Waves and sound frequency

  • Thread starter MozAngeles
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Homework Statement


Two identical strings with the same tension vibrate at 630 Hz.If the tension in one of the strings is increased by 2.20% what is the resulting beat frequency?


Homework Equations



fbeat=⎮f2-f2
f=v/λ
f1=v/2L
v=√(Tension/μ)
μ=m/L

The Attempt at a Solution


I don't know where to start because with every equation i want to start off using there is something missing.
 

Answers and Replies

  • #2
collinsmark
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I don't know where to start because with every equation i want to start off using there is something missing.
Start off by keeping everything in terms of variables (don't try plugging in any numbers yet). Combine your equations to determine a relationship between a string's tension and its fundamental frequency.

Once you have that, you can treat all the other variables as one big constant. And you'll be able to calculate what change in frequency corresponds to a given change in tension.
 
  • #3
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so what i did was:
since f=v/λ
v=√(T/(m/L))
λ=2L
f1=√(T/(m/L))/2L
solving for Tension,
T=m4Lf12
then I plugged that into
f2=v/λ
since T2 is 2.2%, i used .022T
f2= √(.022T/μ)/2L
f2=√(.022∗m*f2*4L/(m/L))/2L
mass cancels
, then L eventually does too
and i get 93.4 for f2
then subtracting f1-f2 to get f beat i get 537, which isnt right...
So i do not know where i am going wrong? do i have the right idea going?
 
  • #4
collinsmark
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Gold Member
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1,477
so what i did was:
since f=v/λ
v=√(T/(m/L))
λ=2L
f1=√(T/(m/L))/2L
solving for Tension,
T=m4Lf12
You didn't really need to solve for T. :tongue2: But there's nothing wrong with doing so. Anyway...

The important point to gather from the above equations is that the fundamental frequency of a string is proportional to the square root of the tension. In other words,

[tex] f_1 \propto \sqrt{T} [/tex]
then I plugged that into
f2=v/λ
since T2 is 2.2%, i used .022T
Your above approach would work if the problem statement said, "If the tension in one of the strings is decreased to a value that is 2.20% of its original tension..."

But that's not what it says. It says, "If the tension in one of the strings is increased by 2.20%." That means the new tension of the string is what it was before, plus an additional 2.20%. In other words, the new tension is 1.022T.
 
  • #5
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That worked.. Thank you soooo much for the help :)
 

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