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Continuous String; Initial Conditions; Normal Modes

  1. Dec 1, 2006 #1
    1. The problem statement, all variables and given/known data
    Discuss the motion of a continuous string when the initial conditions are q'(x,0) = 0 and q(x,0) = Asin(3πx/L). Resolve the solution into normal modes. Show that if the string is driven at an arbitrary point, none of the normal modes with nodes at the driving point will be excited.

    2. Relevant equations
    i. q(x,0) = Asin(3πx/L)
    ii. q'(x,0) = 0
    iii. q(x,t) = Σ[μsin(nπx/L)cos(ωt)]
    iv. ω = (nπ/L)[(T/μ)^(1/2)]
    <the sum is from n=1 to infinity, and the ω is different for each n>
    <μ is a function of the amplitude>

    3. The attempt at a solution
    At t=0, for iii, cos(ωt)=1. Therefore, the string is oscillating with n=3. I assumed that this is the 3rd normal mode. Therefore, by iv, the string is oscillating with a frequency of ω = (3π/L)[(T/μ)^(1/2)].
    From this point, I do not know which method to take to find the normal modes.
    I will be working on the question in the mean time. Please give any helpful suggestions. Thank you.
  2. jcsd
  3. Dec 2, 2006 #2


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    But you just did !
  4. Dec 2, 2006 #3
    Well, what I think is that's the frequency at t=0; but how about at some other time? Are there any other modes? And how can I find them?

    Also, I am completely stumped about the second part of the question, which refers to the driving force.
  5. Dec 2, 2006 #4


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    What does driven mean? From the statement about not exciting any modes that have nodes at the driving point, I think it means an additional time dependent boundary condition, as opposed to some additional force being applied at some point. At some point x_o the motion is determined, possibly for all time (such as with a mechanical clamp moving sinusoidally in time, or perhaps as a square, triangular, or sawtooth vibrator), or possibly just as an initial condition such as plucking the string.

    You might find the simulators at this site interesting

    http://www.kw.igs.net/~jackord/bp/n2.html [Broken]
    Last edited by a moderator: May 2, 2017
  6. Dec 2, 2006 #5
    Before I can discuss the driven part... how do I do the first part with the normal modes?
    At t=0, the mode is n=3... how about some other time? How can I find it?
  7. Dec 3, 2006 #6
    Just to let everyone know, I still don't have solutions for my above questions.

    Any help is appreciated, thank you.
  8. Dec 3, 2006 #7


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    I thought you had gotten past the first part. Let's go back to your series

    iii. q(x,t) = Σ[μsin(nπx/L)cos(ωt)]
    iv. ω = (nπ/L)[(T/μ)^(1/2)]
    <the sum is from n=1 to infinity, and the ω is different for each n>
    <μ is a function of the amplitude>

    Each sin(nπx/L)cos(ωt) term in the series is one possible mode of vibration of the string. The use of μ in your series is an unusual symbol for represnting the coefficients because it was also used for the mass density. Whatever letter you use, it also needs a subscript. This just represents the amplitude of the contribution from one mode of oscillation. I am going to use a more conventional representation of b_n for the coefficient. What this eqation represents is a superposition of all possible modes of oscillation of the string, and should be written as

    [tex] q(x,t) = \sum\limits_{n = 1}^\infty {b_n \sin (n\pi x/L)\cos (n\pi vt/L)} = \sum\limits_{n = 1}^\infty {b_n \sin (n\pi x/L)\cos (\omega _n t)} [/tex]

    where v = [(T/μ)^(1/2)] is the velocity. The factors [itex] {b_n \sin (n\pi x/L} ) [/itex] are position dependent amplitudes of the harmonic motion of each little bit of mass dm at position x of the string. At positions where the sine is zero, there is no motion (node) associated with mode n, and where the sine is 1, there is maximum amplitude (antinode) associated with that mode.

