0
$\begingroup$

When there are N coherent sound sources playing the same note at equal loudness, their sound waves add up to make a sine wave of the same frequency but most likely different amplitude (can be anything from zero to N times the amplitude of one wave), depending on their relative phase shifts.

Can you prove that on average, the square of the amplitude of the resultant wave is N times the square of the amplitude of one wave? This would explain why we add intensities of coherent sources (intensity is proportional to the square of pressure) to get the average intensity of the resultant sound.

This page explains what happens for 2 waves, the new amplitude is

https://www.phy.duke.edu/~rgb/Class/phy51/phy51/img396.png

...and the average value of this is 2I_0.

I need to generalise this for N identical randomly shifted sine waves.

Thanks!

$\endgroup$
  • $\begingroup$ To get the Intensity from N sources at any point you add their amplitudes and square it. If you have coherent sources i.e. 'Phase' (not necessarily they should have equal amplitude) among them remain constant you will get peaks and dips depending on where you are listening. But if the phase among them oscillate very rapidly (incoherent sources) then you will just listen to their amplitude sum because the cosine square term in your expression converted into its average and become 1/2 (since k$\Delta$x is rapidly changing) ,you will get 2I. $\endgroup$ – hsinghal Jun 4 '16 at 14:19
  • 1
    $\begingroup$ It sounds like a random/drunkard's walk problem? $\endgroup$ – Farcher Jun 4 '16 at 14:23
  • $\begingroup$ Sorry, when I wrote incoherent, I actually meant coherent. They all have a constant phase difference, but the difference doesn't have to and most likely isn't zero... So we are adding identical sine waves except shifted sideways. $\endgroup$ – mrk1357 Jun 4 '16 at 14:26
  • $\begingroup$ You are thinking of these waves overlapping at the central point. you may think of these waves overlapping at some angle from array. Consider a linear array of N sources and calculate the amplitude at an angle, there will be constant phase difference between each source when you add them you will find an expression something like $sin^2(Nkx)/sin^2(kx)$. $\endgroup$ – hsinghal Jun 4 '16 at 16:17
1
$\begingroup$

I remember working this out in the opposite direction, hoping to get a paradox: that the amplitude of the sound produced by a choir of 100 singers is only 10 times the amplitude of the sounds produced by one of them.

Each individual sound can be represented as a vector in 2-dimensional space, with length 1 (since all are equally loud) and random phase. The resultant sound heard will be the sum of all those vectors.

Adding the vectors one at a time is thus effectively undertaking a random walk in 2 dimensions. The following statements are therefore equivalent:

  1. The expected distance travelled in an $n$-step random walk is $\sqrt{n}$.

  2. The expected amplitude of $n$ coherent equally loud sound sources is $\sqrt{n}$ times the amplitude of one of them.

It is entirely up to you to decide which one of them you prove from which, but in this case you can assume the truth of 1 (it is a standard result) and deduce the truth of 2, which, squared, is the result you asked for in your question.

(And my paradox wasn't a paradox because I had forgotten that loudness is proportional to the square of amplitude, so 100 choristers are 100 times as loud as one of them).

Incoherent sound sources yield essentially the same answer because you can think of them as coherent sources with varying phase - so again you end up with a total of random phases.

$\endgroup$
1
$\begingroup$

Yes. Let's consider the phase shifts to be random. Then we can consider each amplitude to be an independent random variable $A_i(t)$. Each one has a variance of $$\text{var}(A_i) = \langle A_i^2 \rangle - \langle A_i \rangle^2 = \langle A_i^2 \rangle \propto I_i$$ because the average amplitude should be zero, and intensity is proportional to amplitude squared. The variances of independent random variables add, so $$I = \text{var}(A) = \sum \text{var}(A_i) = \sum I_i$$ as desired.


You might not like the 'random variable' phrasing, but it's not essential. At heart, the idea is that the $A_iA_j$ cross terms in $(\sum A_i)^2$ must average to zero. I call that property 'independence', you can call it something else.

$\endgroup$
1
$\begingroup$

You cannot prove that because it is not generally true, or because it depends on your definition of "equal loudness". Consider the case where the sound sources are inside a flaring pipe with a standing wave (say, a modified trumpet).

Sound sources in tube

A standing wave is typically drawn as a standing wave of the air displacement in the pipe (illustrated), but note that the nodes of the air displacement amplitude coincide with the antinodes of the pressure amplitude.

The sound sources (say, small piezo loudspeakers) are pressure transducers and they are placed at the antinodes of the pressure amplitude. If each of them contributes to a fixed pressure amplitude $\Delta p$ at the position of the listener, then $N$ of these pressure transducers ($N=7$ in the picture) will produce $7\Delta p$ of pressure amplitude and $N^2=49$ times more acoustic energy compared to a single transducer.

(Update: I assumed that the sound source primarily act on pressure. It might be more appropriate to consider sound sources as acting on air displacement, in which case they should be placed at the antinodes; the reasoning is otherwise similar.)

How is this possible? Where does the energy come from? Rest assured: the law of energy conservation is not violated. In a case like this, where the acoustic sources can feel each other, the sources have to work harder in order to achieve the same pressure. Imagine that you are driving a damped harmonic oscillator. If you put in twice as much amplitude, you get twice as much amplitude out, but you need to put in four times more power.

Now, if you say that each sound source adds a fixed amount of acoustic power, rather than a fixed amount of amplitude, then your statement is automatically true because of energy conservation. For a case like an ensemble of singers or musical instruments, the coupling between the sound sources is so weak that the difference between sources of fixed amplitude and sources of fixed power is negligible. But if you design a loudspeaker line array for use in churches or train stations, with the purpose of radiating the sound horizontally and not vertically, then this distinction will not be negligible and must be taken into account.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.