# What does it mean to play a note for half a second?

This is my first serious look at the science side of music, as I am getting started with audio programming, and I am greeted by an awkward bouncer first up.

I know that a note is simply a certain frequency for which we have a name. For example, a sound wave with a frequency of 440 Hz is recognized by Western music as A4. Now, this means that in order to hear an A4, the molecules/atoms/whatever of the medium must vibrate 440 times within one second. So, playing an A4 for half a second doesn't make sense to me because you'd have only 220 vibrations, and that is far from being an A4. I am sure humans are sensitive to that kind of difference in frequency. Are we not?

Does playing an A4, or any note, for half a second (or in under a second in general) make sense? If it does, please explain.

If I am wrong, and I have an inkling that I might be, what have I missed in my understanding that is leading to this paradoxical situation?

(I find this situation strange because I can hear a note when I strum my guitar and damp the string immediately. I haven't timed it but am sure I can damp it under a second. Yet I hear the note. May be a few Hz lost in the final microseconds doesn't make much difference.)

• Your car travels at 60 miles per hour. You drive it for 1/2 of one hour. Do you mean to tell me that the car has not traveled any distance at all, because it failed to travel for a full hour? Of course not. Your car has traveled 30 miles. Similarly, your note vibrates at the rate of 440 vibrations per second. You play the note for 1/2 second. Do you mean to tell me that your note made no sound? Of course not. Your note made a sound for 0.5 seconds. I think your problem is that you do not understand maths. – user1044 Apr 25 '15 at 19:59
• @WheatWilliams I think what I don't understand is the physics of sound, maths I do. – vin Apr 25 '15 at 20:03
• in order to hear an A4, the ...whatever of the medium must vibrate 440 times within one second. That's where you go wrong. They must vibrate at the rate of 440 cycles per second, not necessarily vibrate 440 actual times in an actual second. It's the difference between actually driving 60 miles in an hour, and driving at 60mph. – topo Reinstate Monica Apr 25 '15 at 23:25
• @topomorto aah, so its the little word rate that was missing in my understanding. Got it. – vin Apr 26 '15 at 6:07
• This thread has been revived, otherwise I would not have commented, but I think one feature that is missing in all of the answers is that a human ear does not take a single point sample over time, but rather multiple point samples with spatial separation simultaneously (and in parallel). I think this matters in a way I cannot quite articulate, and probably reduces the time needed to identify a waveform. – Yorik Aug 30 '17 at 21:23

You only need 1 cycle to actually state the pitch, so minimum 1/440s - it would take a human much longer to recognise that pitch, perhaps with a good ear a 1/100s, but a computer, given a pure sine wave, could get it in one cycle [or technically half that, as the 2nd half is a mirror of the 1st].
…or, as has been mentioned in comments, shorter than that if guaranteed a sine wave input.

• I guess that I learned a rule of thumb that is slightly more stringent than you. – Dave Apr 25 '15 at 19:54
• @Dave - you made me have to think for a minute… is it 440 complete cycles, or 220 'halves'… but it is complete cycles == 440/s so 1/880s for a computer algorithm to be able to be certain [assuming pure sines all the way]. – Tetsujin Apr 25 '15 at 19:56
• Surely you only need one cycle to state the pitch as long as you're also given the information that what you've 'seen' is one cycle. Otherwise you have no clue whether what you've 'seen' so far is actually the whole cycle or not..? The first zero crossing doesn't necessarily represent the end of the first half of the waveform, much as the second zero crossing doesn't necessarily represent the end of the waveform. – topo Reinstate Monica Apr 25 '15 at 20:02
• if the input is guaranteed pure sine [which makes it a maths question more than 'music' I admit], you only need to 'see' it pass the same point twice in the same direction for a full cycle, or pass zero twice for anything less. You'd need someone smarter than me to extrapolate from a partial sine with less info, but it can probably be done, computationally. – Tetsujin Apr 25 '15 at 20:05
• Since the ear effectively does frequency analysis an uncertainty relation applies en.wikipedia.org/wiki/Fourier_transform#Uncertainty_principle -- thus a very short short signal does not have a well defined frequency, rather is has a spread of frequencies. – Dave Apr 25 '15 at 20:49

Eggs are one dollar per dozen. But you can have 6 for 50 cents. Same eggs, same price, same deal. The lowest note on the piano, A0, has a wavelength of something over 12m. So can you hear it in a room 6m long? Sure you can. You can even hear it when using headphones - which could be thought of as a "room" measuring only 1cm or so! And you can be travelling at 100 m.p.h. even though you stop after 5 minutes. See what I'm getting at? Just because we measure the frequency of a note in vibrations per second, it doesn't mean we need to hear a full second's worth to recognize it. Music is FULL of notes WAY shorter than one second. (And of ones way longer, of course.)

• Actually, I look at it more like laps in a marathon, so I can't see how it works. – vin Apr 25 '15 at 19:37
• Not sure what you're saying there, but if you've latched on to an analogy that doesn't work, ditch it and find a better one! – Laurence Payne Apr 25 '15 at 19:40
• @thePetProjectProgrammer how are you relating laps in a marathon to frequency? – topo Reinstate Monica Apr 25 '15 at 19:51
• @topomorto If a marathon (indoor) needs N laps then running 0,98N won't be considered as completing the race. Similarly if the necessary freq isn't reached the note shouldn't happen. – vin Apr 25 '15 at 20:06
• @thePetProjectProgrammer that's confusing the definition of a unit with the idea of a value in that unit. I can only restate what others have said - that a car doesn't have to travel a whole hour for you to be able to state its speed in miles per hour. You can state its speed at any instant... – topo Reinstate Monica Apr 25 '15 at 20:20

Yes, it still makes 220 complete oscillations in one half second. As long as the number of oscillations is significantly greater than 1, you'll still hear the pitch. You'll only need to worry about these kinds of effects when the durations of the notes get down below the order of 1/100ths of a second (maybe order of 10ths of a second for bass notes).

