Sound Wave Interference: Investigating Null Results with Sound Level Meter Data

[Rasmus Nielsen]

I collected this data by moving a sound level meter slowly through a total length of 0,75 meter between two speakers pointing towards each other. The wavelength of the sine waves was 0,343 meters and the frequency was 1000 Hz. The sound level meter measured the sound in dbA 20 times a second.

Can anyone explain why there's no constructive interference at all?

 

[sophiecentaur]

What was the signal you used? For a sine wave, the expected signal would be hard to predict because the levels from each speaker could vary a lot, over the distances from the speakers, preventing an ideal interference pattern but I would have expected a sort of standing wave pattern. What happens if you invert the phase of one of the speakers? Do individual runs for each speaker alone and see how the signal level varies with distance. When you get results that you think are really odd, there's often something seriously wrong with the set up so you have to try changing things one at a time and 'follow your nose'.
Fault finding is best done 'hands on'.

 

[Rasmus Nielsen]

Thanks for the answer. Unfortunately, I do not have more time to make the experiment again. Do you have any ideas why I didn't get that standing wave pattern? According to the theory, there should be constructive interference at the center, but nothing happens in the data I've collected... I hope you can help :)

 

[sophiecentaur]

Do you have any ideas why I didn't get that standing wave pattern?

I think the explanation could be partly down to the logarithmic (dB) scale that's being used. You may not b e too familiar with using dB scales and they don't always say what you think they are saying. In addition there is variation of signal levels from the speakers as you move between them.

On the two traces, the cancellation is only around -10dB. That is a reduction in amplitude to 1/3. The ratio of the two amplitudes at the actual point of cancellation would only need to be 3:2 - leaving 1/3. The sum would be 1.67, which corresponds to only +4dB. So, on the dB scale, you wouldn't expect a massive peak (only 6dB and a minimum of several tens of dB if everything went right.

For a normal standing wave pattern, the waves in both directions have a constant amplitude all along the 'string' or tube carrying the waves. Your model of two point sources is not the same. If you were to plot a graph of expected signal levels from each speaker (inverse square law perhaps) as you traverse the path between (did you check this?) and work out the possible sum and difference values at a number of points in between the speakers, you will only get good addition and subtraction in some places. [Edit: actually, only in one place]

Are you sure of the wavelength that you were using? It may be that the humps on either side of the dips are actually 1/4 wavelength from the only dip that is actually occurring because of the varying signal levels.
I wonder who gave you this 'investigation' to do. It's much harder than at first sight. I get cross (not with you but on your behalf and other students) when you guys write in with questions about some of the investigations they have been given when they are just not well enough thought out. I can't blame you if you thought it up but I feel someone should have been helping you at the beginning. Very frustrating for you but I hope I may have given you some pointers in this.

 

[PeterO]

I collected this data by moving a sound level meter slowly through a total length of 0,75 meter between two speakers pointing towards each other. The wavelength of the sine waves was 0,343 meters and the frequency was 1000 Hz. The sound level meter measured the sound in dbA 20 times a second.

Can anyone explain why there's no constructive interference at all?

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I have contemplated this experiment before, but never carried it out. My thoughts were that IF you could find a point of total destructive interference then the sound level should dip to zero - a large drop from nearby readings.(the difficulty being that the input of the sound meter has more than a single point size, so some "not quite destructive" input will get in)
However, IF you could find a point of perfect, constuctive interference, the sound should be "twice as loud" as we expect with just one speaker, which represents a mere 3dB increase. If that point is not in the centre, then the sound from each speaker may be at different intensity, so the increase would be less than 3dB.
If you ever get a chance to repeat this experiment, I would suggest you establish where a pair of nodes are (low sound level), indicating the antinode should be midway between them.
Then, at each node and the (assumed) antinode, take a reading with only the left speaker, then the right, then both connected.
I have always imagined the results should be of the type 60dB, 60dB, 0dB at the nodes, but 60dB, 60dB, 63dB at the antinode. ie. a large drop at the node, but a minimal rise at the antinode.

 

[sophiecentaur]

point of perfect, constuctive interference, the sound should be "twice as loud" as we expect with just one speaker, which represents a mere 3dB increase

The pressure level would be twice, for coherent addition, which means that the sound Power would increase by 6dB. In fact, the result of a non-ideal situation would be more in the direction of 'filling in the nodes' than visibly reducing the antinodes. The easiest way to prove that to yourself is to try a few numerical examples and plot some graphs. A spreadsheet is an easy way to do this sort of thing and the Maths involved is very basic (an advantage for many of us).
But, as I have already mentioned, the basic idea of using two speakers in free space is flawed if you want to get the well know standing wave pattern.
Using a wide tube, with speakers driving each end would be much better. Even a single speaker would be good but it could be difficult to get really deep nulls.
[Edit: Second thoughts. If you want to get a good interference pattern (which is what the standing wave effectively is), the way to do it, which ensures that the received signals from each source are more or less the same as you move around, is to separate the speakers by a few wavelengths and to look at the sound levels at a distance on a line parallel to the line of the speakers. This is the same as the well known Young's Slits experiment and produces good nodes and antinodes. (You need to use a large room to eliminate the effect of reflections from walls.

