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Waves in a conversation

  1. Feb 24, 2008 #1

    ~christina~

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    1. The problem statement, all variables and given/known data
    Suppose that the sound level of a conversation between two men is initially at an angry 70 dB before dropping to a soothing 50 dB.

    (a) Assuming that the frequency of the sound is 500 Hz, determine the initial and final sound intensities.

    (b) What are the initial and final sound wave amplitudes?

    (c)The men dropped the sound level of their conversation because they noticed an “observer” with a small microphone 200 m away. Assume that the sound waves reaching the “observer” were emitted by a point source. What is the sound intensity of these waves?

    (d) The microphone held by the “observer” has a cross-sectional area of 0.750 cm2. Calculate the power intercepted by the microphone.


    2. Relevant equations
    [tex] \beta = 10log (I/I_o) [/tex]
    [tex] \wp= 1/2 \rho v (\omega s_{max})^2 [/tex]
    [tex] I= \wp_{avg}/ A = \wp_{avg} / 4\pi r^2 [/tex]

    3. The attempt at a solution

    a) Assuming that the frequency of the sound is 500 Hz, determine the initial and final sound intensities.

    [tex]\beta_i = 70dB [/tex]
    [tex]\beta_f= 50dB [/tex]
    [tex] I_o= 1.00x10^{12} W/m^2 [/tex]

    [tex] \beta_i = 10log (I_i/I_o) [/tex]
    [tex] 70dB= 10log (I_i/(1.00x10^{-12}W/m^2)) [/tex]
    [tex] I_i= 1x10^{-5} W/m^2 [/tex]

    [tex] \beta_f = 10log (I/I_o) [/tex]
    [tex] 50dB= 10log (I_f /(1.00x10^{-12}W/m^2)) [/tex]
    [tex] I_f= 1x10^{-7} W/m^2 [/tex]

    b) What are the initial and final sound wave amplitudes?

    I'm not sure what amplitude they are refering to is it: A, [tex]s_{max}[/tex], or even [tex]\Delta P_max [/tex]

    which on is it??

    c)The men dropped the sound level of their conversation because they noticed an “observer” with a small microphone 200 m away. Assume that the sound waves reaching the “observer” were emitted by a point source. What is the sound intensity of these waves?
    I don't know how to find this and use which equation...confused as to the amplitude and which one they want.

    (d) The microphone held by the “observer” has a cross-sectional area of 0.750 cm2. Calculate the power intercepted by the microphone.

    I really don't know how to incorperate the cross sectional area and find the power intercepted.


    Please help me.

    Thank you
     
    Last edited: Feb 24, 2008
  2. jcsd
  3. Feb 24, 2008 #2

    Doc Al

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    Looks good.

    I'd say they are looking for the displacement amplitude, what you call [itex]s_{max}[/itex].


    Sound intensity drops off as the inverse square of the distance from the source. (Look at the power vs intensity equation.) What's unclear to me is the distance from the men when the power level is 50 db.

    Intensity is power/area.
     
  4. Feb 24, 2008 #3

    ~christina~

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    Oh okay I got the answer for both but wasnt' sure so I didn't post it.
    [itex]s_{max}[/itex]

    [tex]I= 1/2 \rho v (\omega s_{max})^2 [/tex]
    Ii= 1x10^-5W/m^2
    [tex] 1x10^-5 W/m^2 = 1/2(1.2kg/m^3)(343m/s)(3.1459rad/s*s_{max})^2 [/tex]
    [tex] s_{max}= 7.005x10^-1m [/tex]

    If= 1x10^-7W/m^2
    [tex] 1x10^-7 W/m^2 = 1/2(1.2kg/m^3)(343m/s)(3.1459rad/s*s_{max})^2 [/tex]
    [tex] s_{max}= 7.007x10^-1m [/tex]



    true however I forgot to mention what has me confused was the average power as the equation is [tex]I= \wp_{average} / A= \wp_{average} / 4 \pi r^2 [/tex]

    Is the average power found from the distance? (since I have Ii and If ?)
    As for the distance isn't it the same since the men are not mentioned to move? and the microphone is 200m away? Then I'd find the I ?

    Oh darn..I thought that the A they had in the book was Amplitude ...they didn't specify what it was...now I remember my prof saying something about this.. and it messed me up on the above part as well. (I'm writing that in my book now)

    so I guess it'd be ... I= P/A
    A= 0.750cm^2
    I = ? => I guess I'd find that from above part.
    P = ?




    Thanks Doc Al :smile:
     
  5. Feb 24, 2008 #4

    Doc Al

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    Redo these calculations, paying better attention to the exponents.

    If you knew the distance and the intensity, then you could calculate the average intensity. (Assuming a point source.)
    The problem is that you're not given the initial distance from the men. The sound intensity is given as 50 db. But where? 1 m away? 100 m away? Lacking any info, just for grins I'd assume that they are using a distance of 1 m.

