Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Does a sound at 50dB at 1m have the same intensity of a sound of 51dB at 10m, and also the same intensity of a 52dB sound at 100m?

share|cite|improve this question
up vote 2 down vote accepted

The answer to your question is, sometimes, but it depends on the source and your hypothesised relationship approximately holds in some cases.

In three dimensions, your relationship does not hold. If the sound source is small, then its pressure field outside the source's hardware will be a general multipole scalar field, i.e. a general superposition of spherical waves each fulfilling Helmholtz's equation $(\nabla^2 + k^2)\psi = 0$ where $k = 2\pi/\lambda$ is the wavenumber at the frequency in question in the medium in question:

$$\psi(r, \theta,\phi) = \sum\limits_{\ell=0}^\infty \sum\limits_{\nu= -\ell}^\ell \psi_{\ell,\nu}\,j_\ell(k r) P^{|\nu|}_\ell(\cos(\theta)) e^{i\,\nu\,\phi}$$

where $(r,\theta,\phi)$ are the spherical co-ordinates of the point in question, $P^{|\nu|}_\ell$ are the associated Legendre functions and $j_\ell$ are the spherical Bessel functions of the first kind and order $\ell$. In the farfield, i.e. when $k r \gg \ell$ we have $j_\ell(k\,r)\approx \sin(k\,r - \ell \pi/2) / r$ so that the wave's intensity varies like $1/r^2$ in the farfield.

Therefore, a factor of 10 increase in the distance from the source leads to a factor of 100 decrease in the intensity (radiated power per unit area), which in decibel terms is a loss of $10\log_{10}100 = 20{\rm dB}$. Therefore, to match the intensity of a 50dB source at 1m distance, you're going to need a 70dB source at 10m and a 90dB source at 100m. You can get this result imagining the source is like an isotropic radiator with some, say dipole, radiation pattern. There will be an inverse square intensity dependence on dustance.

However, suppose your source is somehow cylindrical. Maybe it is like a paper cone loudspeaker but the cone is replaced by a very long paper cylinder radiating in and out. If you are near enough to the cylinder that it can be approximated as being infinitely long, then you've essentially gotten yourself a two dimensional problem, the waves are now cylindrical waves:

$$\psi(r, \theta) = \sum\limits_{\ell=0}^\infty \sum\limits_{\nu=-\infty}^\infty \psi_{\nu,\ell}\,J_\nu(k_\ell r) e^{i\,\nu\,\phi}$$

where now $J_\nu$ is the Bessel function of the first kind and order $\nu$. Now in the farfield, $J_\nu(k_\ell r)\approx\sqrt{2/\pi}\cos(k_\ell r -\nu\pi/2 -\pi/4)/\sqrt{r}$ so that the wave's intensity varies like $1/r$ in the farfield and now a factor of ten increase in distance corresponds to a loss of 10dB. Therefore, to match the intensity of a 50dB source at 1m, you would need a 60dB source at 10m way or a 70dB source at 100m away.

share|cite|improve this answer

The scale of sound pressure (decibel) is logarithmic $$ L_p=20\log_{10}\left(\frac{p}{p_{ref}}\right) $$ With $p_{ref}$ a reference pressure, with a commonly used value of $20\mu Pa$ according to Wikipedia, because it is roughly the threshold of human hearing.

Due to this definition the intensity/pressure roughly doubles for every 6 dB ($20\log_{10}(2)=6.020599913...$).

A wavefront of sound, when not obstructed, propagates like a sphere. This means that whenever you double your distance from sound producing source, you will hear the sound half a loud. $$ L_{p2}=L_{p1}+20\log_{10}\left(\frac{r_1}{r_2}\right) $$ From this you should be able so derive yourself that what you are asking will not be true (assuming that all sounds are measured from the same reference distance).

share|cite|improve this answer

This depends on things like the shape of the pressure wave. You're probably thinking of a point-source w/ an expanding spherical pressure wave, in which case the equations for energy per unit area as a function of radius are pretty straightforward (but keep in mind the log function involved in dB). As presented at Wikipedia, [Edit: apologies for the errors induced when copypasting markup. it's correct now]

When sound level Lp1 is measured at a distance r1, the sound level Lp2 at the distance r2 is Lp2=Lp1 + 20 * log10(r1/r2)

However, if for example there were an infinite plane wave, there would be no decrease in sound pressure as it propogates.

PS did you test it on a molpy?

share|cite|improve this answer
No, I didn't test it, mine was only curiosity. – HAL9000 Jan 2 '14 at 12:39

Taking the question exactly as written (what else can you do?)

The decibel level is the sound intensity experienced, in logarithmic form. It indicates the amount of sound energy reaching 1 square meter in 1 second, at the point of measurement.

Therefore, a sound intensity of 50 dB is less intense than one of 51 dB which in turn is less intense than one of 52 dB. The particular combination of location and type of source that produces these sound intensities is irrelevant.

What one could do from the given information is calculate the sound power output of the various sources at their respective distances.

The decibel level is not a property of a sound source; it is a measure of the effect of that source at a particular, specified spot. You cannot refer to a sound source as producing a certain decibel level without specifying exactly where in relation to the source, the dB meter is located.

For example, I have seen references to a jet plane at take-off power as an example of 110 dB. Now, I have no idea how many jet planes are taking off right now, but none of them are producing any decibel level at my ears. But, if you are trying to sleep in a house $50\text{ m}$ from the runway, the decibel level you experience would be very different.

Similarly, a parent may be happy to hear that his child's new toy fire truck (with siren) produces only 40 dB at the parent's eardrum when used in the next room, but a safety council would be more interested in the 135 dB produced at a child's eardrum when the truck is held against the child's head.


The OP seems to describe three events: experiencing a sound level of $50 \text{ dB}$ from a source $1 \text{ m}$ away; experiencing a sound level of $51 \text{ dB}$ from a source $10 \text{ m}$ away; and experiencing a sound level of $52 \text{ dB}$ from a source $100 \text{ m}$ away.

Compared to the source $1 \text{ meter}$ away, the second source,$10 \text{ meter}$ away would need around $126 \times$ the power, and the third, $100 \text{ meter}$ away, would need about $15850 \times$ the power.

share|cite|improve this answer
Well, actually, dB is the log of the ratio of the pressure where you are relative to some standard value or location. For example, dBm is short for log(power/1mW) for electrical stuff, and acoustics often refers against 20 micropascals pressure. – Carl Witthoft Jan 2 '14 at 18:39
@CarlWitthoft: $ \times 10 $ to convert Bels to decibels – DJohnM Jan 2 '14 at 18:55

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.