If a thousand people whisper inaudibly, will the resulting sound be audible?

If a thousand people whisper inaudibly, will the resulting sound be audible? (...assuming they are whispering together.)

I believe the answer is "yes" because the amplitudes would simply add and thus reach an audible threshold. Is this right?

if possible, please provide an explanation simple enough for non-physics people

• I think this question carries potential ambiguity in the meaning of "audible". The simple interpretation is simply "a noise was heard". The higher level interpretation is "a message was understood". Wolpertinger john-bentin and Farcher are addressing the former. @stafasu and rob are addressing the latter. In the former the signals can be analyzed using linear signal processing. In the latter not so. In particular when stefasu talks about "constructive interference" I believe he is talking about the message and not the carrier signal. – Craig Hicks Aug 22 '17 at 20:17
• – Floris Aug 22 '17 at 21:03
• Possibly depends on your definition of inaudible, but a single bee is inaudible beyond a very short distance, but a swarm or hive can be heard many tens of metres away. – long Aug 22 '17 at 22:00
• How are the whisperers arranged in space, relative to the listener? If some whisperers are further away than others, do the further ones get to whisper louder? – user2357112 supports Monica Aug 22 '17 at 22:45
• I know this is serious stuff, but I can't believe no one has referenced Dr. Suess's "Horton Hears a Who!" – user129544 Aug 24 '17 at 12:59

Yes, always.

I would like to disagree with stafusa's answer here, expanding on Rod's comment. Interference will not occur, since for whispering the sources of sound will be statistically independent.

For demonstration, let us look at two people. Person 1 produces a whisper that can be characterized by a propagating sound field $$E_1(\vec{r},t)$$, where $$\vec{r}$$ is the position in space and $$t$$ is time. Similarly person 2 produces a whisper $$E_2(\vec{r},t)$$. The overall field at a point in space is then simply

$$E_\mathrm{tot}(\vec{r},t) = E_1(\vec{r},t) + E_2(\vec{r},t)$$

since sound waves are approximately linear (at least for wave amplitudes achievable by voices).

What you perceive as 'volume' (I will call it $$I$$ for intensity) is the time average of the magnitude of the total signal

$$I = \langle E^*_\mathrm{tot}(\vec{r},t)E_\mathrm{tot}(\vec{r},t)\rangle.$$

That is, your ear is averaging over very short fluctuations in the signal. We can then expand this in terms of the two people's signals to get

$$I = \langle E^*_{1}(\vec{r},t)E_{1}(\vec{r},t)\rangle + \langle E^*_{2}(\vec{r},t)E_{2}(\vec{r},t)\rangle + 2\langle E^*_{1}(\vec{r},t)E_{2}(\vec{r},t)\rangle.$$

So far, this is completely general. Now we assume statistical independence of the sources, which makes the last term zero:

$$I = \langle E^*_{1}(\vec{r},t)E_{1}(\vec{r},t)\rangle + \langle E^*_{2}(\vec{r},t)E_{2}(\vec{r},t)\rangle.$$

So the overall intensity is simply the addition of the two whisper intensities.

• We can understand whispers, which means there are not random sources. And I did assume that, besides being audible, it's desired the whisper is also understandable. – stafusa Aug 22 '17 at 13:41
• But is a well-defined statistics enough to transmit a signal? The correlation between the different people whispers is very high, if they are whispering the same words. I don't see how we can model that as independent sources. – stafusa Aug 22 '17 at 14:35
• @stafusa people whispering the same thing at the same time will have similar spectral characteristics, but unless they whisper at the exact same frequency and they can synchronize their whispers with 0.01s accuracy (not even that is enough), their phases will be statistically independent. You are more likely to observe destructive interference by wiring one speaker backwards, playing a mono sound and turning your head sideways. And that's something you can try in practice using a sound editor. Try it. I recommend using a pure tone so that you can catch the side nodes if you miss the middle one – John Dvorak Aug 22 '17 at 16:20
• @stafusa a whisper is an unvoiced sound created by turbulence in the vocal tract (generally near the vocal folds) so reasonably modeled as incoherent despite a possibly coherent amplitude modulation introduced by gross changes to the vocal tract. – StrongBad Aug 23 '17 at 13:05
• @stafusa think about the correlation between a 1 kHz and 2 kHz sine waves that are both sinusiodally amplitude modulated at 7 Hz or even better the correlation between two IID noises that are then both sinusoidally modulated at 7 Hz. – StrongBad Aug 23 '17 at 13:13

