Standing waves between two speakers in phase

[MaherJ]

Homework Statement

Two identical loudspeakers are driven in phase by a common oscillator at 800 Hz and face each other at a distance of 1.25 m. Locate the points along the line joining the two speakers where relative minima of sound pressure amplitude would be expected.

Relevant Equations

in phase: constructive interference , y = y1 + y2

The solution provided in the manual poses that the point halfway between the nodes at each speaker is an antinode of pressure (node of displacement) but isn't that a contradiction to the fact that the speakers are in phase? My first thought was that they must interfere constructively and have maximum amplitude (double that of the waves) but apparently this is not the case and I didn't understand why.

 

[TSny]

Assume the two speakers lie on a horizontal line with one speaker on the left and the other speaker on the right. When the diaphragm of the speaker on the left is at maximum displacement to the right, describe the displacement of the diaphragm of the speaker on the right.

 

[hutchphd]

When the diaphragm of the speaker on the left is at maximum displacement to the right, describe the displacement of the diaphragm of the speaker on the right.

I don't see where you are going here.
I believe the OP is correct and there will be pressure node midway between. This is why you "phase" your stereo speakers (antiphasing produces a null plane midway between for all frequencies) The displacement direction is not what matters...it is a pressure wave.
Bad problem?

 

[Steve4Physics]

Edit.
I've removed my reply as I think it was wrong/misleading.

Updated answer is now in Post #10.

Hi @hutchphd - I won't be offended if you remove your 'Like'!

 

[TSny]

I don't see where you are going here.

At the instant that the diaphragm of the speaker on the left has maximum displacement toward the right, it displaces the air toward the right. At the same instant, in which direction has the speaker on the right pushed the air? What happens when these two disturbances meet at the midpoint?

It is key to understand what it means for the speakers to be "driven in phase" and to have the speakers face each other.

 

[MaherJ]

I don't see where you are going here.
I believe the OP is correct and there will be pressure node midway between. This is why you "phase" your stereo speakers (antiphasing produces a null plane midway between for all frequencies) The displacement direction is not what matters...it is a pressure

I tend to agree with you.

A displacement-node is a pressure-antinode.
A displacement-antinode is a pressure-node.
This doesn’t depend on the positions or relative phases of the 2 speakers. It’s an intrinsic feature of the (longitudinal) standing wave.

The expression “between the nodes at each speaker ” is confusing and suggests beleiving that each speaker is at a node. This is wrong – we are probably not dealing with a resonance situation. What are the actual words from the solution manual?

Whenever 2 (otherwise identical) waves with opposite directions overlap, we get a standing wave. It doesn't have to be associated with a resonance.

In this question the point midway between the speakers must be a displacement-antinode because of the equal path lengths and the fact that the speakers are in phase. It is therefore a pressure-node.

If you are given the speed of sound, you can find the wavelength and then find the positions of the other pressure-nodes.

I completely agree with what you’re saying, assuming the speakers are in phase with the same frequency and facing each other the constructive interference midway must result in a displacement antinode and thus a pressure node while respecting the boundary conditions (the opposite of what the manual assumes).

 

[hutchphd]

This is true everywhere along the plane equidistant from each speaker. Again the interesting case is when they are out of phase: the plane becomes a null space for all frequencies. Very noticeable should you phase your $1000 speakers wrong. I once spent an entire evening not pointing this out to a rather snobbish technophile whose hospitality I was enjoying (or perhaps abusing)

 

[TSny]

If the speakers are driven in phase, then when the diaphragm of the speaker on the left is pushed "outward", the diaphragm of the speaker on the right is also pushed "outward".

The diaphargms are shown in blue at an instant when they are both pushed "outward". Thus, one is pushed toward the right while the other is pushed toward the left. Their horizontal displacements are in opposite directions. So, they are displacing the air out of phase.

 

[hutchphd]

I agree, but for any region except extreme near field this is irrelevant. The speaker moving "out" creates high pressure and the speaker moving "in" creates low pressure. Sound is fundamentally a pressure wave and these sources are therefore "in phase"

 

[Steve4Physics]

I take @TSny's point about displacements

The midpoint will be a displacement-node because the displacements due to each speaker have opposite directions and cancel. This make the midpoint a pressure-antinode.

I think the sound intensity is determined by pressure (not displacement) variations. So the midpoint is where the sound is loudest.

Referring back to Post#1, this means the manual's solution is correct.

 

[hutchphd]

Referring back to Post#1, this means the manual's solution is correct.

Yes I agree completely.

 

[haruspex]

I agree, but for any region except extreme near field this is irrelevant. The speaker moving "out" creates high pressure and the speaker moving "in" creates low pressure. Sound is fundamentally a pressure wave and these sources are therefore "in phase"

Yes, and saying they are in phase refers to the pressure. But because the waves are traveling in opposite directions, they are 180 degrees out of phase wrt displacement.

@MaherJ, I am curious about the book answer to the question as given in post #1. Do they take attenuation into account? Looks quite tricky to do it correctly.

 

[hutchphd]

If everything is symmetric, then it is exact.

 

[MaherJ]

Update : Taking into consideration the fact that sound waves are primarily pressure waves , I can admit I missed that point for a second. I guess the speakers being in phase refers to pressure and not to displacement, moreover I can just agree with @TSny 's statement about the diaphragm which leads to the pressure wave being formed at the same instant between the speakers (left and right extension of the diaphragm) being in phase and thus the displacement halfway being a node.

 

[Steve4Physics]

I guess the speakers being in phase refers to pressure and not to displacement

The speakers 'being in phase' in this question does not refer to the air-pressure or the air-displacement. It is a description of the relationship between the speakers' positions over time.

As highlighted by @TSny, because of the symmetry, an air-element at the midpoint (P) does not move. This gives a displacement-node at P.

P is a pressure-antinode because of the movement of the air around P. The air around P moves like this:
→P← during a compression (increasing pressure at P)
←P→ during an expansion (reducing pressure at P)
but the air at P remains stationary.

 

[haruspex]

If everything is symmetric, then it is exact.

Taking attenuation into account, I get the following simulation:

This shows the middle three wavelengths from two speakers placed five wavelengths apart. The 11 curves represent the pressure pattern at equal time intervals across half a period (endpoints included). For the other half period, reflect about the X axis.

No point is a perfect node, and the further from the centre the less perfect. Even so, it does appear that the relative minima are at the same places as for no attenuation.
@hutchphd , do you have an easy way of proving that? The algebra looks messy to me.