4
$\begingroup$

This question already has an answer here:

At night, we hear weak and far sounds approximately clear, while during the day we cannot.

My high school physics teacher was saying that “this is because of interference of sound waves. During the day, there are a lot of sounds and they cancel each other due to interference. But, during the night, there are few sounds and they can reach to our ears without canceling each other”.

But, this doesn’t make sense because even in silent days (according to my personal experience), we don’t hear those sounds that night are heard clearly.

As I am not familiar with waves so much, I will appreciate if someone clear me by simple explanation.

$\endgroup$

marked as duplicate by Rob Jeffries, user36790, Floris, John Rennie, Qmechanic Jun 24 '16 at 9:04

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • 1
    $\begingroup$ this question is a little dubious; there may be many circumstances in which we hear better at night, or the premise may be altogether false. I can't think of any being specifically caused by darkness/nighttime which aren't otherwise explainable $\endgroup$ – anon01 Jun 23 '16 at 20:40
  • 3
    $\begingroup$ In general, what our senses tell us and what is physical reality do not scale linearly. So I am actually questioning whether this is an effect of how our senses work or an actual physical effect ... $\endgroup$ – Sanya Jun 23 '16 at 21:14
  • 5
    $\begingroup$ Possible duplicate of Why do I always hear remote train horn at night? $\endgroup$ – Rob Jeffries Jun 24 '16 at 0:14
  • 1
    $\begingroup$ The answer is the change in refractive index of air with temperature. And it is a multiple duplicate. See also physics.stackexchange.com/q/52269 and physics.stackexchange.com/q/255844 and physics.stackexchange.com/q/128144 $\endgroup$ – Rob Jeffries Jun 24 '16 at 0:15
  • 1
    $\begingroup$ lucas - the update pretty much ruins the question. If you had left it as "why do we hear better at night" it's a perfectly good question. Adding in an unfounded assumption on EM radiation into the question means you get a downvote from me. $\endgroup$ – Rory Alsop Jun 24 '16 at 7:35
6
$\begingroup$

My high school physics teacher was saying that “this is because of interference of sound waves. During the day, there are a lot of sounds and they cancel each other due to interference. But, during the night, there are few sounds and they can reach to our ears without canceling each other”.

You need a better high school physics teacher.

Temperatures tend to decrease with altitude above ground during daytime. This acts to curves sound upward. Thus in turn means you cannot hear the sound of a nearby train (a kilometer away or so) blowing it's crossing whistle. The sounds of that train are directed upwards into the atmosphere, where they dissipate.

At night, the atmospheric boundary layer tends to develop a marked temperature inversion, up to over a kilometer high. This acts to curve sound downward. This in turn means that at night you can hear the crossing whistle of a train that is from several kilometers from you. You can hear the train's progress along its track as it blows it's whistle at one crossing, and then another, and then yet again another. Even if the train was the only noisy object in the daytime, you could not hear that remote whistle in the day. You can only hear it at night.

The reason for this upward diversion of sound in the daytime versus the downward diversion at night is the strong dependency of the speed of sound in the atmosphere on temperature. The atmosphere acts like a lens that focuses sound energy upwards during the day, but keeps it at ground level during the night.

$\endgroup$
  • $\begingroup$ Am I right that waves propagating across the gradient won't be refracted in theory (although because of inhomogeneities they would)? $\endgroup$ – Andrii Magalich Jun 24 '16 at 7:15
  • $\begingroup$ Some linked an answer that links to these animations: acs.psu.edu/drussell/Demos/refract/refract.html . If they are correct, what exactly enhances the hearing? Naively, this would influence only the timing of the waves, but not the loudness $\endgroup$ – Andrii Magalich Jun 24 '16 at 7:19
  • $\begingroup$ @AndriiMagalich -- Any change in the speed of propagation of a wave phenomenon results in the wave being refracted. Here the change is continuous rather than discontinuous (e.g., light entering water), but the same result happens: The sound rays are bent in the direction of lower propagation velocity, upwards in the case of temperature decreasing with altitude (typical daytime), or downwards in the case of temperature increasing with altitude (typical nighttime). $\endgroup$ – David Hammen Jun 24 '16 at 15:52
  • $\begingroup$ I think you imply that at nighttime the "soundrays" bend to the ground and more of them hit us. Which does not seem right and animations I linked agree with my feeling $\endgroup$ – Andrii Magalich Jun 24 '16 at 15:55
  • $\begingroup$ @AndriiMagalich -- Forget the animation. You aren't seeing it correctly. (It does not help that this is not a well-done animation.) Look instead at the two images on the left of your linked page. $\endgroup$ – David Hammen Jun 24 '16 at 16:01
3
$\begingroup$

I would tend to agree that background noise is a factor, but rather than reducing, adding to the sound you are trying to make sense of. So part of that may be how your brain is able to filter the information from the background noise.