The amplitude of the sum of $1000$ equally loud uncorrelated noises will be about $\surd1000$, or approximately $32$, times the amplitude of a single noise. That might be enough to make an inaudible whisper just audible. Consider the practicalities, though. The people cannot all occupy the same spot. If dispersed, most of them will be too far away to hear. Even if crowded together, their bodies and clothing will make an excellent sound-absorbing medium. In probability, all you will hear is the involuntary sound occasionally emitted by a single individual.

• Put differently: 1000 uncorrelated noises of $x$ dB will be like noise of $x+30$ dB (because we refer to power, not amplitude, IIRC). So if $x$ is just slightly below the $0$ dB hearing limit, the sum might just be audible, but less than any "normal" 30 dB sound – Hagen von Eitzen Aug 22 '17 at 10:10
• Not sure it's true for any boundary conditions. In my edited answer I mention people praying softly in a church, where it appears they aren't absorbing that much of the sound. But I may be wrong. – stafusa Aug 22 '17 at 10:34
• @LordFarquaad This answer avoids talking about the notion of n times as loud, which is ill defined and largely subjective. What is well defined are amplitude and power density, where the latter is proportional to the square of the former. The decibel scale (side note: why doesn't anybody every just use the basic bel unit?) is logarithmic both with respect to amplitude as to power density, but not in the same way: 10 dB (1 bel) corresponds to a factor 10 in power density, or to a factor $\sqrt{10}$ in amplitude. – Marc van Leeuwen Aug 23 '17 at 7:10
• Isn't this answer in contradiction with the accepted answer which seems to suggest that the intensities simply sum up? – thermomagnetic condensed boson Jun 13 '19 at 20:54
• The square root is the key here and it was well worth emphasising. Supposing that everyone whistles the same note instead of whispering, they will do so with random phase, so the total amplitude will be the same as the distance travelled in a random walk, which is the square root of the number of steps. Amplitude therefore increases as the square root of the number of people, or, equivalently, power linearly with the number of people. – Martin Kochanski Jun 14 '19 at 6:53

I think it depends on what you mean by "whispering inaudibly." There are stage whispers, which are actually meant to be heard from far off, and real whispers, meant to be heard by the person next to you but not the person next to them, and inaudible whispers, which are not even audible to the person who is emitting them.

I just got back from a choir rehearsal in which the conductor had the chorus do some moderately vigorous stretch, then took a break and said, "Take five deep breaths." Followed immediately by "Take five deep breaths so that I can't hear them." The instant difference in the sound in the room was remarkable.

I have certainly been in crowds of a thousand people where many were whispering, and the result was audible --- but I can't recall a crowd where everyone was "inaudibly" whispering the same thing. There's a place near the end of Mahler's second symphony where a large chorus --- like, 100--150 singers --- enters as quietly as possible, hopefully quieter than a single coughing audience member. I can tell you from experience that the way to accomplish that is for everyone in the chorus to sing "inaudibly" --- but that's not whispering. And I've also been in crowds of more than a thousand where there was complete silence, where I felt compelled to whisper "inaudibly" to myself just to make sure that I hadn't gone deaf --- but I have no way of knowing how many others were doing the same.

So, my anecdotal experience is that "whispering inaudibly" is defined murkily enough that it's possible for corporate quiet noise-making to fall above the threshold of hearing, and also possible for it to remain below the threshold of hearing, even for very large crowds. It depends on what you mean by "inaudible."

The answer is "maybe". 1000 inaudible whispers may still be inaudible; the question you probably meant to ask is "would the sound of 1000 people whispering at the same time be louder than the sound of 1 person whispering?"

