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Pressure amplitude/sound level

  1. Sep 15, 2006 #1
    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
     
  2. jcsd
  3. Sep 15, 2006 #2

    Andrew Mason

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    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
     
  4. Sep 15, 2006 #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
     
  5. Sep 16, 2006 #4

    Andrew Mason

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    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
     
  6. Sep 16, 2006 #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
     
    Last edited: Sep 16, 2006
  7. Sep 16, 2006 #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
     
  8. Sep 16, 2006 #7

    Andrew Mason

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    The POWER carried by the pressure wave is proportional to the square of the amplitude of the pressure wave.

    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
     
  9. Sep 16, 2006 #8
    ohhh ok i get it now!--as I was looking through my text book 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
     
    Last edited: Sep 16, 2006
  10. Sep 16, 2006 #9

    Andrew Mason

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    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
     
  11. Sep 16, 2006 #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
     
  12. Sep 16, 2006 #11

    Andrew Mason

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    ?? 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
     
  13. Sep 16, 2006 #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!...
     
  14. Sep 16, 2006 #13

    Andrew Mason

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    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
     
  15. Sep 16, 2006 #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
     
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