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The intensity of a wave is proportional to the square of the amplitude of the and the frequency of the wave given by the equation:

$$I = \frac{\overline{P}}{S} = 2\pi^2\rho vf^2A^2$$

Now when talking about sound waves the sound level of a noise (loudness), it is proportional to the intensity of the source. If intensity of a wave is proportional to frequency, why doesn't sound level, in general, depend on frequency. i.e. Higher frequency, higher intensity, higher sound level.

Why is this not the case?

For example if you change the frequency of a sound wave from 500Hz to 15000Hz, this won't necessarily be perceived as a change in loudness. But hasn't the intensity of the sound increased, because the frequency has increased? The loudness of a sound does not seem to depend on frequency. But according to the equation above, intensity DOES depend on frequency. So why does it not change perceived loudness?

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    $\begingroup$ Loudness is not proportional to intensity. hyperphysics.phy-astr.gsu.edu/hbase/sound/loud.html $\endgroup$
    – Farcher
    Feb 27, 2016 at 15:32
  • $\begingroup$ Not sure what you are asking. If you have the same displacement, you will have greater pressure at higher frequency. Sound level depends on the energy in the wave, which depends on the pressure; and pressure is a function of (displacement) amplitude and frequency. $\endgroup$
    – Floris
    Feb 27, 2016 at 16:33
  • $\begingroup$ @Farcher yes it does... but loudness is not SIMPLY intensity as your link points out. $\endgroup$ Feb 27, 2016 at 17:41
  • $\begingroup$ @JaneJacobson Sorry, I should have written "not just proportional". $\endgroup$
    – Farcher
    Feb 27, 2016 at 17:54

1 Answer 1

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You make several assertions in your "For example" paragraph which simply aren't true. I'm curious as to your sources for these statements, or are they simply a statement you've formed yourself?

If you change only the frequency of a source (constant amplitude), you most definitely change the intensity of the wave, so the level will change. Plus, you are confusing loudness (a psychoacoustic perception) with level (a logarithmic ratio of intensities). To answer your final question: changing frequency with constant amplitude does change the loudness.

What you may seeing in the physics relationships is $$\beta=\log_{10}\left(\frac{I}{I_0}\right)$$ where $\beta$ is the intensity level in units called bels (decibels would be $10\times$ this), $I$ is the intensity of a wave, and $I_0$ is some reference intensity to define a level of $0$ bels.

While you don't explicitly see the frequency in this formula, it's there. It is true that if you have two sources of equal intensity and different frequency, they will have the same level, $\beta$.

Loudness is affected not only by amplitude and frequency, but also by the ear mechanism and the brain. Equal level sounds of differing frequencies have differing loudnesses. You can research Fletcher-Munson curves.

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    $\begingroup$ thanks for the reply. I am just reading a lot about sound and several sources are saying the frequency will change the "pitch" of the sound and the amplitude will change the energy of the wave, thus changing the "loudness." Say if I create a tone using a computer program, changing the frequency from 500Hz to 15000Hz doesn't change the intensity of the sound, only the pitch? Shouldn't a 14500Hz increase significantly change the intensity and "loudness"? $\endgroup$ Feb 28, 2016 at 0:26
  • $\begingroup$ If the speaker you have has constant efficiency at all frequencies, you should be able to measure an increase in intensity and possibly loudness. If you use a regular microphone, it will measure amplitude, not intensity. There is extra processing needed to get the intensity (and hence level). For most adults, the loudness drops to close to nothing at 15000 Hz because of age-related stiffness in the auditory pathway. Children can hear up to around 20000 Hz, but the loudness drops rapidly above about 12000 Hz. Efficiency of the ear at high frequencies is low. It's best around 2000-3000 Hz. $\endgroup$
    – Bill N
    Feb 28, 2016 at 3:20

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