The answer to that question is an emphatic "yes". How much louder will they be - and will that result in an audible / understandable message?

For this you need to understand the concept of interference and coherence. Two sources (of sound) are coherent if they produce the same waveform. In the real world, coherence is usually limited in time: if I have two tuning forks that are producing a nominal 440 Hz, one of them might produce a frequency of 440.1 Hz and after 5 seconds the two waveforms would have gone out of step by 180 degrees (this is the cause of "beats"). Any sound you make is comprised of many frequencies - see for example this question and associated answers - that together make up a recognizable phoneme (sound that a letter or group of letters makes). When two people "talk at the same time", they will be producing a phoneme, but not at the same frequency. Yet, when two people both say "A", our ears are pretty good at picking up the fact that they are saying "A" even when they are using a different fundamental frequency.

When two waveforms are incoherent (as is the case for multiple people speaking) we can add together the power of the individual voices, which goes as the square of the amplitude of individual voices. The actual amplitudes will sometimes add in phase (double the amplitude - four times the instantaneous power), at other times they will interfere destructively (zero amplitude, zero power). The time average is still the same as the sum of the power of the two sources.

The same is true for "many" sources. So if you have 1000 voices whispering, you can expect the amplitude on average to increase by about 30 x ($\sqrt{1000}$); if that amplitude is sufficient to exceed the threshold of hearing for you, you might be able to hear them; and if their voices are "quite similar" in pitch, you may be able to understand what they are saying. But the latter is not at all certain - the ability to distinguish phonemes becomes trickier when more frequencies are present. In fact, if everyone speaks "at their own pitch", the resulting sound will become like white noise and you will not understand what is being said.

UPDATE

I decided to do an experiment. I recorded myself saying a certain phrase 19 times, in approximately the same tempo and volume. I reduced the amplitude of the recording, and added some noise. This resulted in an "inaudible message".

Next, I cut the sound track into 19 segments that I aligned with the help of some signal processing (there was a distinct "th" sound at the start of the message). Adding these signals (remember - these are "different" recordings of the same message - a bit like having 19 different people trying to whisper the same thing at the same time), with the same amount of noise added, resulted in an audible messsage.

Finally, I messed around with the delays. Assuming that people would stand no closer than 1 m apart, you can assume that a large "chorus" of people will have some amount of relative delay in their whispering; I added a shift of "1 m delay" between each of the 19 signals before adding them up, and while the signal gets a little bit less crisp, it's still clearly audible.

Of course a group of 1000 people would be arranged to try to minimize this delay - if you arrange a large group of people in a series of concentric (semi) circles, the delay in arrival of the voices need not be much worse than in my example.

If you are interested in the Python code I used to do the image processing (note - there are a number of other experiments and plots in this code... feel free to play with it):

# read the whisper file
import scipy.io.wavfile as WVF
from scipy.signal import argrelextrema

import numpy as np
import matplotlib.pyplot as plt
import wave

# convert mp3 to wav:
# ffmpeg -i ~/Desktop/170826_0080.mp3 ~/Desktop/longwhisper.wav"

# attenuate the sound wave so I have some dynamic range for adding later
soundWave = 0.1*A[1].astype('float')

N = len(A[1])
timeAxis = np.arange(N).astype('float')/A[0]

# visualize sound wave
plt.figure()
plt.plot(timeAxis, soundWave)
plt.title('original sound wave')
plt.show()

# do some filtering
tt1 = np.linspace(-5,5,1000)
filt1 = np.exp(-tt1*tt1/2)
filt1 = filt1 / np.sum(filt1)

tt = np.linspace(-5,5,50000)
filt = np.exp(-tt*tt/2)
filt = filt / np.sum(filt)

baseline = np.convolve(soundWave, filt1, mode='same')
# high frequencies only:
hf = soundWave - baseline
plt.figure()
plt.plot(timeAxis, hf)
plt.plot(timeAxis, baseline, 'r')
plt.title('after subtracting baseline')
plt.show()

soundPower = hf*hf

soundPower = np.convolve(soundPower, filt, mode='same')
plt.figure()
plt.plot(timeAxis, soundPower)
plt.title('smoothed sound power')
plt.xlabel('time (s)')
plt.show()

