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The Tiny Vibrating Source

  1. Feb 14, 2012 #1
    1. The problem statement, all variables and given/known data

    A tiny vibrating source sends waves uniformly in all directions. An area of 3.25 cm2 on a sphere of radius 2.50 m centered on the source receives energy at a rate of 4.20 J/s.
    (a) What is the intensity of the waves at 2.50 m from the source?
    (b) What is the intensity of the waves at 10.0 m from the source?
    (c) At what rate is energy leaving the vibrating source of the waves?
    2. Relevant equations

    I = Power/Area

    Area (sphere) = 4(3.14)r2

    3. The attempt at a solution

    a) Power = 4.20 J/s

    Radius = 2.5 m

    Area = 4(3.14)(2.5)2= 25(pi)

    I= 0.0534 W/m2

    The actual value of I is 1.29x104 W/m2

    If I say Area= 3.25cm2 then I get I = 129.23 W/m2
     
  2. jcsd
  3. Feb 14, 2012 #2

    PeterO

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    This is very close to what you should have been doing.
    The area of the whole shpere was unimportant, since you were told the powere reaching just 3.25cm2 of it.

    While there are 100 cm in a metre, I think you overlooked that that means 10000 cm2 in each m2
     
  4. Feb 14, 2012 #3
    Okay but how does the 2.5m radius factor into that?
     
  5. Feb 14, 2012 #4

    gneill

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    The "area of 3.25 cm2" is located on the r = 2.5m sphere. So it's where it has to be in order to answer question (a). No further need for that radius :wink:
     
  6. Feb 14, 2012 #5

    PeterO

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    For the calculation here - it doesn't.

    It is common for such a question to be just one part of a bigger situation, and to go on to ask "what is the total power output of the sound source?". That is where the full sphere would come in.

    EDIT: just noticed! - that is part (c)
     
  7. Feb 15, 2012 #6
    So if the 3.25cm^2 area is only part of the sphere at 2.5m, how can I go about using that to find the area at 10m? Like unless I'm supposed to just times it by four or something I don't know.
     
  8. Feb 15, 2012 #7

    PeterO

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    You should already have studied the effect of distance on intensity of sound waves as they spread out.

    Inverse-square-Law should mean something.

    EDIT: Alternately, answer part (c) before part (b) this time.
     
  9. Feb 15, 2012 #8
    It was something that came up on mastering physics only once. So I don't remember how to link 1/d12 and 1/d22
     
  10. Feb 15, 2012 #9

    PeterO

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    If you do part (c) you work out the total amount of energy given out each seocd.

    Work out the area of the 10m sphere, and that same amount of energy is spread over that new area.

    Post your answers
     
  11. Feb 15, 2012 #10
    But the issue is that I don't know that equation either.
     
  12. Feb 15, 2012 #11

    gneill

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    Google: surface area of a sphere
     
  13. Feb 15, 2012 #12
    I mean I don't know the equation to find out "total energy given out each second"
     
  14. Feb 15, 2012 #13

    gneill

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    The small area of 3.25 cm2 caught a portion of it at its location. What fraction of the total did it intercept?
     
  15. Feb 15, 2012 #14

    PeterO

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    You know how much energy per second was striking each 3.25 cm2. how many lots of 3.25 cm2 make up an entire sphere of radius 2.5 m [Hint: you worked out how many square metres make up that sphere in your original post; - be careful with cm2 and m2.]
     
  16. Feb 15, 2012 #15

    jambaugh

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    This advice may or may not be helpful to you but...

    I would solve this problem by first calculating the power per spherical angle.
    [Spherical angle in steradians = surface area on the unit sphere cut out by that spherical angle. This gives surface area of a spherical region as [itex]A = r^2 \Theta[/itex] where [itex]\Theta[/itex] is the spherical angle in steradians. Note also that the spherical angle of an entire sphere is 4pi, so spherical angle is 4pi times the percentage of the whole sphere.]

    Since the sound is radiating uniformly from a point the power per spherical angle is constant. So figure the power through an area at a given radius and figure the spherical angle of that area (A/r^2) and divide. Use that to solve the other problems.

    If you dislike spherical angle, then use instead the percentage of the whole sphere and it works the same way. You'll realized it comes to exactly the same thing when you work out the details.
     
  17. Feb 15, 2012 #16
    C)
    Total Area at 2.5m = 4(pi)(2.5m)2 = 25(pi)m2

    Sub-Area = 3.25cm2 = 3.25x10-4m2

    Sub area/total area = 4.14x10-6

    Energy at sub-area = (total energy)(fraction)

    total energy = 1.015x106 W

    Correct Answer.

    B)

    I = P/A

    P = 1.015x106 W

    A = 4(pi)(10m)2 = 400(pi)m2

    I = 807.69 W/m2

    Correct Answer
     
  18. Feb 15, 2012 #17

    PeterO

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    Well done!

    btw - the inverse square law tells us that if you increase the radius by a factor of 4, then you reduce the intensity by a factor of 16.
    The question setters obviously thought you would do it that way - which could explain why the 10m sphere was part (b) while the total energy was part (c)
     
    Last edited: Feb 15, 2012
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