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Fundamental frequency and changes to it

  1. Jun 3, 2009 #1
    1. The problem statement, all variables and given/known data

    A stretched wire vibrates in its fundamental mode at a frequency of 384 Hz. What would be the fundamental frequency if the wire were half as long, its diameter were doubled, and its tension were increased five-fold?

    2. Relevant equations

    F= [tex]\frac{1}{2L}[/tex][tex]\sqrt[]{\frac{T}{\mu}}[/tex]

    3. The attempt at a solution
    ok i know that this F= [tex]\frac{1}{2L}[/tex][tex]\sqrt[]{\frac{T}{\mu}}[/tex] is the formula for the one where F1= 384 Hz

    i know that length is cut in half and that tension increases by 5and when the diameter is doubled that means mass is quadrupled my problem is that i dont know how to show that wrt [tex]\mu[/tex]
    for the second frequency would my equation look like
    F2= [tex]\frac{1}{2(L/2)}[/tex][tex]\sqrt[]{\frac{5*T}{4*\mu}}[/tex]
    and then if we cancel off the 2 in the l would it look like F= [tex]\frac{1}{L}[/tex][tex]\sqrt[]{\frac{5*T}{\mu}}[/tex] *[tex]\frac{1}{2}[/tex]
     
  2. jcsd
  3. Jun 3, 2009 #2

    LowlyPion

    User Avatar
    Homework Helper

    Looks like you have it then right?

    384*√5
     
  4. Jun 3, 2009 #3
    thanx i wasnt sure if it looked right
     
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