# find the actual peaks
pks = argrelextrema(soundPower, np.greater)
pkVals = soundPower[pks[0]]
pkSort = np.argsort(pkVals)

# time points corresponding to the 40 largest peaks... this includes the "pops"
# at the start of each phrase
timePoints = np.sort(pks[0][pkSort[-40:]])

# look at the spacing between pops - we know it should be roughly 82000 samples
makeSense = np.diff(timePoints)

startPoints = []

currentTime = makeSense[0]
lastTime = currentTime
for ii in makeSense[1:]:
if abs(currentTime - 82000 - lastTime) < abs(currentTime + ii - 82000 - lastTime):
startPoints.append( currentTime)
lastTime = currentTime
currentTime += ii

# shift back a bit - we need to start just before the pop:
startPoints = np.array(startPoints)+timePoints[0]-8000

plt.figure()
for ii in range(len(startPoints)):
temp = soundPower[startPoints[ii]:startPoints[ii]+78000]
plt.plot(temp/np.max(temp)+0.1*ii)

plt.title('sound power after aligning')
plt.show()

# sum the blocks:

# high frequency filter on the noise - make it a bit more "pink":
tt2 = np.linspace(-5,5,20)
filt2= np.exp(-tt2*tt2/2)
filt2 = filt2 / np.sum(filt2)

noise = np.convolve(np.random.random_integers(-noiseAmp, noiseAmp, size=np.shape(waveIn)), filt2, mode='same')
return waveIn + noise

def writeFile(block, fileName):
pv = block.astype(np.int16).tobytes()
sound = wave.open(fileName, 'w')
sound.setparams((1,2,44100, 0, 'NONE', 'not compressed'))
sound.writeframes(pv)
sound.close()

def hpFilter(block, f=filt1):
return block - np.convolve(block, f, 'same')

# for noiseAmplitude in [0, 100, 200, 500, 1000]:
# stagger the sounds: 1 m = 1/300th second = 130 samples
# a crowd of 1000 people could be placed in a semicircle of 50 people, 20 deep
# that makes the delta x about 10 m if they are "optimally aligned"
noiseAmplitude = 500
for spacing in np.arange(0,2,0.5):
stagger = int(spacing * 44100 / 340.)
duration = 78000
start = startPoints[0]-10*stagger
catblock = np.copy(sumblock)

for ii in range(1,19):
ti = startPoints[ii] +(ii-10)*stagger
temp = hpFilter(soundWave[ti:ti+duration])
sumblock = sumblock + temp;

writeFile(sumblock,'/Users/floris/Desktop/onewhisper_%d_s=%.1f.wav'%(noiseAmplitude, spacing))
writeFile(catblock, '/Users/floris/Desktop/evenwhisper_%d_s=%.1f.wav'%(noiseAmplitude, spacing))

plt.figure()
plt.plot(sumblock)
plt.title('sum signal: noise = %d'%noiseAmplitude)
plt.show()


With a "thank you!" to AccidentalFourierTransform who suggested using Archive.org as a possible place to host the audio files.

• Wow! Now, that's definitely the best answer. – stafusa Sep 4 '17 at 21:44

Yes, and if done carefully, not only audible, but also understandable.

[Update: indeed, see the answer from Floris - include audio files to prove it!]

For example, they should whisper exactly together only if they are equidistant from the hearing point - for an arbitrarily chosen point, they should whisper with small delays among them, so that the sound reaches the point in sync, interfering constructively.

Edit: That's so if, besides being audible, it's desired the sound is also understandable. As many pointed out, even whispered random noise will lead to increased volume.

Also, "done carefully" can be achieved in ways other than above, which is just an example. Another way is whispering/talking slowly: like when students greet in unison an incoming teacher, or people in a auditorium respond to an entertainer request.

And, lastly, probably it really has to be only "somewhat carefully", since the increase in volume also happens (and the words can often be understood) when the public in concert, a choir, or a group of churchgoers sing together.

