For two waves, if it can be seen that after an interval of time no sound is heard, which option is liable for this incident? One option is interference of sound, and another is beat. I know that beat is one kind of interference. But sometimes I fail to make difference between them. If interference is liable for that incident, how will you explain the action of interference behind this? If I say that beat is not liable for this, what is the reason behind this?
[Physics] What are the differences between beat and interference
acoustics
Related Solutions
The reason you are hearing the train farther away is more consequence of the the geometry of different spaces than anything else.
It starts with an inversion layer of cold air clinging close to the ground. Just as glass bends light by making light move more slowly through it, an inversion layer of cold air bends sound because sound moves more slowly through cold air (the molecules move more slowly is why). So, this inversion layer behaves like the audio equivalent of a large sheet of glass covering the ground and guiding the sound away from the air above it. This is the same principle that allows a fully transparent optical fiber to capture light moving through it and transmit that light for many kilometers with very few losses. (@hwlau already noticed the inversion in the second of his three possible answers.)
On the ground side of the inversion layer, a fresh fall of snow further helps confine and preserve sound because it looks smooth to the long wavelengths of sound. So, even though snow thoroughly jumbles up the much shorter wavelengths of light, which is why it looks white, it looks very different and much more mirror-like to sound.
Put those two together -- diffraction on the top and reflection on the bottom -- and you have an example of two-dimensional sound dispersion. By way of contrast, sound dispersion from a train on a summer day that lacks any inversion layer and has sound-absorbing grass on the ground is an example of largely three-dimensional sound dispersion.
So, why is the dimensionality of the sound dispersion important?
Because sound (or any other radiation) disperses at a rate that is dependent on the number of dimensions of the space into which into which it is dispersing. If $L_n$ is the perceived loudness of the sound, $s$ is the distance to the sound source, and $n$ is the number of space dimensions into which sound is dispersing, the general equation for how loud the train will sound is:
$L_n = 1/s^{n-1}$
Notice that the lower the number of dimensions, the more slowly the sound disperses. (I discussed this same issue from a slightly different perspective a few months ago in my answer to this question about why objects look smaller when they are farther away.)
This equation explains why optical fibers can transmit light many kilometers without loss any significant of intensity ("loudness"). The dispersion space $n$ for optical fibers is $n=1$, so $L_1 = \frac{1}{s^{1-1}} = \frac{1}{s^0} = 1$. That is, there is no diminution of intensity. The audio equivalent would be a long tube, like the ones they used to use as intercoms in old houses (and still use in some playgrounds).
Now for your inversion layer case, $n=2$ and:
$L_2 = 1/s^{n-1} = 1/s$
But because your perception of train distance was tuned to $n=3$ space, you expected the sounds of the train to diminish at the much faster rate of:
$L_3 = 1/s^{n-1} = 1/s^2$
The analysis of how much farther away the train really is turns out to be trickier than it might seem. That's because the model I just described assumes that the energy of a 3D sound source can be compressed into a mathematically precise 2D plane. The physical world just doesn't work that way, since the sound energy in a 3D volume cannot be forced into a true 2D plane without creating infinitely high energy densities in the plane. Why? Well, pretty much for the reason that you cannot compress a 3D volume of air into an infinitely thin 2D plane without creating infinite mass densities. Crossing dimensionalities is often done a bit casually in physics, but one need to be careful with it.
So, in this case, instead of assuming a simple 2D plane, what you have to do is model the problem by using a "pancake" that more realistically represents the thickness of the inversion layer confining the sound. That allows sound intensities that "look" 3D in the immediate vicinity of the train, but then fade off more according to the dimensionality diffusion rules as distances increase to many times the thickness of the inversion layer.
So, everything from this point is obviously guesswork about what happened in your case, but a nice ballpark height for your inversion layer might be 10 meters. Approximating again, that 10 meters also becomes the "unit of equality" for distance from the train at which the sound of the train is perceived as the same in both cases. This approximation should work reasonably well for any more-or-less point source of sound coming from the train, in particular its whistle. So, call this unit of distance $s_u$ for hearing a similar loudness for the whistle $s_u = s_w = 10 m = 0.01$ km.
