Submarine Distance - Hard Doppler effect

AI Thread Summary
The discussion revolves around a homework problem involving two submarines moving towards each other and the Doppler effect of sonar signals. The French submarine emits a sonar signal at 1100 Hz, which is detected at 1222 Hz by the French submarine after reflecting off the U.S. submarine. Participants are trying to determine the speed of the U.S. submarine, with calculations leading to confusion and errors in unit conversions and mathematical steps. The correct approach involves using the Doppler effect equations for both the source and observer, with the expected speed of the U.S. submarine being approximately 65 m/s. The conversation highlights the challenges of solving the problem and the importance of accurate assumptions in calculations.
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Homework Statement



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 traveling towards the French submarine? Assume the velocity of sound in sea water is 1500 ms-1.

Homework Equations





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.
 
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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 because when I sub the first into the second Vus cancels out...
 
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 doesn't cancel)
Thats what i did and everybody at uni i talked to... should come out at about 65ish
 
cool, thanks Pat666
 
that formula you used is the one out of the textbook for both the source and observer moving isn't it??
 
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.
 
what did you end up getting for the speed
 
just doing it now but be buggered if I can figure it out! keep getting 0.33km/hr or 0.0905m/s
 
did you get up to here the same??
 

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  • #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

How have you been finding it all? You seem to have a fairly good grasp. Are you on campus?
 
  • #11
yeah group A for eneg skills... you get around 65m/s.... Have you done the rest of the questions? I am 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??
 
  • #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
 
  • #13
Im pretty sure your Q 5 is wrong and i think my Q 9 is wrong.
 
  • #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
 
  • #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
 
  • #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)
 
  • #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
Do your term 1 classes come up in cqucentral i can't get any of my cover sheets for my portfolio or physics assignment.
 
  • #20
No it's only showing term 2?
 
  • #21
yah I fured it out there's a tab that says "other academic" click on it and go to assignments and there all there... ps did you end up changing your sound velocity assumption?
 
  • #22
hehehe yep :blushing:
 

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