An example of a choir whispering might convince the naysayers ;-) https://youtu.be/yaNeIgBZSUE?t=89

• Comments are not for extended discussion; this conversation has been moved to chat. – rob Aug 22 '17 at 16:26

This is an interesting question which cannot be answered exactly but here are some things to think about.

For the "standard" ear according to Wikipedia the auditory threshold of hearing at a frequency of $1\, \rm kHz$ is taken to be $0\, \rm dB$ which corresponds to a sound pressure of $2 \times 10^{-5} \, \rm Pa$. So I will use this figure for the whole range of frequencies which the human voice and assume that this is the whisper from one source that arrives at ear of the person listening to the thousand.
If this is the sound level of whispering at source then a correction would have to be made for the reduction in intensity of the sound due to the sound having to travel a distance between source and receiver.

Now one has to think about the nature of the sounds which are coming from each of the sources.
I will assume that the sound intensity due to each source is the same.
If the sound sources are coherent then one needs to add the pressures (amplitudes) and then square them to get the intensity.

So for one source $0 \, \rm dB = 10 \log_{10} \left ( \dfrac {I_{0\, \rm dB}}{I_{\rm reference}}\right )$ where $I$ is the intensity.

For $1000$ coherent sources the sound level is

$10 \log_{10} \left ( \dfrac {1000^2\times I_{0\, \rm dB}}{I_{\rm reference}}\right ) = 10 \log_{10} \left ( 10^6\right ) + 10 \log_{10} \left ( \dfrac {I_{0\, \rm dB}}{I_{\rm reference}}\right ) = 60 \, \rm dB + 0 \, \rm dB = 60 \, \rm dB$

which according to Wikipedia is the sound from a television or normal conversation.

At the other extreme is having the sound sources as completely non-coherent.
In this case it is the intensities which must be "added" and the intensity for 1000 such sources would be

$10 \log_{10} \left ( \dfrac {1000\times I_{0\, \rm dB}}{I_{\rm reference}}\right ) = 10 \log_{10} \left ( 10^3\right ) + 10 \log_{10} \left ( \dfrac {I_{0\, \rm dB}}{I_{\rm reference}}\right ) = 30 \, \rm dB + 0 \, \rm dB = 30 \, \rm dB$

which according to Wikipedia is the sound level in a very calm room which of course one could hear.

It is likely that the crowd would tend towards being a set of non-coherent sources?

So depending on how for away from the crowd it appears that you are (very?) likely to hear a "buzz" from a crowd of 1000 people.

• In the ideal case. I think @JohnBentin made the critical point that, in a crowd of 1000 people the attenuation due to absorption through the crowd and in reflection from other surfaces would produce a result considerably lower than the 30dB purely incoherent case. Possibly still audible, of course, but this I think 30dB is pretty much the theoretical upper limit. – J... Aug 22 '17 at 13:07
• @J... Thank you for your comment. My aim was not to give a detailed answer but to do a sort of order of magnitude calculation to see if there is a possibility of a crowd being heard. – Farcher Aug 22 '17 at 14:14
• I can't imagine how the noise could be coherent. If it's inaudible they can't hear each other to coordinate even if they were skilled enough to do so. – Loren Pechtel Aug 23 '17 at 3:44
• @LorenPechtel even more than that, while the amplitude modulation of the waveform might be correlated since it is controlled by gross movements of the vocal tract, I dont see how we could control the fine structure of unvoiced sounds like a whisper. – StrongBad Aug 23 '17 at 13:00

Whether you can hear a sound depends on several factors:

1. How intense a sound is when it reaches your ear. This, in turn, also depends on several factors, among which of the most important are:

1. What radiant intensity the sound is emitted with in your direction by the source of the sound,

2. How far away you are from the source of the sound, as the sound intensity is inversely proportional to the distance squared, and

3. From what direction the sound propagates to you and its spectral composition (i.e. how the sound intensity is distributed in the sound spectrum), since your head and ears will block or amplify a specific part of the sound spectrum differently depending on the frequency and the direction, as explained briefly in this SmarterEveryDay video.

