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

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Homework Help Overview

The discussion revolves around a physics problem involving the Doppler effect, where a 600 Hz sound generator is tied to a rope and whirled in a circle at 100 RPM. Participants are exploring the calculations related to frequency shifts as the sound source moves towards and away from an observer.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the correct application of the Doppler effect formula and the calculations for frequency shifts. There are attempts to clarify the use of velocity and the correct signs in the equations. Questions arise about converting RPM to a usable velocity for the calculations.

Discussion Status

The discussion has seen participants correcting each other’s equations and clarifying misunderstandings about the Doppler effect formula. Some have provided guidance on how to properly substitute values into the equations, while others are still working through the calculations and seeking confirmation of their results.

Contextual Notes

Participants are addressing potential errors in calculations and assumptions about the speed of sound and the relationship between RPM and linear velocity. The original poster expresses uncertainty about their approach and seeks further clarification.

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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
 
  • #10
Check your messages in your inbox on this site!
 
  • #11
hey Draco, do u go to UTSC?
 

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