Pressure amplitude/sound level

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In summary: It is all related. If you divide the distance by the pressure amplitude, that is how loud the sound is. But the distance is not the amplitude.
  • #1
physics_06er
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Hi there

Just wondering I have been given a problem asking me for the pressure amplitude of sound level of a flute 53dB...I am able to do this...then it asks me what the pressure amplitude and sound level would of the flute would be if it the flute were only one-half of the distance away?

I am sure it has something to do with Pressure is proportional to 1/r--but am unsure on how to work this out...can someone please explain what formuals/how to go about this problem

Thanks
physics_06er
 
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  • #2
physics_06er said:
Hi there

Just wondering I have been given a problem asking me for the pressure amplitude of sound level of a flute 53dB...I am able to do this...then it asks me what the pressure amplitude and sound level would of the flute would be if it the flute were only one-half of the distance away?

I am sure it has something to do with Pressure is proportional to 1/r--but am unsure on how to work this out...can someone please explain what formuals/how to go about this problem

Thanks
physics_06er
How does power vary with distance (think of the energy per unit time of the source being distributed evenly over the wave front and then determine how the size of that wave front varies with time)?

What is the relationship between amplitude and power?

Answer those two questions and you will have the answer to the question.

AM
 
  • #3
does the power decrease with distance?...if that is correct and amplitude is a mease of sound's power then it too must decrease right?...then if the pressure amplitude is 8.934x10^-03N/m^2 and the distance is only half-sized now then pressure amp. is itncreased to double and seeing as amplitude is proportional to intensity hence the sound level must also double to (53x2=106dB)...? is this correct or am I COMPLETELY off track??

thanks
Physics_06er
 
  • #4
physics_06er said:
does the power decrease with distance?...if that is correct and amplitude is a mease of sound's power then it too must decrease right?...then if the pressure amplitude is 8.934x10^-03N/m^2 and the distance is only half-sized now then pressure amp. is itncreased to double and seeing as amplitude is proportional to intensity hence the sound level must also double to (53x2=106dB)...? is this correct or am I COMPLETELY off track??

thanks
Physics_06er
Power decreases with distance from the source, but the question is how does it decrease as a function of distance?

Amplitude is the measure of the maximum pressure of the sound wave during a vibration. In graphical terms, the sound wave pressure as a function of time is sine wave. The amplitude is the maximum height of that sine wave. There is a particular relationship between amplitude and power. You have to find out what that is. (1)

The sound level is a measure of the sound intenisty or power per unit area of the sound wave. So in order to determine how the level changes as a function of distance from the source, you have to figure out how power per unit area changes with distance. Assume the sound wave is a spherical wave front with the source at its centre. That front moves outward from the source. As the wave front gets bigger, the area gets bigger. The total power is distributed over a larger area, so the power per unit area gets smaller. You have to figure out how it changes as a function of the distance from the source. That is a pretty simple calculation.(2)

Put (1) and (2) together to get your answer.

AM
 
  • #5
OK..the relationship between pressure/amplitude is that the pressure is proportional to amplitude squared

and the relationship between sound level is sound level=power/area, so if area decreases the sound level increase.

Therefore if the distance (hence amplitude) is halved then the amplitude is 0.25 of the original

and if area decreases by a half then sound level increases by a half..

I this correct?

physics_06er
 
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  • #6
I also have another problem which I'm unsure about-

a trombone played fortissimo in the lower part of its range, around C3 (131Hz), has a sound level of about 100dB at a distance of 3m. A violin played fortissimo in the lower part of its range, around C4 (262Hz), has a sound level of about 80dB at the same distance. How many violins would need to play together to reach the same loudness level (in phons) as the trombone?

all I know is that I have to use the fletcher-munson graph but no idea how to actually interpret it!


Thanks for the help

physics_06er
 
  • #7
physics_06er said:
OK..the relationship between pressure/amplitude is that the pressure is proportional to amplitude squared
The POWER carried by the pressure wave is proportional to the square of the amplitude of the pressure wave.

and the relationship between sound level is sound level=power/area, so if area decreases the sound level increase.

Therefore if the distance (hence amplitude) is halved then the amplitude is 0.25 of the original
The distance from the source is not the amplitude of the sound wave. The distance from the source gives you the radius of the spherical wave front. The power is distributed over the area of that wave front. The power per unit area gives you the intensity or sound level.

The total power of the wave front is determined by the power of the source which is constant. Since that power is distributed over the spherical wavefront and the area of that wavefront increases with distance from the source ie [itex]A = 4\pi r^2[/itex], power/area = intensity, falls off as 1/A. This means that [itex]I = P/4\pi r^2[/itex]. So if you halve r, what happens to I?

Given that I is proportional to P and P is proportional to the square of the amplitude, if I changes by that factor, what is the factor in the change in amplitude?