Alas, it gets messier. The sound of the train itself is anything but a point source, since you may be able to hear wheel-on-rail sounds for very long lengths, such as a kilometer for a long train. That also messes up the model and adds even more complexity in the form of orientation and sound delays. So, I'm going to wrap all of that complexity up into a single huge approximation and say that for a long train, the sound of the all the train wheels on all the track sounds "about the same" for anyone within a kilometer of the train as it passes by, inversion layer or not. So, the length unit for assessing how train track noises changes over distance becomes $s_u = s_t = 1$ km.
The equation now has to be altered slightly so that these "sounds the same" units $s_u$ are factored in:
Actual: $L_2 = s_u/s^{n-1} = s_u/s$
Perceived: $L_3 = s_u/s^{n-1} = s_u/s^2$
Solving for the $s$ distances in terms of loudness:
Actual: $s_2 = s_u/L$
Perceived: $s_3 = \sqrt{s_u/L}$
The error factor $e$ for how far off your distance estimate was then is:
$e = \text{(actual)}/\text{(perceived)} = s_2/s_3 = \frac{s_u/L}{\sqrt{s_u/L}} = \sqrt{s_u/L}$
For the train whistle, $s_u = s_w = 0.01 km$. With $L$ in km:
$e_w = \sqrt{s_u/L} = \sqrt{0.01/L} = 0.1/\sqrt{L}$
For the track noise from the entire train, $s_u = s_t = 1 km$. With $L$ in km:
$e_t = \sqrt{s_u/L} = 1/\sqrt{L}$
So, finally, a couple of very rough estimate examples are possible.
Assume the train is actually about $L = 16$ km away. In that case, the whistle sounds like it is $e_wL$ km away, or:
$e_wL = 16e_w = 16(0.1/\sqrt{16}) = 0.4$ km away.
In the same case, the train track sound will appear to be $e_tL$ km away, or:
$e_tL = 16e_t = 16/\sqrt{16} = 4$ km away.
So, not only are sounds moving through a winter inversion layer highly deceptive for estimating distances, they can be deceptive in different ways at the same time! A point source such as the train whistle may well sound like it is even closer than the train as a whole -- and both perceptions will sound way, way closer than the actual distance.
Here are some topics to read about:
- Frequency doubling, also called second-harmonic generation as Johannes mentions. Here, you put one wave into a medium, and some fraction of it is converted to a wave with a different frequency. By carefully engineering the medium you can get quite a high conversion percentage.
- Other nonlinear optical processes, not just frequency doubling, but includes difference frequency generation, optical parametric oscillators, N-wave mixing, etc. These involve putting one or more waves with different frequencies into a medium and, getting an additional wave out with a frequency that you didn't put in.
- Raman scattering which is spontaneous, and its stimulated equivalents CARS and CSRS. In Raman scattering you don't really get a coherent wave out; you just put one wave into a medium and a tiny portion of the scattering has a different frequency. CARS and CSRS use the same mechanism (a molecule absorbing a photon, using its energy to transition to a different level, then re-emitting a photon with the excess energy) but they are types of four-wave mixing and really belong in the previous point.
Best Answer
Beats cannot extinguish sound completely, but destructive interference of two waves of the same frequency and intensity, however in opposition of phase, yes can. Beats are the effect of superposition of waves of different frequencies. Thus, while you listen, sometimes they fit in phase, which increases the total amplitude and thereby increases much the intensity, and sometimes they are in opposition of phase, which diminishes the total amplitude and thereby reduces the intensity down to zero.
The arrow in the pictures indicates the direction of propagation of the sound. The two superposing waves are colored, one in orange, and one in blue. The pictures represent the amplitudes along the propagation direction, at a given instant of time. One can see that for equal frequency and intensity the amplitudes of the two waves are always opposite, s.t. the total will be zero. For different frequencies, the amplitudes are at some positions, opposite and cancel mutually, and at other positions they are equal and interfere constructively. For obtaining the evolution in time at a certain point in space, just consider the two pictures moving in the indicated direction, and passing through the desired point.