What is the smallest frequency difference that can be detected by the human ear?

In summary: If both the frequencies appear separately, then we need to obtain the minimum change in frequency the ear can measure.If both the frequencies appear separately, then we need to obtain the minimum change in frequency the ear can measure.
  • #1
burningbend
31
0

Homework Statement


Consider two sound sources S1 with a frequency of

f1 = 3367 Hz

and S2 with a frequency of

f2 = 3362 Hz.

Determine the smallest difference in frequency from 3367 Hz that the average person can detect?

Homework Equations


The only thing I can think of is approximating the ear as a closed tube, so f=nv/4L. Also, beat f = 5Hz.

The Attempt at a Solution


I'm honestly completely clueless here. I've never heard of a formula that deals with this and I can't find anything online. Is there an actual formula that I can use here? Maybe something related to the beat vs the fundamental frequency of the ear?
 
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  • #2
burningbend said:
Determine the smallest difference in frequency from 3367 Hz that the average person can detect?

Homework Equations


The only thing I can think of is approximating the ear as a closed tube, so f=nv/4L. Also, beat f = 5Hz.

The Attempt at a Solution


I'm honestly completely clueless here. I've never heard of a formula that deals with this and I can't find anything online. Is there an actual formula that I can use here? Maybe something related to the beat vs the fundamental frequency of the ear?

If two sources of sound having slightly different frequency are sent say before an observer with say his ears as detectors
then he can find the difference in frequencies - say n1=480 and n2 =484 by hearing the superposition produced by those waves which are waxing and waning of sound per sec - these are called ' beats' - i know your frequencies are in audible range ;
for details you see
http://hyperphysics.phy-astr.gsu.edu/hbase/sound/beat.html
 
  • #3
burningbend said:
I'm honestly completely clueless here
Same with me. Is your problem statement the litteral rendering ? It seems so strange and the last sentence almost disconnected from the first part.

Does this part of your text talk about beats ? Then the 'two sources' part suggests you should interpret this as asking what is the lowest beat frequency you can distinguish.

But if the material is about physiology, the term 'pitch resolution' is much more logical. Also to be found at hyperphysics . But I don't know how to translate the resolution from 1 Hz at 440 Hz to 3367 Hz...
 
  • #4
OK here is a factual piece of information. The human ear can differentiate between two frequencies if their difference is greater than 16Hz.

So our hero here cannot differentiate between the two given frequencies. Thus the minimum difference in frequency will be 16-(3367-3362)= 11Hz.
 
  • #5
CrazyNinja said:
factual piece of information
Any references ? Is 60 - 76 Hz equally distinguishable as 11000 - 11016 Hz ?

And what is this 11 Hz ?

Link in post #3 "The normal human ear can detect the difference between 440 Hz and 441 Hz" .
 
  • #6
CrazyNinja said:
The human ear can differentiate between two frequencies if their difference is greater than 16Hz.

So our hero here cannot differentiate between the two given frequencies. Thus the minimum difference in frequency will be 16-(3367-3362)= 11Hz.

There is a small elaboration : The human ear can differentiate between two frequencies which occur simultaneously if their difference is greater than 16 Hz. This implies that the human ear can detect the phenomenon of beats only if the difference between the two frequencies is 16 Hz.

Thus, whether our hero can distinguish between these two frequencies depends on whether they are occurring simultaneously or separately. The 440 Hz and 441 Hz example is correct because they occur separately .

If in the question, the frequencies appear simultaneously, then the hero's ear cannot detect them separately. If f2 is increased by 11 Hz, then the difference between f1 and f2 (new) is now 16 Hz and the ear can now detect the two as different.

If both the frequencies appear separately, then we need to obtain the minimum change in frequency the ear can measure.
 
  • #7
Yeah, that's the whole thing. This is for a student I'm helping, so I didn't have access to the text while we were working, but there was absolutely no context in the questions around it. Not a clue here.
 
  • #8
CrazyNinja said:
If in the question, the frequencies appear simultaneously, then the hero's ear cannot detect them separately. If f2 is increased by 11 Hz, then the difference between f1 and f2 (new) is now 16 Hz and the ear can now detect the two as different.

well in our experience in the labs the no. of peaks of intensity of sound after superposition could be counted per sec if the difference is a few only say 2 or 3 - that we did by applying a bit of wax on the tuning fork of one frequency out of a pair of identical freq. tuning forks.This is waxing and waning of sound Intensity and is not related to exact frequency observations. to hear sixteen changes is two much.
 
