Some good info here
The above link has some good info, I think it was designed for kids, but still good.
The above link has some good info, I think it was designed for kids, but still good.
50hz means the cones move p-p 50 times in one second, 1/50 is how long each p-p takes to occur. So from rest position the cone moves one inch in 1/50 of a second so its velocity is 1"/(1/50sec) = 50 in/sec.
Sound moves at ~1125 ft/sec --->13500 in/sec
(13500 in/sec)/270in = (50 in/sec)/1 in (Speed/wavelength alway equals the frequency/displacement)
The cone is transferring its low speed and low displacement energy to air which "communicates" the pressure at the speed of sound and therefore creates longer waves than the source is providing.
Another way to word it is it is all time dependant, the speaker and the air both oscillate at the same frequency but the air moves faster and therefore cover more "ground".
Hopefully that was written well enough to make sense, if not I'll try again later.
Theli said:
The SPEED at which the driver moves in and out gives the TONE or frequency of the sound.
The AMOUNT by which the driver moves in and out gives the loudness, or amplitude, of the sound.
So, by saying the driver moves 1" in and out - this only gives an idea of the amplitude or how loud the sound "is".
Hope this helps!
Cheers
Wavelength:
Think of it this way: imagine a little red dot that travels horizontally at a speed of 1120 feet/sec.
As the red dot moves (fast!), it also goes UP and DOWN, at a rate of 50 times per second.
Question: How far apart are the PEAKS in position of the red dot?
Answer: The dot moves up and down 50 times in a second - so the amount of time between PEAKS in position is 1/50 of a second - if the dot is traveling at 1120 feet per second, it would have traveled
1120/50 = 22.4 feet
There's your wavelength (50Hz).
Cheers
Think of it this way: imagine a little red dot that travels horizontally at a speed of 1120 feet/sec.
As the red dot moves (fast!), it also goes UP and DOWN, at a rate of 50 times per second.
Question: How far apart are the PEAKS in position of the red dot?
Answer: The dot moves up and down 50 times in a second - so the amount of time between PEAKS in position is 1/50 of a second - if the dot is traveling at 1120 feet per second, it would have traveled
1120/50 = 22.4 feet
There's your wavelength (50Hz).
Cheers
Re: Re: basic question about sound.
No that doesnt help, SPEED is the change in position per unit time, two identical woofers playing the same frequency while one has p-p exc. of 1/2" and the other 1", the one moving 1" has higher speed. Frequency is cyclic and independant speed, and amplitude.
The AMOUNT the driver moves and the resulting loudness are functions of frequency, if two drivers are moving 1"p-p but one is playing 20hz and the other is playing 40hz, the one playing the 40hz tone will be 4 times louder.
His question is about the nature of sound and how a cone moving some small distance(ex. 1") can cause long waves(ex. 20') in air. And Im not sure if the red dot thing was an analogy, but it is generally confusing to reference vertical sine waves when discussing sound especially when the above link illustrates exactly how the air particles move.
clem_o said:
No that doesnt help, SPEED is the change in position per unit time, two identical woofers playing the same frequency while one has p-p exc. of 1/2" and the other 1", the one moving 1" has higher speed. Frequency is cyclic and independant speed, and amplitude.
The AMOUNT the driver moves and the resulting loudness are functions of frequency, if two drivers are moving 1"p-p but one is playing 20hz and the other is playing 40hz, the one playing the 40hz tone will be 4 times louder.
His question is about the nature of sound and how a cone moving some small distance(ex. 1") can cause long waves(ex. 20') in air. And Im not sure if the red dot thing was an analogy, but it is generally confusing to reference vertical sine waves when discussing sound especially when the above link illustrates exactly how the air particles move.
Hi nunayafb,
Yes, your points are correct, I was just trying to make the explanation as 'simplistic' as possible, i.e. in terms of how (what I perceive) a layman would think about it.
If one wants to get really technical then even the statement "4 times as loud at 40Hz than at 20Hz for the same 1 inch movement" would have to be qualified - i.e. this happens only in half-space with the backwave canceled... and is also dependent on the acoustic loading of the cone...
Cheers!
Yes, your points are correct, I was just trying to make the explanation as 'simplistic' as possible, i.e. in terms of how (what I perceive) a layman would think about it.
If one wants to get really technical then even the statement "4 times as loud at 40Hz than at 20Hz for the same 1 inch movement" would have to be qualified - i.e. this happens only in half-space with the backwave canceled... and is also dependent on the acoustic loading of the cone...
Cheers!
Thanks guys, I appreciate the effort to explain a simple question. I feel a bit silly now. After all, the following statement sums it all up:
So once the driver has reached it's max excursion (compression of the air) and then cycled to the next peak, the previous peak has travelled 20ft.(50hz tone) The amount the excursion is umimportant except in relation to volume.
I just had it messed up in my head. And on a side note, it explains some of the difficulties of creating a fullrange driver ( especially at the extremes)
sreten said:Hi,
Sound propogates at ~ 330 metres per second.
Frequency = number of times per second.
Wavelength = 330 metres per second / number of times per second.
So once the driver has reached it's max excursion (compression of the air) and then cycled to the next peak, the previous peak has travelled 20ft.(50hz tone) The amount the excursion is umimportant except in relation to volume.
I just had it messed up in my head. And on a side note, it explains some of the difficulties of creating a fullrange driver ( especially at the extremes)
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