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Sound Intensity and Sound Waves

  1. Jul 2, 2006 #1
    I am having trouble with these two problems..
    1.By how many decibels do you reduce the sound intensity level due to a source of sound if you quadruple your distance from it? Assume that the waves expand spherically.
    For this one do I have to find ratios of I1/I2 and then substitute it in beta= 10 log (I/Io). I found the ratios to be 16 and I dont know if that's right.

    2. Two identical loudspeakers are 3.4 m apart. A person stands 5.8 m from one speaker and 3.6 m from the other. What is the lowest frequency at which destructive interference will occur at this point? The speed of sound in air is 343m/s
    For this one I tried getting the lowest frequency by dividing longest wavelength with speed in the air...like 343/5.8=59. I am not sure if this is right
    Please any hints will be appreciated
     
  2. jcsd
  3. Jul 2, 2006 #2

    nrqed

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    Yes, the intensity is 16 times smaller if you quadruple the distance.

    Now, notice that you cannot calculate the sound level at the initial or at the final positions since you don't have the initial or final intensities. However, you *can* calculate the difference of sound level between the initial and final positions.

    Notice that
    [itex] \beta_f = 10 log ({ I_f \over I_0}) [/itex] and [itex] \beta_i = 10 log({I_i \over I_0 }) [/itex]. Now using algebra and the properties of logs, you should calculate an expression for [itex] \beta_f - \beta_i [/itex]. Notice that the [itex] I_0 [/itex] will cancel out so that it won't appear in your expression, only the ratio[itex] { I_f \over I_i } [/itex] (which is equal to 1/16). Your answer for [itex] \beta_f - \beta_i [/itex] will be negative (which makes sense since the sound level decreases with distance) and will of course be in dB.

    I am a bit puzzled by the question. Do they actually mean that the 5.8 and 3.6 m are measured from each speaker (each along a straight line)? If so, the 3.4 m separation is not needed to do the problem.

    What you need is to consider the *difference* of the two distances, 5.8m - 3.6m. This is the additional distance the wave must travel from the farthest speaker compared to the distance from the closest speaker.
    (It's sometimes called the "path difference" in books). What is the condition on this difference of distance to get destructive interference? What condition gives you the longest possible wavelength for destructive intereference (in order to get the lowest possible frequency)?

    PAtrick
     
  4. Jul 3, 2006 #3
    Thanks for your help...I definitely understood the first question and I got my answer as -12.04 but I am still confused with 2nd question. Should I use pythagorean theorem to get that additional distance for 2nd question. Can you like explain me what exactly does "path difference".Maybe that can clarify my confusion and from there I maybe able to understand this question

    Thanks a lot!!!
     
  5. Jul 3, 2006 #4

    nrqed

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    Hi. You are welcome.

    By "path" difference I mean the following. You take your point, measure its distance [itex] r_1 [/itex] from the first speaker, measure its distance [itex] r_2 [/itex] from the second speaker and take the absolute of the difference, [itex] | r_2 - r_1 | [/itex]. That's the path difference. For two sources in phase, in order to have destructive interference one must have [itex] |r_2 - r_1 | = (2 n +1) \lambda /2 [/itex] (right?).

    The only question here is what is meant by the 5.8 and 3.6 m (a drawing would make things completely clear). If those are the actual distances to the two speakers then that's all that is needed to use the above formula and then finish the problem. No need for Pythagora's thorem or for the 3.4 meters. On the other hand, if some of those distances are a and y distances, one might need to do a calculation to find r_1 and r_2. That's the part that I am not sure about. Again, a drawing would make things completely clear. The *wording* of the question seems to suggest that what they give are directly r_2 and r_1.

    Pat
     
  6. Jul 3, 2006 #5
    I am trying to comprehend what you are trying to say here..but for destructive interference, we use L2-L1= n lambda/2 (right)?? Where did you get 2n+1 from? if I use this formula I get the answer to be 77.95...but I have one question n would be 1, right
    I hope I am making some sense but if we are talking about two identical speakers that would be in-phase,right and we should use L2-L1= n lambda/2 to get our wavelength and from there we substitute that value in v=f lambda and get our answer...right
    I hope this is right....Thank you once again for taking time to explain me this stuff.
     
  7. Jul 3, 2006 #6

    nrqed

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    Your equation is incorrect if by "n" you mean any integer. But your equation *does* give the correct answer for *n=1*. It would not give the correct answer for any other n.
    The key point is that destructive interference occurs (when two sources are in phase) for [tex] r_2 - r_1 = {\lambda \over 2}, {3 \lambda \over 2}, {5 \lambda \over 2} ...[/tex]
    in other words for any odd multiple of lambda/2. The general formula can therefore be written as either [itex] r_2-r_1 = n_{odd} \lambda/2 [/itex] OR [tex] r_2-r_1= (2 n+1) { \lambda \over 2} \, \, {\rm where } \, \,n =0,1,2,3....[/tex]
    You can verify that setting n=0,1,2... will give back the odd multiples of lambda/2. Sometimes people use instead
    [tex]r_2-r_1= (2 n -1) {\lambda \over 2} \, \, {\rm where } \, \,n =1,2,3,4..[/tex]
    which again gives back the odd multiples of lambda/2.
    Your answer sounds right (always include the units, btw). But again, your formula would not work for other values than the longest wavelength (*unless* by "n" you meant just the odd integers!).

    And you are very welcome.

    Patrick
     
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