What is the formula for the sound pressure of a pure tone of 500Hz, ex-pressed as a function of time?

  • Might be okay here, but might be better off over on Physics.SE
    – Rory Alsop
    Commented May 6, 2013 at 7:29
  • I think this is relevant here - managing SPL is part of audio studio work.
    – Warrior Bob
    Commented May 6, 2013 at 16:05

4 Answers 4


Sound pressure level is directly related to the amplitude of the waveform.

A pure tone is a sine-wave and sine-waves are defined by ω (omega) and t (time)

amplitude = sin(ωt) --- "sin" is the mathematical operator you did in trigonometry at school and t is time.

ω = 2 * π * f --- π is 3.141592654 (approx) and f is 500

So for 500Hz, ω = 3141.5927 (approx)

You will find that if you recreate the 500Hz sinewave by producing samples at (say) 20kHz you will get a string of numbers (every 50 microseconds) that rise to a peak (+1) at 500µsecs then start to fall through zero at 1000µsecs and go negative to -1 at 1500µsec then fall back to 0 at 2000µsecs. The waveform repeats.

  • @JoshP - if you are going to edit make sure you do it correctly - the character "Ω" is WRONG - it should be "w"
    – Andy aka
    Commented May 6, 2013 at 12:50
  • 1
    How is this answer relates to the question?
    – Eugene S
    Commented May 6, 2013 at 13:47
  • Andy, fixed. My apologies. Also, if you see an incorrect edit, by all means, please correct it.
    – JoshP
    Commented May 6, 2013 at 13:58
  • @EugeneS the OP asked for a formula for a pure tone and the sound pressure is directly related to the amplitude of a sinewave
    – Andy aka
    Commented May 6, 2013 at 14:19
  • The formula you gave here (W=2*pi*f) is an an angular frequency which units are radians per second. I don't really understand how this relates to sound pressure level which is measured in dB..
    – Eugene S
    Commented May 9, 2013 at 2:13

The instantaneous sound pressure of a pure tone equals the ambient pressure (p0) with a superimposed pressure that varies in time as a sine function, i.e.: p(t) = p0 + A sin ωt, where A is the peak amplitude of the pressure variation and ω the angular frequency ( ω = 2πf ).

The amplitude A is related to sound pressure level L in dB by the following equation:

L = 20 log10( prms / pref ), where, in the case of a sine wave, the following equality holds: prms = A / √2. The reference pressure is arbitrary, but a fairly common value is 20 µPa, which would put a SPL of 0 dB around the threshold of human hearing.

Through substitution, we obtain the following for the pressure as a function of time of a pure tone with frequency f:

p(t) = p0 + √(2) 10L/20 pref sin 2πf


This was a question asked in a Digital Sound and Music exam which has been copy-pasted here (even the "ex-pressed" has been left intact). The question only gives two marks. I think this is the answer:

f(t) = Amplitude * sin(2 * π * Frequency * t)

So in this case the Frequency would be replaced with 500.


Sound Pressure Level (SPL) is a dB scale defined relative to a reference that is approximately the intensity of a 1000 Hz sinusoid that is just barely audible.

0dB SPL = 20 micro Pascal

Since sound is created by a time-varying pressure, sound levels computed in dB-SPL by using the average fluctuation-intensity (averaged over at least one period of the lowest frequency contained in the sound).

Generally, the intensity level of a sound wave is its dB SPL level, measuring the peak time-domain pressure-wave amplitude relative to 10^{-16} watts per centimeter squared (i.e., there is no consideration of the frequency domain here at all).

Reference and more complete explanation here.

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