You are capable of doing this kind of problem. You just have to get all the concepts clear in your mind, which is what I am trying to help you with. If you give up on these problems you will never learn physics.

AM
 
  • #8
ohhh ok i get it now!--as I was looking through my textbook I did come across I=P/4pir^2 but didn't think much of it cos I wasn't sure how it was related!...thank you so much...so just to make sure--if i halve r it means that that the sound level now is 0.25 of that hence 53x0.25=13.25dB?

and then the amp. is now 0.25x0.25=0.0625x8.934x10^-3)=5.584x10^-4N/m^2?

Thanks
physics_06er
 
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  • #9
physics_06er said:
..so just to make sure--if i halve r it means that that the sound level now is 0.25 of that hence 53x0.25=13.25dB?

and then the amp. is now 0.25x0.25=0.0625x8.934x10^-3)=5.584x10^-4N/m^2?

Thanks
physics_06er
If you halve r, does the area increase or decrease? If the power is spread out over a smaller area, does the power density (power/area = intensity) increase or decrease?

If power varies as the square of the amplitude, how does amplitude vary with power?

AM
 
  • #10
Hi

if we halve r-the area decreases hence power density decreases

if power varies with the square of amplitude-then amplitude is 4*original P
 
  • #11
physics_06er said:
Hi

if we halve r-the area decreases hence power density decreases

if power varies with the square of amplitude-then amplitude is 4*original P
?? You are not thinking. You can't guess. You have to work it out.

[tex]I = P/4\pi r^2[/tex]

If we halve r, ie r' = r/2,

[tex]I' = P/4\pi r'^2 = P/4\pi(r/2)^2 = 4P/4\pi r^2 = 4I[/tex]

So the power/unit area = intensity is 4 times as great.

If [itex]P \propto \text{Amplitude}^2 \rightarrow \text{Amplitude} \propto \sqrt{P}[/itex]

AM
 
  • #12
nooo...this is from my high school book

If P is pressure and a is amplitude
P is proportional to a2

This can be written as an equation
P = ka2, where k is a constant of proportionality

If we double the amplitude, that is put a=2a, then
P = k(2a)2 = 4ka2 = 4 * original P

hence FOUR times the original pressure!...ok obviously I am wrong if you you are saying its the sqare root of it...that did cross my mind however after reading the book I figured it is the above--obviously I am misreading something there...in another book of mine it says the intensity of a sound wave is proportional to the pressure squared. in other words doubling the sound pressure quadruples the intensity.!(shouldn't this be the sq.rt as well)--ok i know i sound really dumb at the moment:redface: but I am seriously not guessing...i really thought I was doing it correctly!...
 
  • #13
physics_06er said:
nooo...this is from my high school book

If P is pressure and a is amplitude
P is proportional to a2

This can be written as an equation
P = ka2, where k is a constant of proportionality

If we double the amplitude, that is put a=2a, then
P = k(2a)2 = 4ka2 = 4 * original P
But the question is: what is the change in amplitude if the power/area increases by a factor of 4? An increase in amplitude of a factor of 2 or [itex]\sqrt{4}[/itex] will increase the power/area by a factor of 4.

AM
 
  • #14
hmmm ok i think i get this now...ill go find other examples and see if its certain i know,,thank you for all youe help-sorry for all the q's..and once again thanks for the help
 

1. What is pressure amplitude?

Pressure amplitude is the maximum deviation of pressure from the average or ambient pressure in a sound wave. It is a measure of the strength or intensity of a sound wave, and is typically measured in units of Pascals (Pa) or decibels (dB).

2. How is pressure amplitude related to sound level?

Pressure amplitude and sound level are directly related. As the pressure amplitude increases, the sound level also increases. This means that a sound with a higher pressure amplitude will be perceived as louder than a sound with a lower pressure amplitude.

3. What is the difference between sound pressure level (SPL) and sound intensity level (SIL)?

Sound pressure level (SPL) is a measure of the sound pressure amplitude relative to a reference pressure level, usually 20 micropascals. On the other hand, sound intensity level (SIL) is a measure of the sound intensity (power per unit area) relative to a reference intensity level. In practical terms, SPL is a more commonly used measure for describing the loudness of a sound, while SIL is used in more technical applications.

4. How is sound level measured?

Sound level is typically measured using a sound level meter, which measures the sound pressure level in decibels (dB). The meter has a microphone that converts the sound waves into an electrical signal, which is then amplified and displayed on the meter. The measurement can also be adjusted to account for the frequency range of the sound being measured.

5. What are the typical sound levels for various everyday sounds?

The sound level of a typical conversation is around 60 dB, while a vacuum cleaner can reach up to 80 dB. A rock concert can have sound levels exceeding 100 dB, and a jet engine can produce sound levels of 140 dB or higher. It is important to note that prolonged exposure to sounds above 85 dB can cause hearing damage.

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