$\$

Thanks for your response! I cannot relate this to the question. Is there any other explanation? Thanks!

 

[haruspex]

Thanks for your response! I cannot relate this to the question. Is there any other explanation? Thanks!

Strictly speaking, your answers include a phase difference of 3π. Is that equivalent to a phase difference of π? Since the description implies pure tones, I would say yes, making your answer correct.

 

[rsk]

I did it like this. (This is a high school teacher's approach, some of you will look at it differently I am sure)

For waves to meet in antiphase, the path difference must be (n+ 1/2)λ - ie one wave must have traveled exactly half a wavelength further than the other (or a whole nº + a half)Let L be the distance of the point of antiphase from source 1. Then the distance from source 2 is 2 - L

In otherwords, we have antiphase where the wave from source 1 has traveled L and that from source 2 has traveled 2 - L

So the path difference is (2-L) - L or 2 - 2L or 2(1-L)

And, since we know that for antiphase the phase difference must be (n + 1/2) λ I took the simplest case, ie 0.5λ, in this case 0.5m

So 2 (1 - L) = 0.5, giving 1-L = 0,25 and therefore L = 0.75, the distance from the first source.

 

[haruspex]

I took the simplest case,

But is that valid?

Suppose the actual set up is a carrier wave with some signal, amplitude modulation say, being fed to two speakers. So one cycle does not look exactly like the next.
At x=0.75m, the phase difference is π, as required, but at 0.25m it is 3π. Likewise 1.25m versus 1.75m. The correct answer would then be as claimed, 0.75m and 1.25m.

In the actual question it seems it is a pure tone, so the full solution is (0.25+0.5n)m. Since it asks for those points between the speakers, that leaves @hidemi's set of four.

 

[rsk]

But is that valid?

Suppose the actual set up is a carrier wave with some signal, amplitude modulation say, being fed to two speakers. So one cycle does not look exactly like the next.
At x=0.75m, the phase difference is π, as required, but at 0.25m it is 3π. Likewise 1.25m versus 1.75m. The correct answer would then be as claimed, 0.75m and 1.25m.

In the actual question it seems it is a pure tone, so the full solution is (0.25+0.5n)m. Since it asks for those points between the speakers, that leaves @hidemi's set of four.

Again, to stress, this is coming from a HS teacher perspective.

While what you say may be completely valid, I think we have to look at the question itself and think about the course it is from and the level at which it is pitched. Physics is full of simplifications and suppositions and without them we'd never get anywhere. And, when trying to solve a problem, it's always wise to start with the simplest case.

Sometimes, when people try to be helpful pointing out all of the advanced and complex things which could be taken into consideration, it can hinder rather than help.

If the situation were as you suggest, "a carrier wave with some signal, amplitude modulation" then I would expect that to be clear from the question - but in this case there is nothing to suggest that or that it comes from anything other than a basic level topic on wave phenomena.

Some of the complex discussions which arise from these straightforward questions can be very interesting (certainly to the likes of me who long ago forgot anything above high school/undergrad physics level) but are often "what if..." scenarios rather than actually focussed on solving the problem as set out.

And, as a high school teacher, I'd say to you what I'd say to any of my students - you get no credit if you don't answer the question you've been given.

 

[haruspex]

when trying to solve a problem, it's always wise to start with the simplest case.

Sure, but the question wording suggests that all places where the phase difference is π are to be found. This is confirmed by the official answer in that it specifies two points. So taking only the simplest solution, as you did, is invalid.

If the situation were as you suggest, "a carrier wave with some signal, amplitude modulation" then I would expect that to be clear from the question

That is precisely my point. The only way I can justify the book answer omitting the 0.25 and 1.75 solutions is by complicating it in the way I described. Since the question as stated does not suggest any such complication, omission of those solutions is an error.

 

[rsk]

Sure, but the question wording suggests that all places where the phase difference is π are to be found. This is confirmed by the official answer in that it specifies two points. So taking only the simplest solution, as you did, is invalid.

That is precisely my point. The only way I can justify the book answer omitting the 0.25 and 1.75 solutions is by complicating it in the way I described. Since the question as stated does not suggest any such complication, omission of those solutions is an error.

hidemi
I think the 0.25 and 1.75 answers come from setting Path diff = 3λ/2 and that's why they've been omitted.

We were told to look for a phase diff of Pi which corresponds to path diff λ/2
A path diff of 3λ/2 would be a phase diff of 3pi - rather than pi. It sounds pedantic but it's far more likely to be the reason for these being omitted in this case than the carrier wave and modulation suggested above.(Also I think we can say that the 0.75 and 1.25 answers both come from the same calculation, - it just depends which source you measure the 0.75 from - so taking the simplest case, pd = 0.5λ, gives both of these answers)

 

[BvU]

I cannot relate this to the question.

How so ? At the sources the waves are in phase at the instant of emission: $$
\begin{align*} t &= 0, \ x = 0\ \Rightarrow \ \phi_1 = 0\ \\ t &= 0, \ x = 2\ \Rightarrow \ \phi_2 = 0\
\end{align*}$$ Points at e.g. distance $d$ away from one of the sources have to wait $\ d/\lambda\; T$ so they lag by $d/\lambda * 2\pi$ . Blue and red line in #5: the phase.
Green line is phase difference $\ \phi_1 - \phi_2\$ , NOT modulo $2\pi\ \$ (*)
At x = 1 $\ \phi_1 - \phi_2\ = 0$ : same phase for all time
At x = 0.75 and at x = 1.25 you see $\ \phi_1 - \phi_2\ = \pm \pi$

(*) This is the core of the discussion: we are so used to this modulo $2\pi$

 

[Steve4Physics]

I know that at the antinodes, the diffrence is pi radians ...

You mean 'nodes', not 'antinodes'. The phase diffference (between two waves) at a node is (2n+1)π where n = 0, 1, 2, 3, etc.

At x=0.25m, Δϕ=3π
At x=0.75m, Δϕ=π
At x=1.25m, Δϕ=π
At x=1.75m, Δϕ=3π

All these points are nodes but the question does not ask for node positions – it asks for the points where the phase difference is π. Only x= 0.75m and x=1.25m satisfy this.

EDIT: corrections made.

 

[hidemi]

Thank you so much. I learned a lot.