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Transverse vibrations

  1. Sep 25, 2006 #1
    I'm stuck on question in where I need to set up a system of equations, can someone please help me with it? There are various things in this question which I just cannot figure out so any assistance would be great.

    Q. A light string stretched to tension T has its ends fixed at a distance 9L apart. Masses 2m, m are attached to the string distances L, 5L respectively from one of its fixed ends. The system undergoes transverse vibrations.

    Show that the equations of motion are

    8\mathop x\limits^{ \bullet \bullet } + n^2 \left( {5x - y} \right) = 0
    4\mathop y\limits^{ \bullet \bullet } + n^2 \left( { - x + 2y} \right) = 0

    where [tex]n^2 = \frac{T}{{mL}}[/tex].

    Note: Presumably they mean that the 2m and m masses weigh 2m kg and m kg respectively.

    Sorry for the lack of a diagram but the situation is rather simple so I'll try to describe it. Basically, take a string, stretch it till it reaches a length of 9L and then fix each of its two ends so that the string is horizontal. This is defined by the question as the equilibrium configuration of the string.

    The mass 2m is attached to the string at a distance L from the left end of the string while the mass m is attached to the string at a distance of 5L from the left end of the string.

    When the system is set in motion the masses oscillate up and down. x(t) denotes the vertical displacement of the mass 2m while y(t) denotes the vertical displacement of the mass m.

    This has been a rather frustrating problem for many reasons. I just can't figure out what needs to be done. I'm used to questions where the masses oscillate in the longitudinal direction. From what I can gather, there will probably be square roots all over the place. But here are my thoughts anyway.

    When the string is in equilibrium (ie. when y(t) = x(t) = 0), the tension in the string is T as given in the question. In order to maintain static equilibrium of the two masses in the horizontal the tension in the string on either side of them must be equal to T. So the spring constant of the string is k = T/(9L)

    When the masses start moving up and down the string is stretched further. So for example the distance between the mass 2m and the left most point increases from the original distance L to sqrt(L^2+x^2). But does this really mean anything in the context of the problem?

    I can't get anywhere with this question nor can I figure out how n^2 could possibly be equal to T/(mL)...it just doesn't make sense because n^2 = T/(mL) implies that the spring constant of the string is T/L even though the length of the string is 9L and not L. Any helpful advice would be really good thanks.
  2. jcsd
  3. Sep 25, 2006 #2
  4. Sep 25, 2006 #3
    So following the example from the link I posted before (which looks wrong to me anyway - but I only really skimmed trough it):

    First of all let me state the problem in this way - the total length of the string is [itex]L[/itex], with fixed points at [itex]x=0[/itex] and [itex]x=L[/itex] and the tension in the string is [tex]T[/itex]. A mass [itex]m_1[/itex] is attached at position [itex]x=a_1[/itex] while a second mass [itex]m_2[/itex] is attached at position [itex]x=a_2[/itex].

    We now want to find the equations of motion for small amplitude transerve oscillations of masses [itex]m_1[/itex] and [itex]m_2[/itex].

    What do we mean by small-amplitude vibrations? If we denote the respective transerve positions of masses [itex]m_1[/itex] and [itex]m_2[/itex] as [itex]y_1[/itex] and [itex]y_2[/itex] then we will say that the amplitude of oscillation is small if:

    y_1 \ll a_1 \\
    |y_2-y_1| \ll a_2-a_1 \\
    y_2 \ll L-a_2 \\

    If you draw a diagram of this then these conditions should make sense (I hope). So this gives us the following small parameters:

    \frac{y_1}{a_1} \ll 1 \\
    \frac{|y_2-y_1|}{a_2-a_1} \ll 1 \\
    \frac{y_2}{L-a_2} \ll 1 \\

    We want to find set up our equations in the lowest order of these parameters. Write the tranverse position as [itex]y_{1/2}[/itex], and find the force acting on each particle.

