How come speed of sound is not releated to frequency and more?

[sameeralord]

Hello everyone

I'm learning sound at the moment and I would really appreciate if you people can fine tune my knowledge.

My understanding of the sound wave. An object is vibrating

The forward motion of the object: The particles closer are compressed and they push the adjacent particles and they push the other and so on passing the compression and return to their mean positions.

Backward motion of the object : The particles closer are pulled apart causing a rarefaction.The particles that are pulled apart pull the adjacent particles and they pull the other and so on hence passing the rarefaction. Then they return to mean positions again.

Is my understanding right. After the forward vibration do the particles return to their mean position and then pulled back or is the backward motion what that causes them to be pulled back to their original position .I'm confused here

How is speed of sound not related to frequency

If the particles in a solid are close together that means the vibrations would be passed more rapidly. Wouldn't this mean that the time it takes for one wavelength is shorter hence shorter period and higher frequency. I saw in another question they divided the wavelength by the period to get the speed. How can frequency not be related.

Any help would be appreciated. Thanks in advance

IMPORTANT EDIT: I just had a think about this again. This is my new reason. tell me if this is correct. So the speed of sound in air is 340 ms-1 and this is a constant. Let's a musical note or an object want to create an frequency 240 Hz. Then the sound wave takes this speed (340 ms-1) into consideration and alters the wavelength so that it matches with the speed of sound in air. So actually what happens is frequency and wavelength inversely change so that all sounds travel at 340 ms-l.Or else it would sound horrible. Am I right?

 

[Astronuc]

Some notes on waves and wave properties, specifically sound. The speed of sound varies with temperature and molecular/atomic mass of the gas in which is propagates, i.e. it is a consequence of the characteristics and properties of the gas.

http://hyperphysics.phy-astr.gsu.edu/hbase/wavrel.html

http://hyperphysics.phy-astr.gsu.edu/hbase/sound/souspe.html

http://hyperphysics.phy-astr.gsu.edu/hbase/sound/souspe3.html

It is interesting to compare this speed with the speed of molecules as a result of their thermal energy. For the specific example of dry air at 20°C, the speed of sound in air is 343 m/s, while the rms speed of air molecules is 502 m/s using a mean mass of air molecules of 29 amu.

Sound is related to the variation in local pressure of a gas as the sound wave passes through. One's understanding of a vibrating source (surface) is correct.

 

[cesiumfrog]

It seems like a good bet that the speed of sound in typical air does depend on frequency (and amplitude) at least to some degree.

(On the other hand, perhaps one can show generality of a linear wave equation analogously with how every extremum may be approximated by a parabola..)

 

[Topher925]

It seems like a good bet that the speed of sound in typical air does depend on frequency (and amplitude) at least to some degree.

I wouldn't take that bet. The speed of sounds is directly dependent of the mechanical properties of the median in which it propagates, not the properties of the wave itself. The frequency and amplitude can determine how much energy is in the wave but not its speed.

 

[LURCH]

I think your "important edit" got it right on the money. Let a tuning fork be struck, and set to vibrate in dry air at sea level at about 20^o C. The fork is tuned to vibrate at concert "E-sharp"; a rate of vibration of roughly 340 Htz. The peak of one compression wave travels 1m before the next wave leaves the tuning fork. So the wave-peaks are about 1m apart, and 340 of them pass through any given point over the course of 1 second. That is to say, the sound travels at 340m/s, and the the sound waves are 1m apart, so you hear "E-sharp" (340 Htz).

Now, place the base of the tuning fork against a piece of metal, like a steal rod, and put the other end of the rod against your ear. Sound travels much faster through metal (I don't know the exeact speed, but let's say 100 times faster). So the peak of one wave travels 100m before the next peak starts out. Now the waves are 100m apart, and traveling at 34,000 m/s, so they still arrive at your ear at a rate of 340 Htz, and you still hear "E-sharp."

 

[Mapes]

I wouldn't take that bet. The speed of sounds is directly dependent of the mechanical properties of the median in which it propagates, not the properties of the wave itself.

You should take that bet; http://en.wikipedia.org/wiki/Dispersion_relation" is very real.

 

[Topher925]

You should take that bet; http://en.wikipedia.org/wiki/Dispersion_relation" is very real.

Excuse my ignorance, but what does the dispersion affect have to do with the speed of sound being dependent of frequency?

 

[Mapes]

Well, that's the definition of dispersion...

 

[Topher925]

Huh? I don't understand. Are you sure your not confusing the actual wave velocity with the phase velocity? I found nothing in on that web page that suggested what you are claiming.