    Right.
     
  6. Feb 24, 2008 #5

    ~christina~

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    fixed.

    [tex]I= 1/2 \rho v (\omega s_{max})^2 [/tex]
    Ii= 1x10^-5W/m^2
    [tex] 1x10^-5 W/m^2 = 1/2(1.2kg/m^3)(343m/s)(3.1459rad/s*s_{max})^2 [/tex]
    [tex] s_{max}= 7.007x10^-5m [/tex]

    If= 1x10^-7W/m^2
    [tex] 1x10^-7 W/m^2 = 1/2(1.2kg/m^3)(343m/s)(3.1459rad/s*s_{max})^2 [/tex]
    [tex] s_{max}= 7.007x10^-6m [/tex]

    [tex]I= \wp_{average} / A= \wp_{average} / 4 \pi r^2 [/tex]

    But I'll say it again..isn't the distance from the microphone the same? (200m) from the begining and thus since I have Ii and If from when they went from 70dB to 50dB while staying in the same spot thus I'm not understanding what your saying....

    thus in my view...

    [tex] (Ii + If)/ 2= I_{average} [/tex]

    thus use that here since I can find A

    [tex]I= \wp_av / A [/tex]

    I'm guessing that I'm incorrect but I'm just not getting what you're saying because the speaking men do not move their own position and the microphone is a stationary 200mm away. The men just lower their voice...=(

    Thank you
     
  7. Feb 24, 2008 #6

    Doc Al

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    Much better.
    It's certainly true that the distance to the microphone doesn't change. What we need to know is how far away is the microphone compared to the where the 50 db level was measured.

    For example, if the intensity was I_1 at a distance of 10 m, the intensity at 200 m would be I_2 = (10/200)^2*I_1. Make sense?
     
  8. Feb 24, 2008 #7

    ~christina~

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    Oh...well it sort of makes sense but I don't get how you got that equation...but it looks familiar..(for some reason or another)

    but how would I find the sound intensity at 1 m ? would I use the fact that it was 70dB and then find A using 1m and then use that to find [tex]I_1[/tex]? I assume so and then use that to find the [tex] I_2[/tex] by the ratio..

    so I assume that it was measured at 1m?

    Thanks
     
    Last edited: Feb 24, 2008
  9. Feb 24, 2008 #8

    Doc Al

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    It comes from the power equation. The average power doesn't change as you get further away from the source, but it's spread out over a greater area ([itex]4\pi r^2[/itex]). And intensity is power/area.

    Assuming that is better than doing nothing. But you are given the intensity (or at least the dB level); it's the same numbers that you used in part 1. We are just assuming that it was measured at 1 m. I'd use the intensity for the 50 db sound level.

    Just out of curiosity: What textbook are you using?
     
    Last edited: Feb 24, 2008
  10. Feb 24, 2008 #9

    ~christina~

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    oh my goodness...I'm mixing up all the intensities and powers.....basically getting confused here...lets see..

    the intensity I found for 50dB
    [tex]I_f= 1x10^-7W/m^2 [/tex]
    and I'm lost.....on where I'd use this since I need intensity at r= 200m

    and I know that

    [tex]I= \wp_{average} / A = \wp_{average}/ 4 \pi r^2 [/tex]


    need I but don't have Pav ...basically I'm not sure where I use the I_f = 1x10^-7 N/m^2 to find the I for what they ask for....

    I really need to learn how to find Pav since that' what getting me confused here

    basically where do I use the intensity for 50dB? since I need intensity..for 200m

    I'm using or attempting to use...Physics for Scientists and Engineers 7th edition by Serway & Jewett


    Thanks
     
    Last edited: Feb 25, 2008
  11. Feb 25, 2008 #10

    Doc Al

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    Read this: Inverse Square Law for Sound

    [tex]P = IA = I_1 4\pi R_1^2 = I_2 4\pi R_2^2[/tex]

    R_1 = 1 m (we're just guessing); R_2 = 200 m (we're given this).
    I think I have the 6th edition floating around somewhere. Not a bad book.
     
  12. Feb 25, 2008 #11

    ~christina~

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    Originally I thought: average usually mean that you divide by 2? thus I thought that you would find P for I_i and P for I_f and then add the 2 P's then divide by 2 and that would be your average...

    but looking at that ratio I'd deduce that P is the same for both (why do they label it Pav :bugeye:)?!?!
    thus the average power is just:

    [tex]P = IA = I_1 4\pi R_1^2 = I_2 4\pi R_2^2[/tex]

    [tex]P_{average}= I_2 4\pi R_2^2=(1.0x10^-7N/m^2)(4 \pi)(200m)^2 = 0.05026W [/tex]

    But if this is true then the intensity at 200m is the same as I originally calculated (I_f= 1x10^-7N/m^2)

    hopefully I get it now...