2. The spectral composition of the sound that enters your ear, as your ear will pick up on different parts of the spectrum differently much and requires different sound intensities for two different monochromatic sounds with two different frequencies to be perceived equally loud (some frequencies are hard to perceive or cannot be perceived at all, for example).

If the loudness of a sound exceeds a certain theshold, it can be heard.

Assuming that all thousand people are whispering approximately equally loud and with approximately the same spectral composition in their voices, and facing you approximately equally much, and that point 1.3 has a negligible effect, out of the listed points we only have to consider point 1.2.

Besides, as some people point out, the sound pressure of a sound (roughly equivalent to the sound "amplidute" for monochromatic sounds) that consists of multiple sounds will just be the sum of the different sound pressures contributed by the different sound sources.

Since all sound waves can be assumed to be parallel as they enter one of you ear channels, the velocity of the air particles will be proportional to the sound pressure, and the intensity of the sound will be proportional to the sound pressure squared.

Since the sound pressure averaged over time equals zero, the average sound intensity will be proportional to the variance of the sound pressure. If all the sounds from all the thousand whispering people can be assumed to be uncorrelated, the variance of the sum of the different sounds pressures equals the sum of the variances of the different sound pressures.

Hence, the average sound intensity of the total sound is equal to the sum of the averages of the sound intensities of the different sounds, if that makes any sense.

Or in other words, the loudness increases with the number of (uncorrelated) sound sources.

However, if the fact that you increase the number of people from one to one thousand means that they have to stand farther away from you, this additional fact will decrease the loudness of the sound and may cancel out the effect of increasing the number of people, or even make the sound less loud than it would be with only one person, depending on how the people are placed, as the sound intensity will be proportional to

$$\sum_i^N \frac{1}{d_i^2} = N\left<d^{-2}\right>,$$

where $d_i$ is the distance to the $i$th person and $N$ is the number of people.

There is a simple way to test this by adding a bunch of sine waves with different phases.

If we take a random set of phases on the interval $[0, 2 \pi]$ then we can get constructive and destructive interference. Using Mathematica one can set this up as:
xx := RandomReal[{0,2 \[Pi]},20]
mysin[t_] := Sum[Sin[t + xx[[i]]],{i,1,20,1}]
Plot[mysin[t],{t,0,2 \[Pi]}]

The net result will be the noisy waveform shown below. Notice that the amplitudes exceed ~10 but the maximum magnitude of sine is 1.0. The larger amplitude results from constructive interference.

If we only let the phases vary on the interval $[0, \pi]$ then we get almost entirely constructive interference seen as the "fuzzy" sine wave below. yy := RandomReal[{0,\[Pi]},20]
mysin2[t_] := Sum[Sin[t + yy[[i]]],{i,1,20,1}]
Plot[mysin2[t],{t,0,2 \[Pi]}]

If a thousand people whisper inaudibly, will the resulting sound be audible? (...assuming they are whispering together.)

The answer is basically yes precisely because of the effect seen in the first example above. This is also why a swamp full of frogs or crickets can sound almost deafening even though each individual is not very loud.

I believe the answer is "yes" because the amplitudes would simply add and thus reach an audible threshold. Is this right?

Some add yes, but some "subtract," which is what I meant by destructive interference. This is why the first sine wave example above looks like a mess.

The second example waveform would be an extremely idealized result of an orchestrated crowd whispering in unison. However, the sound produced by speaking is almost never a nice, single sine wave like this but rather a lot of sine waves that have a modulated envelope.

Think of it like speakers. If you have one speaker at a specific volume and then you add a second speaker in the range of the first the sound volume will increase.

• Not necessarily. See active noise control en.wikipedia.org/wiki/Active_noise_control – user1583209 Aug 22 '17 at 8:34
• @user1583209 I tried destructively interfering with my friend's whispering by whispering 180° out of phase, but it didn't work. Can you show me you doing it? – Sneftel Aug 22 '17 at 8:40
• @Sneftel Neither did whispering in phase work, I presume; which is exactly my point. – user1583209 Aug 22 '17 at 8:53