 

[Rasmus Nielsen]

On the two traces, the cancellation is only around -10dB. That is a reduction in amplitude to 1/3. The ratio of the two amplitudes at the actual point of cancellation would only need to be 3:2 - leaving 1/3. The sum would be 1.67, which corresponds to only +4dB. So, on the dB scale, you wouldn't expect a massive peak (only 6dB and a minimum of several tens of dB if everything went right.

I don't really understand this. Why wouldn't I expect the intensity of a constructive interfernce to be equal to the destructive? Could you perhaps elaborate it?

If you were to plot a graph of expected signal levels from each speaker (inverse square law perhaps) as you traverse the path between (did you check this?) and work out the possible sum and difference values at a number of points in between the speakers, you will only get good addition and subtraction in some places. [Edit: actually, only in one place].

I don't see how the variation of signal levels through out the path between the speakers can cause the problem? Doesn't the sound level only affect the amplitude of a wave? And could sound that reflects from the walls be a part of the problem?

I'm currently making a discussion of the experiment in my project, and I'm looking for things that could have caused the unexpected results.

 

[sophiecentaur]

I don't really understand this. Why wouldn't I expect the intensity of a constructive interfernce to be equal to the destructive? Could you perhaps elaborate it?
I don't see how the variation of signal levels through out the path between the speakers can cause the problem? Doesn't the sound level only affect the amplitude of a wave? And could sound that reflects from the walls be a part of the problem?

This is not an ideal situation and you need to realize that your ideas about a well designed experiment will not apply.

When two signals interfere destructively, the sum can be Zero. When they interfere constructively the result will be twice the amplitude. On a dB scale, this will look like -60db (or whatever the sensitivity is) to +6dB. So the trace will never be symmetrical on a log (dB scale). You will only get perfect cancellation when the signals have equal amplitudes. The two signals will only be equal in just one place when using two loudspeakers. A point source will not produce a uniform level and, on the way in between, one is going up and one is going down.

Reflections from elsewhere can have a big effect (which is why it's best to do this sort of thing in a 'dead room' or outside on a quite day. Better still, as I suggested, it should be in a tube or trunking so the levels from each loudspeaker. do not change with distance. When you are away from a node, the resulting amplitude will be somewhere in between the difference and the sum. The maximum is much wider than the minimum and, as you say, there can be reflections from walls to mess it up further.

Your results show a minimum that is not in the centre. That is probably because the output levels from the two speakers are not equal. Away from that point, only one speaker will dominate.

 

[Rasmus Nielsen]

On the two traces, the cancellation is only around -10dB. That is a reduction in amplitude to 1/3. The ratio of the two amplitudes at the actual point of cancellation would only need to be 3:2 - leaving 1/3. The sum would be 1.67, which corresponds to only +4dB. So, on the dB scale, you wouldn't expect a massive peak (only 6dB and a minimum of several tens of dB if everything went right.

How did you find out that a cancellation of -10dB is a reduction in amplitude to 1/3? And what do you mean by the ratio of the two amplitudes at the actual point of cancellation being 3:2 - leaving 1/3? I don't know why, but I just don't understand why the contructive interference only can raise by 6 dB, while destructive can go all the way down to zero. And that still doesn't explain why there's no constructive at all.
I'm really sorry if I'm wasting your time :(

 

[Nugatory]

How did you find out that a cancellation of -10dB is a reduction in amplitude to 1/3?... but I just don't understand why the contructive interference only can raise by 6 dB, while destructive can go all the way down to zero.

The decibel scale is nonlinear. Doubling the amplitude increases the the decibel level by 6dB, reducing the amplitude by a factor of 3 lowers the decibel level by 10dB. The Wikipedia article on "decibel" will explain this in more detail.

 

[sophiecentaur]

I just don't understand why the contructive interference only can raise by 6 dB

Do you know the dB scale you have been quoting on your diagrams? I wonder who gave you the equipment without telling you how to use it.
The maximum resultant from two signals (when they are co-phase) is 2A. Twice the amplitude corresponds to 6dB. If the signals are not in phase and not equal in amplitude, the resultant will be less than that. You should read about these dB things. They are universally used in signal measurement. This wiki link has all the information and has the formulae for conversion. I really can't think of a non mathematical way into this. Do you understand about the way waves add in a situation where they are of different amplitudes and an arbitrary phase?
I don't mind pointing you in the right direction. If anyone has wasted time, it could be your teacher or whoever suggested this task was wasting your time.

 

[Rasmus Nielsen]

Thanks to both of you, It makes a bit more sense now. I now understand why constructive interference only can go up 6 dB.
If I reduced the sound in dB all the way to zero by making destructive interference, how much would I then have reduced the amplitude? And no, I don't understand the way waves add in a situation where they are of different amplitudes and an arbitrary phase.

 

[sophiecentaur]

would I then have reduced the amplitude?

-10000s of dB. (what's the Log of zero?)

different amplitudes and an arbitrary phase

The resultant A is the sum of the two waves.
A=A₁cos(ωt) + A₂cos(ωt + θ)
Where A₁ and A₂ are the two amplitudes and θ is the phase difference. ω is the angular frequency (2πf)