 

[Mapes]

Huh? I don't understand. Are you sure your not confusing the actual wave velocity with the phase velocity? I found nothing in on that web page that suggested what you are claiming.

Even the first sentence, viz.: "Dispersion relations describe the ways that wave propagation varies with the wavelength or frequency of a wave."? You say "The speed of sounds is directly dependent of the mechanical properties of the median in which it propagates, not the properties of the wave itself." This isn't strictly true; the speed of sound is (weakly) dependent on frequency. The effect is small but increases with increasing frequency until the phonons form a standing wave and the group velocity goes to zero.

 

[billiards]

I believe that sound will be dispersive unless it is propogating in a uniform medium. This is because sound travels at a speed which is determined by a halo around the ray path. (Believe it or not, the sensitivity to the speed of sound exactly on the ray itself is zero!)

The lower the frequency the greater this halo of dependency is, and therefore, at increasingly low frequencies the speed of sound will be affected by increasingly more of the medium through which it is travelling. It should therefore go without saying that if the medium is not uniform then sound will be dispersive. Take the example of an earthquake which propagates seismic waves through the earth, imagine rays ricocheting around the crust. The low frequency waves will travel faster than high frequency waves simply because they can sample deeper, where the rocks are (generally) faster.

 

[sameeralord]

I think your "important edit" got it right on the money. Let a tuning fork be struck, and set to vibrate in dry air at sea level at about 20^o C. The fork is tuned to vibrate at concert "E-sharp"; a rate of vibration of roughly 340 Htz. The peak of one compression wave travels 1m before the next wave leaves the tuning fork. So the wave-peaks are about 1m apart, and 340 of them pass through any given point over the course of 1 second. That is to say, the sound travels at 340m/s, and the the sound waves are 1m apart, so you hear "E-sharp" (340 Htz).

Now, place the base of the tuning fork against a piece of metal, like a steal rod, and put the other end of the rod against your ear. Sound travels much faster through metal (I don't know the exeact speed, but let's say 100 times faster). So the peak of one wave travels 100m before the next peak starts out. Now the waves are 100m apart, and traveling at 34,000 m/s, so they still arrive at your ear at a rate of 340 Htz, and you still hear "E-sharp."

Thanks! So my understanding was right

Thank you everyone who has contributed to this thread. There is still one question remaining thought. Is it the backward vibration that brings back the particles to their mean position?

 

[Topher925]

This isn't strictly true; the speed of sound is (weakly) dependent on frequency. The effect is small but increases with increasing frequency until the phonons form a standing wave and the group velocity goes to zero.

I did not know this...well, I guess I stand corrected.

Is it the backward vibration that brings back the particles to their mean position?

I don't really understand what you mean. Can you define "backward vibration"? The fact that the particles are vibration shows that they will have a mean position. Thats part of the definition of a simple harmonic motion.

 

[billiards]

Sound travel is understood in terms of continuum mechanics. This means we don't necessarily have to think about particles interacting, we can understand it by thinking of the medium as being divided up into cells of equal volume. When sound passes through a cell, the cell will respond by changing its volume; whether the volume compresses or expands depends on the polarity of the first arriving energy. An explosion would produce an outward blast that would propagate a compression, whereas some kind of implosion would propagate a rarefaction first arrival.

I think your "important edit" got it right on the money. Let a tuning fork be struck, and set to vibrate in dry air at sea level at about 20o C. The fork is tuned to vibrate at concert "E-sharp"; a rate of vibration of roughly 340 Htz. The peak of one compression wave travels 1m before the next wave leaves the tuning fork. So the wave-peaks are about 1m apart, and 340 of them pass through any given point over the course of 1 second. That is to say, the sound travels at 340m/s, and the the sound waves are 1m apart, so you hear "E-sharp" (340 Htz).

Now, place the base of the tuning fork against a piece of metal, like a steal rod, and put the other end of the rod against your ear. Sound travels much faster through metal (I don't know the exeact speed, but let's say 100 times faster). So the peak of one wave travels 100m before the next peak starts out. Now the waves are 100m apart, and traveling at 34,000 m/s, so they still arrive at your ear at a rate of 340 Htz, and you still hear "E-sharp."

I don't think this really counts as an "understanding".
What you are setting out to prove is this:

1. Speed (as a function of Frequency) = Constant

and we know that by definition:

2. speed = frequency x wavelength

so you justify 1. with:

3. frequency x wavelength = Constant

The argument is circular, all you have really achieved is to rephrase the question.