1. Limited time only! Sign up for a free 30min personal 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!

How to find f0 when given highest and lowest frequency

  1. Nov 25, 2014 #1

    HHH

    User Avatar

    < Mentor Note -- thread moved to HH from the technical physics forums, so no HH Template was used >

    How would you determine f0 (frequency of the source) and Vsource when given highest and lowest frequency

    Highest Frequency = 640 Hz
    Lowest Frequency = 50 Hz

    I cant use the Doppler effect equation because there are two unknown variables
     
    Last edited by a moderator: Nov 25, 2014
  2. jcsd
  3. Nov 25, 2014 #2
    Your question is very unclear.
     
  4. Nov 25, 2014 #3

    HHH

    User Avatar

    Basically we did a lab, in which our teacher had a device emitting a sound. He attached the device to a string and started swinging it above his head. He told us to use an app, to determine the highest and lowest frequency, then based on that he told us that we should be able to find the speed at which he was spinning the device and the original frequency of the device (f0).

    We got 640 Hz for the highest frequency (moving towards)
    and we got 50 Hz for the lowest frequency (moving away)
     
  5. Nov 25, 2014 #4
    Why do you think the Doppler formula won't work?
     
  6. Nov 25, 2014 #5

    davenn

    User Avatar
    Science Advisor
    Gold Member

    which variables did you think were unknown ?
     
  7. Nov 25, 2014 #6

    HHH

    User Avatar

    I dont have f0 and Vsource, which are two variables

    Fobs = ((Vsound + Vdetector) / (Vsound+Vsource)) f0

    i have 2 values for Fobs (640 and 50 - high and low frequency)
    i have local speed of sound = 344.12 m/s
    I have Vdetector = 0 m/s
    I dont have f0 and Vsource so i dont know how to solve this with two unknown variables
     
  8. Nov 25, 2014 #7
    to solve two unknowns, you can use two equations, or you can use one equation at two data points.
     
  9. Nov 25, 2014 #8

    HHH

    User Avatar

    Do you mean by using substitution? Also doesn't the frequency difference seem to large (between high and low)?
     
  10. Nov 25, 2014 #9

    davenn

    User Avatar
    Science Advisor
    Gold Member

    my maths is shockingly bad..... it helps to draw pic's

    doppler.GIF

    the Vsource at points 1 and 3 are constant just have + or - values

    For an instant of time at point 2 that forward/reverse velocity relative to the observer is zero
    and at that point Fo is heard

    with the velocity being constant, I would expect the Fo at point 2 to be half way between 50 and 640 ?

    some one please correct me if I am wrong.

    if so you now have Fo and can work out the Vs at points 1 and 3 which = the overall Vs of the spinning source


    Dave
     
  11. Nov 25, 2014 #10

    HHH

    User Avatar

    So basically i can find f0 by finding the average between the two frequencies?
    f0 = fhigh + flow / 2
    f0 = 640 +50 /2
    f0 = 690/2
    f0 = 345Hz

    Then what would i do next with that value?
     
  12. Nov 25, 2014 #11
    Don't ask questions before you try to solve it with the help people have given you.
     
  13. Nov 25, 2014 #12

    davenn

    User Avatar
    Science Advisor
    Gold Member

    so now you have everything except Vs

    what are you going to do with your Doppler formula to solve for Vs instead of the usual Fobs ?
     
  14. Nov 25, 2014 #13

    HHH

    User Avatar

    I used the highest frequency and plug it in as fobs and solved for Vsource, however when i do this i get 158 m/s. When i try the same thing with the lowest frequency (plugging it in as fobs) i get 2000 m/s. Both these values seem to high and both these values should also be the same, it doesnt make any sense
     
    Last edited: Nov 25, 2014
  15. Nov 25, 2014 #14

    davenn

    User Avatar
    Science Advisor
    Gold Member

    show us the formula and working you used to solve for Vsource
    and we can see if it is correct

    there is a reason why you got a difference.

    Hint one of your answers is correct, but for the other answer you forgot to take something into account
    do you know what that was ?


    Dave
     
    Last edited: Nov 25, 2014
  16. Nov 25, 2014 #15

    HHH

    User Avatar

    1. Using highest frequency
    fobs = ((Vsound + Vdetector) / (Vsound+Vsource)) f0
    640= ((344.12 + 0) / (344.12- Vsource)) 345
    640/345 = 344.12/344.12-Vsource
    1.885 = 344.12/344.12-Vsource
    344.12 = 638.367 - 1.885Vsource
    344.12 - 638.367 = -1.885Vsource
    -294.247= -1.885Vsource
    Vsource = 158.6 m/s

    1. Using lowest frequency
    fobs = ((Vsound + Vdetector) / (Vsound+Vsource)) f0
    50 = ((344.12 + 0) / (344.12 + Vsource)) 345
    50/345 = 344.12/344.12+Vsource
    0.1449 = 344.12/344.12+Vsource
    344.12 = 49.87 + 0.1449Vsource
    344.12 - 49.87 = 0.1449Vsource
    294.24 = 0.1449Vsource
    Vsource = 2030.64 m/s
     
    Last edited: Nov 25, 2014
  17. Nov 25, 2014 #16

    HHH

    User Avatar

    I am not sure what i did wrong.... it seems right to me
     
    Last edited: Nov 25, 2014
  18. Nov 25, 2014 #17

    davenn

    User Avatar
    Science Advisor
    Gold Member

    OK to lessen the complication you only need to work out the Vs for the increase in freq ( as the source is coming towards you

    you got the right answer 158 m/s

    taking into account that the velocity is constant, the velocity of the source as it is moving away from you is still 158 m/s its just going to be negative

    ( was trying myself to justify the same result and haven't been able to do that haha)
    someone may show us both how to use the 50Hz and get the -158 m/s ... as I said my maths suck LOL

    here's an example graphic I took out of a physics paper that showed the rearranged formula to solve for Vs


    doppler2.gif
     
  19. Nov 25, 2014 #18

    HHH

    User Avatar

    Isnt the speed of 158 m/s way to fast... for some swinging something above their head?
    Also, can anyone explain why it is different when moving towards and moving away... i still don't understand it.
     
  20. Nov 25, 2014 #19

    davenn

    User Avatar
    Science Advisor
    Gold Member

    we can only go on the observed sound freq's given in the problem ( they may or may not be realistic) its irrelevant, we both were able to come up with a correct solvable answer :)

    your moving away speed of ~ 2000m/s is incorrect. As I said, because the source velocity is constant the moving away speed is the same, just negative

    Dave
     
  21. Nov 25, 2014 #20

    HHH

    User Avatar

    Thanks for the help
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted