Doppler effect with 600 Hz generator on rotating rope

[neoking77]

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?

 

[Dick]

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.

 

[neoking77]

omgosh ... i got the equation wrong...lol it's fo /(1 +/- vs/v)
my bad. thanks

 

[Dick]

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.

 

[meghamala]

Oh! but how do we find the velocity (V) ? It's 100 rpm, so do we convert that to revolution per second? Thanks!

 

[Dick]

One minute has sixty seconds.

 

[meghamala]

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?

 

[Draco]

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

 

[meghamala]

Draco! Yes! Thanks a lot! :D:D:D

f+ = 618.9
f- = 582.2

 

[Draco]

Check your messages in your inbox on this site!

 

[strikingleafs01]

hey Draco, do u go to UTSC?