How sound is represented in a graph?

Question

• Let me imagine that air nearby has structure like this, meaning that pressure variation without any disturbance is almost equal.

  

  Now if I speak, I set some molecules of it in motion. They move with a velocity $v$ and sets motion to the molecules next to them. With every collision, either their amplitude must decrease or they are not losing energy. But in a graphical representation of the same, they are shown as

  

  In each of the representation above, the crests and the troughs at any point lie on the same line. That is the amplitude, even after much time is equal. Does this mean that amplitude will always be equal no matter how much time passes? That doesn't seem real as the molecules set next molecules in motion with 'some of its energy' and not complete energy. If they provided complete energy they wouldn't have get back to their mean position.

• Also the difference between two crests and two rarefactions (that is the wavelength) is equal. How is this possible? at the very start the particles move at a different speed, then they lose some energy to move the next particle. So when the next particle will move the third one, the speed will not be exactly same to the first one and after collision second one will also lose some energy taking even more time to get back to it's mean position. Let me explain it like an equation.

  Let's say particle 1 has a speed of V when it is vibrated. Then it moves and vibrates another particle, losing some of it's energy hence resulting in lesser speed. Now it has a speed U.

  Particle 2 got some of the energy of particle 1 and got a speed of $v$. It moves and vibrates Particle 3, again losing some of it's energy hence less speed. now its speed is $u$.

  So, V > v & U > u

  Also V > U & v > u

  During oscillation 1st, wave will move with speed V, U

  During oscillation 2nd, wave will move with speed v, u

  The crest thus formed will be in distance => Ut + vt

  and trough thus formed will be in distance => vt + ut

  at the same time, distance of crest will be less, as Ut > ut (vt is common) and the distance of trough will be more. Then how come wavelength can be equal, which is defined as the distance of two crests or two rarefactions?

• Another question is how the graph is formed of a sound wave. If we make graph, taking some point as a mean position then there will be only one crest and one trough and the rest will be a straight line. If we are change our point every second then which point are we taking, see the graph of a longitudinal wave,

  

  As I can see, there is always a point where air is compressed and there is always a point where air is forming rarefaction. If I follow wave, I can always show the air to be compressed or having rarefactions and if I don't follow then there will be a graph with only 1 crest and 1 compression.

Summarizing the questions, they are as follows:

1. Does this mean that amplitude will always be equal no matter how much time passes? if not, why they are shown like that in graph?

2. How is the difference between between two crests and two troughs are equal?

3. How is a sound wave represented with graphical method?

Any edits or clarification about some topic which I'm confused on will be welcomed.

Answer 1

You are over-thinking this.

The simple sine-wave graphs represent an ideal sound wave that has a single frequency (i.e. a pure tone with no harmonics) and a constant amplitude. What is being plotted in the graph is the displacement of a typical molecule of air against time at a fixed position. You can also think of the graph as plotting density against time or pressure against time - you will get the same shape graph in either case.

To get an approximation to this simple sound wave graph in real life, place a loudspeaker emitting a pure tone at a constant volume at point A and place a microphone at point B some fixed distance from the speaker. If you graph the current in the microphone (which is proportional to the displacement of the microphone's membrane) against time you will get a sine wave. The sine wave has constant amplitude because the speaker is feeding energy into the system so the average energy of the sound wave at the microphone (which is proportional to the square of the graph value) will be constant.

Remember that these sine wave graphs represent the very simplest ideal sound wave. The graph of a real sound wave will be very much more complex, may show attenuation, and may not even be periodic, especially if the sound is random noise rather than a combination of harmonics.

Answer 2

Sound waves definitely experience attenuation in media which depends primarily on the density of the medium and other physical properties. Attenuation means the the amplitude decreases with time.

Sound waves are longitudinal density waves. They require a medium to propagate, and do so via periodic "bunching and unbunching" of the air molecules.

To plot a sound wave, you may want to plot density vs. position for your medium. If the waves spread out radially from a point source, energy will fall off as 1/r^2, which is one cause of attenuation (decreasing amplitude). Another cause of attenuation is due to the properties of the medium and nonlinearities in the dispersion relation.

Answer 3

When sound is emitted from a source there is not really a loss of energy rather it a spreading out of the energy over a larger volume which results in the sound intensity dropping.
However there a dissipative process including heat being produced from the energy associated with a sound wave because of the viscosity (fluid friction) of the air and heat is also produced due to the fact it takes time for the air molecules to respond to the passage of a sound wave, a process which is called relaxation.

As to the visual representation of longitudinal sound waves there are two ways of doing this. A sound wave can be represented by a variation of local pressure from atmospheric or a variation in the "displacement" of molecules.

Consider this very simplistic "photograph" of a sound wave at an instant of time.
At the top is representation of the state of the air without a sound wave being present and at the bottom with a sound wave present.

You will note that the "air molecules" are bunched together in some regions (compressions $C$) where the pressure is above atmospheric and in other regions the "air molecules" are further apart (rarefaction $R$).
Not that the pressure wave and the displacement wave are $90^\circ$ out of phase.

I feel that a description of the pressure wave is relatively easy to understand whereas for the displacement wave it is not so simple as one must first define how the displacement is measured.
The measurement of the displacement cannot focus on just one molecules as without a sound wave being present the gas molecules are moving around at random and are in a state of dynamic equilibrium.

Thus consider a small volume of the gas.
Although the actual molecules in within that small volume will constantly be changing, on average the centre of mass of that small volume of gas will not change.
One can then consider the position of the centre of mass, without a sound wave present, as an equilibrium position.
The passage of a sound wave changes the position of the centre of mass of the small volume of gas and that change of position can be taken to be the displacement produced by the sound wave.

I deconstructed the gif animation and four frames are shown below with the "displacements" shown in frame $1$.

When you view the gif your eye follows the pressure wave as the displacement are very difficult to follow unless one of the vertical lines was of a different colour.

Here is an animation with some "molecules" (centre of mass of a small volume of gas) coloured red.

isvr, Institute of Sound and Vibration Research at the University of Southampton, England.