But at night the temperature is lower and according to this tutorial on sound propagation (which does cite reliable references), air has an energy absorption factor that is a function of temperature: $$\alpha = 869 f^2 \left\{1.84\cdot 10^{-11} \left(\frac{T}{T_0} \right)^{1/2} \!\!+ \left(\frac{T}{T_0} \right)^{-5/2}\left[\frac{0.01275 e^{-2239.1/T} }{F_{r,0}+f^2/F_{r,0}}+\frac{0.1068e^{-3352/T}}{F_{r,N}+f^2/F_{r,N}} \right] \right\} $$

and you can see here that a reduced temperature, $T$ reduces the absorption factor by the square root of $T$ in one component and by an exponent of -5/2 in another. So by reduction in the absorption of energy (by air molecules) in the path of the sound, more energy will reach your ear in the colder temperature.

$\endgroup$
  • 1
    $\begingroup$ Nice post and a very thorough tutorial as a reference. $\endgroup$ – Bob Bee Jun 24 '16 at 3:24
  • $\begingroup$ Absorption is a secondary effect. The temperature gradient acts as a focusing mechanism, as explained in David Hammen's answer (and the suggested duplicate "Why do I always hear remote train horn at night?") $\endgroup$ – Floris Jun 24 '16 at 4:30
-1
$\begingroup$

If we suppose that the phenomenon you describe is related with wave interference. A wave is a kind of mechanical disturbance in the medium through which it is travelling. A sound wave consists of areas of relatively high and low energy, in the form of relatively high and low pressure. To understand how sound is produced, consider a speaker. The cone or diaphragm of a speaker vibrates inwardly and outwardly in response to an electrical signal. These vibrations are typically very small, only visible with larger speakers. However, they all impart energy to the air in the same way. When the cone moves outward, it pushes the air forward that originally occupied that space. This air becomes locally compressed, forming a region of relatively high pressure. When the cone moves inward again, it recoils from the space that it occupied and leaves behind a partial vacuum, a region of relatively low pressure. The frequency and amplitude of the vibrations change the characteristics of the wave that is formed, and hence the sound that we perceive. When we hear the sound, our ears are being bombarded with air molecules of rapidly varying pressures. The signal is sent to the brain where it is interpreted.

As the sound wave progresses through the air, its energy slowly dissipates. This is why sound is louder closer to the source and quieter further from the source.

Wave interference occurs when two or more waves disturb the same air molecules. If a relatively high energy part of one wave combines with a relatively low energy part of another, the result is a region of air with an average of the two. In the most extreme case, the resulting pressure is indistinguishable from that of the undisturbed air, and is therefore undetectable by the ear. This situation is known as total destructive interference. In practice, however, interference is almost always partial. Similarly, if two high- or two low-energy parts of a wave combine, they can be summative. This opposite process is known as constructive interference.

To visualise this process, you may want to look at a video or two on-line. Note that water waves illustrate the concept extremely well, but the mechanisms by which they function are very different to the process described above.

Both constructive and destructive interference occur frequently wherever there are multiple sounds. However, their effects are generally minor in the natural world. Some of the commentaries above probably provide a more accurate explanation of the phenomenon you describe.

$\endgroup$
  • $\begingroup$ This has nothing to do with interference. Suppose the only noisy object is a train blowing it's crossing whistle several kilometers away. You oftentimes can easily hear this during the night but rarely can hear it during the day. The answer lies in refraction rather than interference. $\endgroup$ – David Hammen Jun 24 '16 at 4:11
  • $\begingroup$ You may have noticed from the wording of my answer that I was not implying that it was. Rather, I was try to explain the concept which the post had directly addressed, especially since the person posting the question stated that he had limited understanding of waves. $\endgroup$ – POD Jun 24 '16 at 4:15

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