• This answer looks promising, but I still don't understand how 220 Hz could sound like 440Hz. A bit of explanation would help. – vin Apr 25 '15 at 20:15
• I suspect you're trolling, but here goes anyway :-) 220 vibrations in half a second IS 440 in a whole second (assuming the pitch holds constant). Both are 440Hz. – Laurence Payne Apr 25 '15 at 20:22
• @thePetProjectProgrammer I never said 220Hz -- I pointed out the 220 oscillations to emphasize that there are still a large number of complete cycles even in one half of a second. The only time when you'll run into a problem is when the duration of the note is so short that it can't complete even a few of them. – Dave Apr 25 '15 at 20:55
• @LaurencePayne Its amazing how you conclude that I am trolling, I obviously missed the half when reading the first sentence. I don't care what you accuse me of, but nonetheless its amazing. – vin Apr 26 '15 at 6:02
• @Dave ok, got it. – vin Apr 26 '15 at 6:08

The unit of frequency, Hertz (Hz), is defined as the reciprocal of the unit of time, the second.

A vibrational frequency given in Hz, like your 440Hz A note, can be converted to a vibrational period in units of seconds by taking the reciprocal. 1/440 = 0.00227, so a tone with a frequency of 440Hz has a period of 0.00227 seconds.

In that amount of time, your guitar string has completed a single cycle of vibration. Within two or three of those periods, your ear and brain can "lock on" to those vibrations and comprehend the pitch with reasonable accuracy.

Time and frequency have an "Unschärferelation" which mean that you need a certain time to be able to recognize a frequency with a particular certainty for a given precision and a given noisiness. A lot of other answers suggest that getting a whole period will magically allow you to recognize a frequency exactly while less won't. That's nonsense.

Assuming an exact measurement of a sine wave of unknown frequency, phase, and amplitude, three points will determine the sine wave. Those points may be almost arbitrary close.

But that does not answer the question of whether some frequency is part of a complex mixture with noise, and there is no sharp borderline.

As a rule of thumb, the frequency difference you want to be able to distinguish and the time span for examining the signal are indeed inverse, but there is a factor of proportionality coming in as well and also the overall expectation of noise.

The most compact probability distribution for a given energy is a Gaussian, and $\exp(-\pi t^2)$ has as its Fourier transform $\exp(-pi f^2)$. So we have here $\sigma_t^2 \sigma_f^2 = (2\pi)^{-2}$. This is a theoretical lower limit for compactness in time/frequency-space. Distinguishing frequency peaks of close frequencies requires suitably narrow distribution of the information in the frequency domain, requiring appropriately long measurement in the time domain.

A semitone corresponds to a factor of about 1.06, so telling two adjacent semitones at a bass frequency of 35Hz apart well requires a time window with a hand-waving duration of 300ms. In practice, we get along with a lot less because we do the pitch detection mostly on the overtones which have considerably higher frequency.

But if you use a rather overtone-lacking source of low notes, like an organ subbass windpipe, determining the exact notes in a fast bass run is just no longer possible. Add a mixtur or a reed pipe, and there is no problem whatsoever.

So basically: the higher the notes, and in particular the larger the frequency differences you want to be able to determine, the smaller the time interval you need for being reasonably sure.

A coloratura soprano can stuff as many notes into a phrase as she wants, and you'll hear every single note and how accurate it is (pity that the vowels all sound the same once the fundamental frequency leaves the speech formants behind). A basso profondo singing the same three octaves lower: not so much.

A bass recorder with its lack of overtones: quite a bit worse. Trilling on low notes is pretty pointless for that instrument.

To get back to your original question: half a second is actually sufficient for pretty much all pitch detection tasks. As the time interval shrinks, distinguishing frequency differences becomes harder, and this is spelling trouble first for the low notes where comparatively small frequency differences in Hertz already constitute a semitone of musical difference.

Your guitar or violin string would not exactly vibrate at 440 Hz but rather the oscillations of the air molecules around(up to you ears) would be 440 Hz.

Frequency is nothing but Total Oscillations divided by the Time Period. The SI Unit of Time is seconds so you have to convert the time to it if you want to get the answer in Hz.

440 Hz when written in the form p/q is 440/1 so, simply 440 Hz means that the object oscillates 440 times every second. So far so good I hope.

One should note that this is very similar to the formula for Average Speed, Total Distance/Total Time taken. The Number of oscillations, Time Period and Frequency correspond respectively to Total Distance, Total Time taken and Average Speed.

Pitch of a sound or rather the sound wave depends upon its frequency. 440/1 = 220/0.5 Both give the result 440 Hz. Pitch is independent of the Number of Oscillations or the time for which it oscillates and is only dependent on the frequency of the sound.

Scientific Reason: The disturbance of air molecules around us at specific frequencies at specific atmospheric conditions produce distinct pitches. This is why as temperature changes or pressure changes the pitch produced by the instrument changes. This also varies from instrument to instrument.

Note: Frequency referred to here is Average frequency and not instantaneous frequency similar to average velocity and instantaneous velocity.

Example: The first few oscillations may be really fast and the rest slightly slower as a result different instantaneous frequencies but the average will be the note's pitch. Our human ear can not detect such small changes and hence we usually hear it as one note.

Please check the Physics forum for further questions. Also read more on Harmonics if you are really really interested.