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Homework Help: HELP with Investigation

  1. Oct 31, 2005 #1
    Ive been given the task of investigating what factors will affect the resonant frequency of a piece of string or wire :cry: . I believe it may have something to do with standing or stationary ways. At the moment I believe I have two ways of doing this, one by using string, vibration generator and pulley, the other by using wire, magnet and rheostat.

    Both I believe involve the equation fo = ½L x (T/µ).

    fo = fundamental frequency
    L = length of vibrating wire
    T = tension
    µ = mass per unit length

    Could anyone who has any experience with this formula at all, or knows how to put it into use tell me, particularly the length part, is it node to node or arbitrary lengths. Thanks
  2. jcsd
  3. Oct 31, 2005 #2
    edit - nevermind
    Last edited: Oct 31, 2005
  4. Oct 31, 2005 #3


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    I think you left out a square root.
  5. Oct 31, 2005 #4
    From what I know about waves and diff eq's:
    The equation of a wave is
    [tex]\frac{1}{v^2}\frac{\partial^2 y} {\partial t ^2}=\frac{\partial^2 y}{\partial x^2}[/tex]
    for which a particular solution is the standing wave with say, the fundamental frequency:
    [tex]y(x,t) = A(x)\cos(\omega_0 t)= A\sin(kx)\cos(\omega_0 t)[/tex]
    Where A(x) is the amplitude function, [itex]\omega_0[/itex] is the natural angular frequency, and k is the wave number.
    now since, I'm guessing, that string wave behaves like a harmonic oscilator at any given point along the strong (strictly speaking [itex]y(t,x_0) = A(x_0)cos(\omega_0 t)[/itex]) I'm going to venture a guess that it obeys the differential equation:
    [tex]my'' + cy = 0[/tex]
    Where c is the restoring force proportionality constant or whatever (F = -cx)
    Now, let's impose an external force on it
    [tex]my'' + cy = B(x)cos(\omega t)[/tex]
    Well, we already set the complementary solution as
    [tex]y_c(t) = A(x)\cos(\omega_0 t)[/tex]
    so then the particular solution becomes
    [tex]y_p(t) = B(x)\cos(\omega t)[/tex]
    so that
    [tex] y(t) = \frac{B(x)}{c-m\omega^2} \cos(\omega t) + A(x) \cos(\omega_0 t)[/tex]
    (you can do some trig manipulation and combine this into a product of cosines making the graph look like a wave envoloped in a larger wave)
    Edit -
    but... what happens when [itex]\omega_0 = \omega[/itex] then the particular solution becomes a solution to the associated homogeneous equation.
    Again, complementary solution is
    [tex]y_c(t) = A(x)\cos(\omega t) [/tex]
    but the particular solution becomes
    [tex]y_p(t) = \frac{B(x)t}{2\omega m} \cos(\omega t)[/tex]
    such that
    [tex]y(t) = A(x)\cos(\omega t) + \frac{B(x)t}{2\omega m} \cos(\omega t)[/tex]
    [tex]=(A(x)+ \frac{B(x)t}{2\omega m})\cos(\omega t)[/tex]
    Notice that the total amplitude increases as t increases so what you have is a wave oscillating in between the lines y=+-b/2wmt +- a. However, none of this factors dampening into play but if you do, the equations become a little bit more complicated.
    Disclaimer: I just applied my knowledge of springs and differential equations to my knowledge of standing waves, it may or may not be right.
    Edit 3 - hurrr all that work for nothing, after googlin, the resonant frequency is just another name for natural frequency
    in that case:
    [tex] \sin(kL) = 0[/tex]
    [tex] kL = n\pi[/tex]
    [tex]\frac{2\pi L}{\lambda} = n\pi[/tex]
    [tex]\lambda = \frac{2L}{n}[/tex]
    [tex]v=f\lambda=\sqrt{T \over \mu}[/tex]
    [tex]f_n = \frac{n}{2L}\sqrt{T \over \mu}[/tex]
    Last edited: Oct 31, 2005
  6. Nov 1, 2005 #5
    Thanks, but what does the N stand for in your version, is it just a constant?

    (and yes I did miss out a square root, silly me, thanks for taking the time to do all that)
  7. Feb 24, 2006 #6
    [tex]\sqrt{T}[/tex] ah so thats how it works
    Last edited: Feb 24, 2006
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