    I'd have to disagree for the 7th edition. The 7th edition (I don't know about the 6th) is particularly bad for projectile motion. (They give 2 example which isn't like any problem in the end of chapter questions) I've looked at another physics text and they give numerous useful examples for projectile motion which is better than this book in my opinion.

    Thanks alot
     
    Last edited: Feb 25, 2008
  13. Feb 25, 2008 #12

    ~christina~

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    Okay..well since no one said anything..I'll hope it's correct and use it for the last part.

    I_f= 1x10^-7N/m^2

    since the Intensity is I=P/A

    A of microphone is: 0.750 cm2

    then the power intercepted by it is...

    P= (1x10^-7N/m^2)(7.5 × 10-5 m^2)

    P= 7.5e-12W

    That's pretty small.

    I'm not sure if my thought process is wrong but...is it?

    Thanks Doc :smile:
     
  14. Feb 25, 2008 #13

    Doc Al

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    Yes, it's confusing. They mean for you to assume the power is uniformly emitted in all directions.

    No, you used R_2 (200 m) but assumed that I_2 was 1.0x10^-7N/m^2. But I_2 is what you are asked to find! (You didn't need to calculate the power, just relate the two intensities.)

    No. That I_f is the intensity at some unknown distance close to the men--certainly not 200 m away (that's where the microphone is). We are assuming that unknown distance is 1 m.

    What I wanted you to do is to take that power equation and find the relationship between the intensities as a function of distance from the source:

    [tex]I_1 R_1^2 = I_2 R_2^2[/tex]

    We know I_1 (that's the I_f you had calculated), R_1, and R_2. Find I_2.

    Sorry to hear that! (I've read some of Jewitt's pedagogical articles and thought they were pretty good. But those are meant for teachers, not students!)
     
  15. Feb 26, 2008 #14

    ~christina~

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    I'll remember that.

    Oh...yep I didn't get it before.

    [tex]I_1R_1^2= I_2R_2^2 [/tex]

    (1x10^-7N/m^2)*(1m)= I_2 (200m)

    I_2= 5x10^-10 N/m^2

    d)
    I= P/A

    A= 0.750cm^2 = 7.5x10^-5 m^2
    I= 5x10^-10N/m^2

    P= 3.75x10^-14 W

    Power intercepted by microphone.

    :rolleyes:
     
  16. Feb 26, 2008 #15

    Doc Al

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    Careful. Those distances from the source should be squared. (The famous inverse square law.)
     
  17. Feb 26, 2008 #16

    ~christina~

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    corrected.

    [tex]I_1R_1^2= I_2R_2^2 [/tex]

    (1x10^-7N/m^2)^2*(1m)= I_2 (200m)^2

    I_2= 2.5x10^-12N/m^2

    d)
    I= P/A

    A= 0.750cm^2 = 7.5x10^-5 m^2
    I= 2.5x10^-12N/m^2

    P= 1.875x10^-16W

    Thank you Doc Al
     
  18. Feb 26, 2008 #17

    Doc Al

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    Looks good!

    In my opinion, this problem left out an essential piece of information which we had to supply (R_1 = 1 m). So if your instructor ever reveals the textbook solution, I'd be curious as to what they use.
     
  19. Feb 26, 2008 #18

    ~christina~

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    My instructor posts the answers online after we do the problems so I'll let you know when I find out.

    Thanks for your help Doc. :smile:
     
  20. Mar 20, 2008 #19

    ~christina~

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    this is the part that I had incorrect:
    part c)

    [tex]\frac{I_2} {I_1}= (\frac{r_1} {r_2})^2=> I_2= I_1(\frac{r_1} {r_2})^2 [/tex]
    [tex]I_{2i}= (1x10^-5)(\frac{1} {2x10^2})^2 = 2.5x10^-10W/m^2[/tex]

    and also
    part d)

    [tex]P_{2i}= I_{2i}A_2= (2.5x10^{-10} W/m^2)(.75x10^{-4}m^2)= 1.88x10^{-14} [/tex]

    I apparently didn't go and use the first intensity that was found.

    The thing is I can see why I got c partially incorrect since I didn't use the first intensity (they ask for it) however, I can't see why I got the last part d incorrect because I didn't use the initial sound intensity to find [tex]P_{2i}[/tex].If I'm not incorrect, they do not ask for it or is it just understood that your supposed to find power initial and final from the 2 intensities?

    Thank you Doc Al
     
    Last edited: Mar 20, 2008
  21. Mar 21, 2008 #20

    Doc Al

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    I think it's a bit unfair to mark you off for this. The problem does not specify clearly whether you should have used the initial or final intensity at 1m. Since they had already dropped to the lower intensity in part a, it's quite natural to assume that the new lower intensity is the one you should be using.

    One would have to assume that the power calculation in part d should be based on the intensity used in part c.
     
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