    In equilibrium the the tension in the string was [itex]T[/itex]. But now there is additional stretching of the string due to the transverse oscillations. So how does this effect the tension? If the natural length of the string is [itex]L_0[/itex], then by Hooke's law, the tension [itex]T[/itex] should be:

    T = \lambda\frac{(L-L_0)}{L_0}.

    If we want to make a very rough estimate of how the tension in the string would change with these additional small amplitude oscillations, we could approximate the profile of the string as making a straight line between the fixed point on the LHS and the first mass, and then another straight line between the first mass and the second mass, etc. We then calculate the total length of the string, [itex]L^{\prime}[/itex]

    L^{\prime} &\sim \left(a_1^2+y^2_1\right)^{1/2} + \left([a_2-a_1]^2+[y_2-y_1]^2\right)^{1/2} + \left([L-a_2]^2+y_2^2\right) \\
    &= a_1\left(1+\left[\frac{y_1}{a_1}\right]^2\right)^{1/2} + [a_2-a_1]\left(1+\left[\frac{y_2-y_1}{a_2-a_1}\right]^2\right)^{1/2} + [L-a_2]\left(1+\left[\frac{y_2}{L-a_2}\right]^2\right)^{1/2} \\
    &\approx a_1\left(1+\frac{1}{2}\left[\frac{y_1}{a_1}\right]^2\right) + [a_2-a_1]\left(1+\frac{1}{2}\left[\frac{y_2-y_1}{a_2-a_1}\right]^2\right)+ [L-a_2]\left(1+\frac{1}{2}\left[\frac{y_2}{L-a_2}\right]^2\right) \\
    &= L +\mathcal{O}(\mathrm{all-of-our-small-parameters-squared}).

    So the increased tension in the string due to the oscillations can be neglected in this regime of small-amplitude oscillations if we want the solution to the lowest order in the above small paramaters.

    So we consider the transverse force acting on each particle with the tension in the string constant. For the first particle we have :

    F_1 =-\frac{Ty_1}{\left[y_1^2+a_1^2\right]^{1/2}}+\frac{T(y_2-y_1)}{\left[(y_2-y_1)^2+(a_2-a_1)^2\right]^{1/2}} \\
    \approx T\left\{\frac{y_2-y_1}{a_2-a_1}- \frac{y_1}{a_1}\right\}.


    m_1\ddot{y}_1=-\frac{Ty_1}{\left[y_1^2+a_1^2\right]^{1/2}}+\frac{T(y_2-y_1)}{\left[(y_2-y_1)^2+(a_2-a_1)^2\right]^{1/2}} \\
    \approx T\left\{\frac{y_2-y_1}{a_2-a_1}- \frac{y_1}{a_1}\right\}.

    Perform the same procedure for the second particle to get the equations of motion for the two masses.

    When you're solving these equations you should keep in mind the approximations you made when setting up the equations, and remember not to exceed the accuracy of the approximations you made.
    Last edited: Sep 25, 2006
  5. Sep 26, 2006 #4
    Thanks a lot for your extensive explanation, I'll need to read through it in more detail some time. This question was one one of four from a past exam (the duration of which was only one hour) so I really didn't anticipate that so much working was required. Perhaps the course has changed since then (8 years ago :D).

    Anyway, thanks again, much apprediated.
  6. Sep 27, 2006 #5
    No, probably this much working isn't required. I just put the bit there about Youngs modulus to justify using a constant tension in the string. All the examples I have seen about this kind of problem just breeze over the fact that the tension in the string might actually change, so I thought I'd try and justify it here.

    By the way is your exam in Newtonian mechanics, or are you studying Lagrangian/Hamiltonian mechanics?
  7. Sep 27, 2006 #6
    The exam where I found the question in was on systems of ODEs. It's introductory stuff so it's not specifically related to the more advanced topics you referred to.
  8. Sep 28, 2006 #7
    Alright, so the above (without the Young's modulus bit, and just defining what you consider to be small amplitude oscillations when linearizing your equations) should be enough.
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