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Vibrating string problem

  1. Dec 16, 2007 #1
    1. The problem statement, all variables and given/known data

    A vibrating string vibrates at 100Hz in oxygen.What is the frequency of its vibration in Hydrogen?[assume same temperature and pressure]


    2. Relevant equations
    3. The attempt at a solution

    Equating the velocities of sound gives 400Hz...But what pokes me is that the frequency is the property of the source.Will it change with ambient condition?
    In other words should not the answer be still 100 Hz?
     
  2. jcsd
  3. Dec 19, 2007 #2
    1) The tension in the string is does not depend on the medium.

    2) The work of the medium against the string as it vibrates transversely is a function of (or depends on) the medium. Example: Wave a stick in the air at 100Hz versus waving the stick in water at 100Hz. It takes more work, or energy, to over come the force of drag to achieve the same result (frequency).

    The governing equation for transverse string vibration is:

    [tex]\frac {\partial} {\partial x} \left(T \frac{\partial u}{ \partial x} \right)} - \rho(x) \frac {\partial^2 u} {\partial t^2} + p(x) = 0[/tex]

    This is the work of the "big boys" --> Bernoulli, d'Alembert and Euler --> not vector3

    By letting tension, T, be constant and p(x) = 0 the equation reduces to what we usually see in typical books on physics or boundary value problems:

    [tex]\frac {\partial^2 u} {\partial t^2} = a^2 \frac {\partial^2 u} {\partial t^2} [/tex]

    Which is the well known "Wave Equation".

    The affect you are looking to explore is finding a suitable force of drag [itex]F_d [/itex] function [itex]F_d (x)[/itex] and inserting it into the equation for p(x) subsequently solving.

    [itex]F_d [/itex] comes in a lot of flavors. I suppose I would look for one that made the equation as simple as possible to solve... if even solvable. We have a good equation... how do we solve it?

    Additionally, I submit the analogy of when a kid inhales helium from a balloon and begins to speak using their "normal voice" lung pressure. The vocal chords (the strings) vibrate at a much higher frequency compared to the frequency of vibration when under "normal voice" lung pressure using air. While the tension in the vocal chords is/was the same as when speaking using air the sound is much higher ie. higher frequency.

    I suggesting that the frequency is higher in the hydrogen environment compared to the oxygen environment because of the change in the force drag on the string.

    If you need the derivation of the governing equation I have references you may consult should you so desire.
     
    Last edited: Dec 19, 2007
  4. Dec 20, 2007 #3

    Shooting Star

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    I’ve searched high and low for any indication to the answer, but always landed up with those sites about talking with helium in your voice, and there too, you have widely differing opinions. Now vector3 has come up with a really awesome looking equation.

    One thing I don’t understand about this equation, though. Both in air and Helium we put p(x) = 0. Then how come the results are different, at least for vocal chords? Then we should put in the real drag forces, even though their difference for air and Helium may be small, but the solutions would be different. This seems the only way out. (The frequency difference between the two cases is not that much actually, but to us sounds very different, because our ears are sensitive to that bandwidth.)

    In some of the sensible sites, I found a simple explanation which maybe true, but nobody has derived it. It’s the spectral distribution, that is, the energy distribution in the harmonics that changes, so that in a denser medium, there is more energy in a note of lower frequency, in contrast to air where almost all the energy goes into the fundamental note. Obviously, there is a maximum somewhere. The drag force is important. This whole thing may possibly be derived from the PDE vector3 has given, after he/she solves it for us.

    Any comments from anybody?
     
  5. Dec 20, 2007 #4
    Of the truth, the helium analogy occurred to me without reading other web pages. Seemed a reasonable comparison...

    As I understand the derivation, p(x) is a forcing function. Actually, I was suggesting just substituting in a Fd(x) in lieu of p(x), and evaluating the response of the equation as the "original input energy" disipates as a result of the drag. However, the most appropriate way would be to start at the beginning of the derivation of the governing PDE and inserting a drag function (or damping function) to account for changes in the medium?

    Wikipedia (along with a few other references) relates Fd to both density of the medium and velocity of the object. This would introduce a first derivative (velocity) into the mix.

    I'll take a stab at it as time allows...
     
  6. Dec 20, 2007 #5

    Shooting Star

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  7. Dec 20, 2007 #6
    Thanks for the link... There is a mix of conjecture along with some good points made. I would be interested to see more on the wave energy discussion should you get a chance to post another link...

    I modified the original derivation of the governing PDE to include the drag of the medium.
    It looks like this:

    It all comes from [itex]\sum {F = m a} [/itex]

    [tex]\frac {\partial} {\partial x} \left(T \frac{\partial u}{ \partial x} \right)} - \rho_{gas} C_d A \frac{\partial u} {\partial t}-\rho_{str}(x) \frac {\partial^2 u} {\partial t^2} + p(x) = 0[/tex]

    Writing it like this looks more manageable:

    [tex]Tu_{xx} + p - k u_t -\rho u_{tt} = 0 [/tex]

    [tex]Tu_{xx} - k u_t -\rho u_{tt} = 0 [/tex] (Damped Free Vibration of a String)

    This is looks good for the transverse vibration of the string itself.

    As I mentioned in previous post, I'll see what I can find on a solution to the Damped Free Vibration of a String equation.
     
  8. Dec 26, 2007 #7
    [tex]Tu_{xx} - k u_t -\rho u_{tt} = 0 [/tex] (Damped Free Vibration of a String)

    I've been studying this equation. Ney, rather, I'm returning from a journey through "mathematical hell..." with the following observations and comments:

    1 Using separation of variables will get you where you need to be. ie U(x,t) = A(x)B(t)
    This will lead you to solving 2 Ordinary Differential Equations.

    2 It is not enough to just write the equation. It must be solved in the framework of either intial conditions ie. displacement and velocity or boundary conditions ie U(0,t)=0 and U(L,0)=0

    3 Mathematicians don't like to include the damping affect because the solution does not lend itself to reducing to simply using fourier series superposition. Thus a string that vibrates indefinitely is more manageable mathematically. So much for reality...

    A first approximation to the solution will take the following form:

    [tex] u(x,t)= \{C_1 \exp^{m_1 t}\} sin ({\frac{n \pi x}{L})[/tex] (1)
    with u(0,t) =0 ; u(L,t) =0

    A few comments about u(x,t):

    [itex] m_1 [/itex] is negative, therefore, as [itex] t \longrightarrow \infty \;\;\;U(x,t)\longrightarrow 0 [/itex]

    If U=0 at x=0 then (1) is satisfied
    If U=0 at x=L then (1) is satisfied

    Substituting (1) into the original PDE satisfies the PDE.

    The amplitude of vibration decays with time.

    Using the first approximation to the PDE does not include the other harmonics that may initially on the string. A nice neat fourier series with the "no work or losses" assumption will include the "odd" harmonics on the string.

    Drag, Tension, and string density are included in Constant C1.

    There is so much more work and information available on the undamped free vibration of a string. Not as much on the damped case. The first approx. (1) shown above is enough to gain a good insight on concepts (excepting the additional harmonics).
     
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