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Submarine Distance - Hard Doppler effect

  1. May 26, 2010 #1
    1. The problem statement, all variables and given/known data

    A French and a U.S. submarine are moving directly towards each other during manoeuvres in still water. The French submarine is moving at 50.0 km.hr-1. It sends out a sonar signal at 1100.0 Hz. The frequency detected by the French submarine (reflected back from the U.S. submarine) is 1222 Hz. How fast is the U.S. submarine travelling towards the French submarine? Assume the velocity of sound in sea water is 1500 ms-1.

    2. Relevant equations



    3. The attempt at a solution
    Im thinking that the us submarine becomes a "virtual source" and that i have to solve this in two parts????? I need some help getting started(and finishing probably). Thanks to any helpers.
     
  2. jcsd
  3. May 29, 2010 #2
    I am stumped on this as well...

    towards the us sub I have

    f = (1500 +Vus / 1500 - 50) x 1100

    Then back to french I have

    1222 = (1500 +50 / 1500 - Vus) f

    But I assume this is incorrect cuz when I sub the first in to the second Vus cancels out....
     
  4. May 29, 2010 #3
    1ST put your velocuty into m/s (frequency is in hz) then 1222 = (1500 +50 / 1500 - Vus)*1500 +Vus / 1500 - 50) x 1100 (v_us doesnt cancel)
    Thats what i did and everybody at uni i talked to........ should come out at about 65ish
     
  5. May 29, 2010 #4
    cool, thanks Pat666
     
  6. May 29, 2010 #5
    that formula you used is the one out of the text book for both the source and observer moving isnt it???????
     
  7. May 29, 2010 #6
    yeah I subed the first eq (above post) into second but didn't change km/hr to m/s and made a mathematical error when working it out.
     
  8. May 29, 2010 #7
    what did you end up getting for the speed
     
  9. May 29, 2010 #8
    just doing it now but be buggered if I can figure it out! keep getting 0.33km/hr or 0.0905m/s
     
  10. May 29, 2010 #9
    did you get up to here the same??
     

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  11. May 29, 2010 #10
    :redface: completely missed a step

    I have been doing this assignment all weekend + all friday... I think my brain is officially fried!

    Thank goodness this is my last question :yuck:

    How have you been finding it all? You seem to have a fairly good grasp. Are you on campus?
     
  12. May 29, 2010 #11
    yeah group A for eneg skills.......... you get around 65m/s............. Have you done the rest of the questions???? Im not sure on my answer for the buoyancy 1.
    My answers:
    1)-.892C and an overflow of 0.7cm^3
    2)941Hz and 938Hz and obviously 65m/s for sub
    3)D of 15.6cm and height of 2.2m when the diametre is doubled
    4)150m/s , 1.8*10^8 Nm^2 , Y=1.8*10^9Pa
    5)open 139,278,417Hz and closed 69.7,209.1,348.6Hz and for b) 20 or 30Hz
    6)4.389*10^7 n/m^2
    7) went through on different thread same answer as you
    8)2592grams
    9)a) 2157.89kg/m^3 b) D 4.7kg and A 2.8kg

    you disagree with any of these??
     
  13. May 29, 2010 #12
    I finally get 65m/s :cry: because I am soooo happy!

    Only differences are:

    5. a) (i) 1st Harm. 7.32Hz 2nd Harm. 14.6Hz 3rd Harm. 22Hz
    (ii) F1 = 3.66Hz ; f3 = 11Hz f5 = 18.3Hz

    9. a) 1237kg/m^3
    b) D reads Mblock = 8.2kg E reads Mfluid + Mbeaker = 2.8kg
     
  14. May 29, 2010 #13
    Im pretty sure your Q 5 is wrong and i think my Q 9 is wrong.
     
  15. May 29, 2010 #14
    I though Q5 was asking for individual frequencies of the first 3 harmonics (i) open and (ii) closed

    I used for open f1 = v/2L f2 = 2f1 f3 = 3f1 and closed f1 = v/4L f3 = 3f1 f5 = 5f1
     
  16. May 29, 2010 #15
    Q5)

    Since the pipe is open at both ends all harmonics are present
    f_1 =v/2L=343/(2*1.23) since lambda is half of the pipe length
    f_2=2f_1=278Hz
    f3=3f_1=417Hz
    closed
    In this case only odd harmonics are present
    f_1=v/4L=69.7Hz
    f3=3f_1=209.1Hz
    f5=5f1=348.6HZ

    half your answers arnt audible <12Hz

    Can you show me how you did Q9 pls
     
  17. May 29, 2010 #16
    what you said in your last post is true but you must have screwed up your calculations - what did you use for sound velocity ( that shouldve been an assumption that it is still 343 m/s)
     
  18. May 29, 2010 #17
    Ah... was 343 your assumption? I used a different value.

    Q9.

    Mblock - Mdisplaced fluid - D = 0

    Mblock = 4.7 + 3.5 = 8.2kg

    Or

    D + E = Mbeaker + M liquid + Mblock

    11 = 1 + 1.8 + Mblock

    Mblock = 11 - 2.8 = 8.2kg
     
  19. May 29, 2010 #18
  20. May 29, 2010 #19
    Do your term 1 classes come up in cqucentral i cant get any of my cover sheets for my portfolio or physics assignment.
     
  21. May 29, 2010 #20
    No it's only showing term 2???
     
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