Doppler Effect Homework: Solve 600Hz Sound Generator & 100RPM Rope

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The discussion revolves around solving a physics problem related to the Doppler effect using a 600 Hz sound generator tied to a rope, spun at 100 RPM. Initial calculations for frequency shifts were incorrect due to misapplication of the Doppler effect formula and misunderstanding of the variables involved. Participants clarified that the speed of sound should be used in the calculations, and the correct approach involves adjusting the formula to include proper parentheses and signs for approaching and receding sources. After corrections, the final frequencies were determined to be approximately 618.9 Hz when approaching and 582.2 Hz when receding. The conversation highlights the importance of correctly applying physics equations and understanding the variables in context.
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Homework Statement


A physics professor demonstrates the Doppler effect by tying a 600 Hz sound generator to a 1.0-m-long rope and whirling it around her head in a horizontal circle at 100 rpm.


Homework Equations


f =fo(1/ (V +/- Vsource/V))


The Attempt at a Solution


T = 1/(100r/min)*((1min/60s)) = 0.6seconds
d = 2pi(1.0m)
vsource = d/t = 2pi /0.6 = 10.4719 m/s

f+ = 1.749Hz
f- = 1.749Hz

which is obviously wrong...because they should be hearing different frequencies when source is moving towards or away from the students. can anyone tell me what i am doing wrong?
 
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Sure, one shift is added to 600Hz and the other subtracted from 600Hz. Corresponding to the +/- in the formula. But your frequency shifts look like they are off as well, by something like a factor of 10.
 
omgosh ... i got the equation wrong...lol it's fo /(1 +/- vs/v)
my bad. thanks
 
It's correct if you put another pair of parentheses in f =fo(1/ ((V +/- Vsource)/V)), which then turns into what you just said. Which I thought is what you meant. No problem.
 
Oh! but how do we find the velocity (V) ? It's 100 rpm, so do we convert that to revolution per second? Thanks!
 
One minute has sixty seconds.
 
Ok, so that means 1.67 revolution per second. If I substitute that into the equation
f =fo(1/ ((V +/- Vsource)/V))

f+ = 600(1/ ((1.67 + 10.4719)/1.67))
= 82.39
f- = 600(1/ ((1.67 - 10.4719)/1.67))
= -113.60

Is that right?
 
no, first of all your v is the speed of sound. Your v_s is the speed 10.4719. You only use the period to find the velocity.

so its

f+ = 600/(1- (10.4719/343) ( you got the signs messed up in your equation, - for approaching,+ for receding)

try that see if it works
 
Last edited:
Draco! Yes! Thanks a lot! :D:D:D

f+ = 618.9
f- = 582.2
 
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Check your messages in your inbox on this site!
 
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hey Draco, do u go to UTSC?
 

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