    The transverse velocity at each position x is given by

    [tex] q'(x,t) = - \sum\limits_{n = 1}^\infty {b_n \omega _n \sin (n\pi x/L)\sin (\omega _n t)} [/tex]

    Your conditions

    i. q(x,0) = Asin(3πx/L)
    ii. q'(x,0) = 0

    are saying that at time zero every point of the string is at rest (zero velocity) and that the momentary shape of the string is described by Asin(3πx/L). What you need to do and perhaps already have done is set the initial position function equal to the time dependent series at the particular time (zero) the shape of the string is known

    [tex] q(x,0) = A\sin (3\pi x/L) = \sum\limits_{n = 1}^\infty {b_n \sin (n\pi x/L)} [/tex]

    and use this equality to determine the values of the coefficients in the series. This is particularly easy in this case, because the function on the left that describes the initial shape is identical to one of the functions in the series. It can be shown that in such cases, only one of the terms of the series has a non-zero coefficient, the one in the term that matches the function on the left.

    In general you would have to also consider the initial velocities. If q'(x,0) had been other than zero, then you would not have been able to express the time dependence as a pure cosine function. You would have to add a phase factor to the cosine or use a sum of sine and cosine functions for the time dependence.

    The first part of your problem is essentially done. b_3 = A and all other b_n are zero. The series bcomes a single term. This determines the motion of the string for all time. The string vibrates in the n = 3 mode, and that is all there is to it. You have one product of the position dependent sine and the time dependent cosine with n = 3 multiplied by A.

    When you move on to the driven part, you are going to have to learn how to find all the coefficients in the series. This is what Fourier analysis is all about. You will need to impose some condition on the shape of the string and its velocity at some point x. There are many possible ways to drive a string, so the problem statement is pretty open ended. I suggest that if this is your first look at such a problem you extend the simple first part of this problem to a different shape at time zero, but with every part of the string still at rest. One such shape is a triangle of height A with a vertex at some x value and the other two vertices at x = 0 and x = L where the string is held fixed. This corresponds to plucking a string.

    Take a look at what you have seen to do the Fourier analysis of a periodic function, and see if you can come up wih the equations for determining the coefficients in this case. The problem is trying to get you to prove than no mode that has a node at the driving position can have a non-zero coefficient.
  9. Dec 3, 2006 #8
    Thank you,

    The first part I understand.
    However, the driving force can be at an arbitrary point, which confuses me.
    The driving force I'm used to dealing with is [tex]krcos(\omega t)[/tex], or more simply, I think [tex]cos(\omega t)[/tex] will suffice.

    Since it could be at an arbitrary point, I do not know how to set up my equation for q and q'. I assume that q'(x,t) = cos(wt) since it is the velocity that's being changed with the driving force?

    But, I guess the position should also have that factor of cos(wt)... ?

    Then, from there, I am not sure of the method you speak of. Are you saying I should choose one particular case and then assume it's true for all cases?
    Last edited: Dec 3, 2006
  10. Dec 4, 2006 #9


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    At the risk of oversimplifying, a hamonic driving force leads to a steady state harmonic motion of the string. If you assume that at an arbitrary point x_o the motion is described by

    [tex] q(x_o,t) = A(x_o)\cos (\omega t)} [/tex]

    and use your expansion, generalized to allow for a different time phase you have

    [tex] q(x,t) = \sum\limits_{n = 1}^\infty \sin (n\pi x/L) \left[ {a_n \sin (\omega _n t)} + {b_n \cos (\omega _n t)} \right] [/tex]

    then you have

    [tex] q(x_o,t) = \sum\limits_{n = 1}^\infty \sin (n\pi x_o/L) \left[ {a_n \sin (\omega _n t)} + {b_n \cos (\omega _n t)} \right] [/tex]

    But if

    [tex] \sin (n\pi x_o/L) = 0 [/tex]

    there can be no contribuion from a mode with time dependence of the form

    [tex] {a_n \sin (\omega _n t)} + {b_n \cos (\omega _n t)} [/tex]

    In other words, mode n cannot contribute.
    Last edited: Dec 4, 2006
  11. Dec 4, 2006 #10
    Thank you very much. I ended up solving this earlier today.

    Thank you for all your help.
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