  • #9
There seems to be some confusion in the thread between pitch differences that can be detected in tones played serially and in tones played concurrently. If concurrent, I see no reason for a lower limit of detection to exist. Maybe after a couple of days the listener might be unsure of the orginal loudness.
For serially, at https://en.m.wikipedia.org/wiki/Psychoacoustics it says a difference of 3.6Hz for the octave 1000-2000Hz. Given the logarithmic nature of musical scales, it does seem likely that the discernible difference will vary according to the absolute frequencies, though perhaps not linearly.
As for the question as stated, it is fairly evident that something has been left out.

I agree with BvU that the calculation in post #4 arriving at 11Hz makes no sense.
 
  • #10
My reference is a book titled Concepts of Physics- 1. Its by Dr HC Verma and is quite popular here in India... though I doubt it is internationally recognized.
@haruspex did you go through post #6? I elaborated on how the 11Hz was achieved. Could you ecplain why it didnt make sense?
 
  • #11
drvrm said:
well in our experience in the labs the no. of peaks of intensity of sound after superposition could be counted per sec if the difference is a few only say 2 or 3 - that we did by applying a bit of wax on the tuning fork of one frequency out of a pair of identical freq. tuning forks.This is waxing and waning of sound Intensity and is not related to exact frequency observations. to hear sixteen changes is two much.

Which frequencies did you use for the tining forks?

EDIT: It does not matter.Im sorry. I didnt see that you used identival tuning forks.
 
  • #12
CrazyNinja said:
Which frequencies did you use for the tining forks?

well its about 30 yrs back- as i remember ...but some frequencies like 256, 482...something around 324 were in use.
Moreover there was one Sonometer experiment alongside -which utilized a stretched wire of variable length and when the frequency of a tuning fork matched with stretched wire there was resonance - but near the exact resonance when the frequencies were close but differed by a small number only- we could hear the beats
 
  • #13
CrazyNinja said:
@haruspex did you go through post #6? I elaborated on how the 11Hz was achieved. Could you ecplain why it didnt make sense?
I can see that it makes sense if you interpret the question as "how much would you have to increase the 3367 by so that it can be distinguished from 3362?" But since the question does not say that or anything very much like it, that does not seem a very fruitful avenue.
If we just look at the explicit question
burningbend said:
Determine the smallest difference in frequency from 3367 Hz that the average person can detect?
there is no mention of 3362.

With regard to the discrimination of tones played together, maybe you have it backwards. If two equal amplitude tones differ by 1Hz, reliably, I will hear them beating from full cancellation to full reinforcement and back once a second. That would be very easy to detect. It would be somewhat harder at 16Hz.
For this reason, I feel sure the question is supposed to be in respect of tones played consecutively.
 
  • #14
OK I am sorry. I made a mistake in reading the text properly. The text I mentioned before says

"The beat frequency must not be greater than 16 Hz."

I read it wrong. I hope I'm forgiven. OK, now the question makes no sense. I'm with haruspex on this one. Sorry for bothering you BvU.
 

1. What is sound wave interference?

Sound wave interference is when two or more sound waves interact with each other, resulting in either a reinforcement or cancellation of the waves. This phenomenon is caused by the overlapping of the waves, which can create areas of constructive interference (where the waves amplify each other) or destructive interference (where the waves cancel each other out).

2. How does sound wave interference affect the quality of sound?

The interference of sound waves can greatly impact the quality of sound. In constructive interference, the amplitude of the sound waves is increased, resulting in a louder and clearer sound. In destructive interference, the waves cancel each other out, causing a decrease in amplitude and a decrease in sound quality. This is why sound engineers must carefully consider sound wave interference when setting up audio systems.

3. Can sound wave interference be controlled or manipulated?

Yes, sound wave interference can be controlled and manipulated. This is commonly seen in noise-cancelling headphones, where a second sound wave is generated to cancel out external noise. Sound engineers also use techniques such as phase cancellation to manipulate sound wave interference in a controlled manner.

4. What are some real-life examples of sound wave interference?

One common example of sound wave interference is the "doppler effect", where the sound of a passing vehicle changes in pitch as it approaches and then moves away from the listener. This is caused by the interference of sound waves from the moving vehicle. Other examples include echoes, reverberation in concert halls, and the buzzing sound produced by some insects.

5. How is sound wave interference related to the concept of wavelength?

Sound wave interference is closely related to the concept of wavelength. When two sound waves with the same wavelength overlap, they will either reinforce or cancel each other out. This is because the peaks and troughs of the waves align, resulting in either constructive or destructive interference. The distance between these peaks and troughs is the wavelength of the sound wave.

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