1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Distance needed to walk in order to hear local maximum

  1. Jun 4, 2016 #1
    1. The problem statement, all variables and given/known data
    A person stands in an open space listening to the sound from two speakers. The speakers generate sound with a frequency of 489.5 Hz, the speed of sound in air is 343 m/s. The speakers are 2.00 m apart and the person walks away from one of the speakers along a line that is perpendicular to a line drawn between the two speakers as shown in the diagram. The person starts 1.00 m from the first speaker. What distance does the person need to walk along the straight line in order to hear the first local maximum in sound intensity due to the interference of the two waves?

    2. Relevant equations
    y = Asin(kx - wt)
    v = f*lambda
    c^2 = a^2 + b^2 (Maybe??)

    3. The attempt at a solution
    Since there is an interference between the two waves, i thought you can calculate the addition of the two wave equations. Since both waves have the same equation, it will come out to y = 2Asin(kx-wt), where k = 2pi*343m/s/489.5Hz and k = 2pi*489.5Hz. Since y = 0 at the first wavelength, which is 343m/s/489.5Hz, you can substitute that in to find t, but realised that i was going nowhere because i dont know the amplitude... Am i going in the right direction by adding the two equations?

    Also, how does wave superposition work if one wave comes in at an angle to the other? Are there different equations for the addition of them? Or do the waves just not superimpose?
    Screen Shot 2016-06-05 at 1.05.41 PM.png
    Last edited: Jun 4, 2016
  2. jcsd
  3. Jun 4, 2016 #2
    Please, define your variables. There also seems to be a conflict.
  4. Jun 4, 2016 #3
    Okay so i think i may have been on the wrong track. I think you need to use Path difference to calculate the distance travelled. So that means Absolute Value of (1+d) - (sqrt(4 + (1+d)^2) = n*lambda. But now i need to find another equation in d or n to eliminate either one. Does anyone else think this is the right track or have any other clues?
  5. Jun 4, 2016 #4
    k = 2pi/lambda = 8.96681985
    w = 2pi*f = 979pi.
    But i dont think this is the right way anymore
  6. Jun 5, 2016 #5
    whats the conflict?
  7. Jun 5, 2016 #6
    "k" is defined as being two different values.

    Yes, that is the most straight forward way.
  8. Jun 5, 2016 #7
    Sorry the second k is meant to be w. If you use path difference, how do you know what the value of n is?
  9. Jun 5, 2016 #8
    Ignoring n for the moment, what is the relative phase needed between two waves for their sum to be maximum?
  10. Jun 5, 2016 #9
    there should be no relative phase, if their sum needs to be maximum, since through constructive interference the maximum sum is when they are in phase, i think
  11. Jun 5, 2016 #10
    Right again.
    So what should the ratio of their distances be for them to arrive in phase?
  12. Jun 5, 2016 #11
    d1/d2 = lambda? im not sure
  13. Jun 5, 2016 #12
    Distances being different by one wavelength would certainly do it. What about two wavelengths, or 10 wavelengths? Would that work? What about 10.5 wavelengths?
  14. Jun 5, 2016 #13
    i think it would, but they would be later local maximums, and the question is asking for first one
  15. Jun 5, 2016 #14
    10.5 wavelengths wouldnt work though, since it wouldnt indicate a meeting between two crests, so it wouldnt be a maximum?
  16. Jun 5, 2016 #15


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    What you need to do is to first work at the initial position and set up the equation ## k (d_2 - d_1) = n \lambda## (here you know all the parameters except "n") and solve for that "n". I called it it n but it may not be an integer here (if the initial position is not a maximum, it won't be integer). For the sake of the argument, let's say it comes out to be 2.34 (I just made that up to explain my point). Then you find the position along the line indicated where you have ##k(d_2 - d_1) = N \lambda ## where N is now the nearest integer above the "n" (in my example, N would then be equal to 3). Now ##d_2## and ##d_1## are unknowns (but are related, of course) and you can find the answer.
  17. Jun 5, 2016 #16
    Okay i understand now, so you have to find the first "n" after the intial position which points to a maximum. Also just clarifying, in the equation ## k (d_2 - d_1) = n \lambda## k = 2*pi\lambda right? and d2 and d1 are sqrt(5) and 1 respectively? When you substitute these values into the first equation, you get n, then round up to the nearest integer above n and substitute it back but this time solve for d1 and d2?
  18. Jun 5, 2016 #17


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Yes, everything you wrote is correct. EDIT: as long as you meant ##k = 2 \pi/ \lambda##
  19. Jun 5, 2016 #18
    Replies in reverse order.

    Right, 10.5 would not work.
    True. I was trying to point out the general rule. I think you have already found the wavelength to be less than the speaker spacing. That means the difference in distances from observer to speakers will be greater than one.

    I see nrqed joined.
    Is that 2*pi factor really appropriate?
  20. Jun 5, 2016 #19
    If i substitute the rounded value of N back into the equation, i would get d2 - d1 = Nlambda/k. Are d2 and d1 given by sqrt(4 + (1+d)^2) and (1 + d)? If so the equation gives a value of 0.9999 for d.
    I think so, isnt the value of k given by 2pi/lambda?
  21. Jun 5, 2016 #20
    The problem I see is that d1, d2, λ already have units of linear distance. What units do you get by multiplying linear distance by 2*pi?
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted

Similar Discussions: Distance needed to walk in order to hear local maximum
  1. Maximum Distance (Replies: 4)