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Mass Connected to two springs

  1. Sep 2, 2013 #1
    1. The problem statement, all variables and given/known data
    A mass [itex]m[/itex] rests on a frictionless horizontal table and is connected to rigid supports via two identical springs each of relaxed length [itex]l_{0}[/itex] and spring constant [itex]k[/itex]. Each spring is stretched to a length [itex]l[/itex] considerably greater than [itex]l_{0}[/itex]. Horizontal displacements of [itex]m[/itex] from its equilibrium position are labeled [itex]x[/itex] (along AB) and [itex]y[/itex] (perpendicular to AB).

    *see attached image or link*

    (a) Write down the differential equiation of motion (i.e., Newton's law) governing small oscillations in the [itex]x[/itex] direction.
    (b) Write down the differential equiation of motion governing small oscillations in the [itex]y[/itex] direction (assume [itex]y[/itex]<<[itex]l[/itex]).
    (c) In terms of [itex]l[/itex] and [itex]l_{0}[/itex], calculate the ratio of the periods of oscillation along [itex]x[/itex] and [itex]y[/itex].

    The rest of the question can be found on page 86 at http://www.scribd.com/doc/160672855/Vibrations-and-Waves-a-P-French



    2. Relevant equations
    [tex]
    F = ma \\
    a^{2} + b^{2} = c^{2} \\
    sin(\theta)≈tan(\theta) \text{ when } \theta \text{ is small} \\
    (1+a)^{n} ≈ 1 + na \frac{n(n-1)a^{2}}{2!} +...\\
    [/tex]



    3. The attempt at a solution
    (a) I imagined a small displacement [itex]x[/itex] to the right, which I also declared to be positive.
    [tex]
    -k(l + x - l_{0}) + k(l-x-l_{0}) = ma \\
    -2kx = ma \\
    0 = ma + 2kx
    [/tex]

    (b) I tried going about this using some of the techniques discussed while doing question 3-7 which can be found here: https://www.physicsforums.com/showthread.php?t=694259 But my answer doesn't match the one in the back of the book and I'm not sure why.

    I also imagined a small displacement [itex]y[/itex] so there is some force in each of the springs such that [itex]F_{L} = F_{R} = F[/itex], then the equation of motion in the [itex]y[/itex] direction looks like:
    [tex]
    2Fsin(\theta) = ma \\
    sin(\theta) ≈ tan(\theta) = \frac{y}{l} \text{ *see attached attempt at inkscape*} \\
    2F \frac{y}{l} = ma
    [/tex]
    [tex]
    F = k(l'-l_{0}) \\
    F = k\left(\left(y^{2}+l^{2}\right)^{\frac{1}{2}}-l_{0}\right)
    [/tex]
    [tex]
    2k\left(\left(y^{2}+l^{2}\right)^{\frac{1}{2}}-l_{0}\right) \frac{y}{l} = ma\\
    2k\left(l\left(1+\left(\frac{y}{l}\right)^{2}\right)^{\frac{1}{2}}-l_{0}\right) \frac{y}{l} = ma\\
    \text{using the binomial approx...}\\
    2k\left(l\left(1+\frac{1}{2}\left(\frac{y}{l}\right)^{2}\right)-l_{0}\right) \frac{y}{l} = ma\\
    2k\left(y+\frac{y^{3}}{2l^{2}}-\frac{l_{0}}{l} y\right) = ma \\
    \text{assuming y is small} \\
    2k\left(1-\frac{l_{0}}{l}\right)y = ma \\
    0 = ma + 2k\left(1-\frac{l_{0}}{l}\right)y
    [/tex]
    Also I see in the last step there is something going on with a sign, maybe it's because the way I set it up the displacement is in the negative direction.

    (c)
    [tex]
    \omega_{x} = \left(\frac{2k}{m}\right)^{\frac{1}{2}}\\
    \omega_{y} = \left(\frac{2k\left(1-\frac{l_{0}}{l}\right)}{m}\right)^{\frac{1}{2}}\\
    \frac{T_{x}}{T_{y}} = \left(1-\frac{l_{0}}{l}\right)^{-\frac{1}{2}} \\
    \text{The answer in the back of the book is:}\\
    \frac{T_{x}}{T_{y}} = \left(1-\frac{l_{0}}{l}\right)^{\frac{1}{2}}
    [/tex]
     
  2. jcsd
  3. Sep 3, 2013 #2

    vela

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    Staff Emeritus
    Science Advisor
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    You just made an algebra mistake in the last step. The ratio ##\frac{T_x}{T_y} = \frac{\omega_y}{\omega_x}## has ##\omega_y## on top so it should be proportional to ##(1-l/l_0)^{+1/2}##.
     
  4. Sep 3, 2013 #3
    because
    [tex]
    T_{x} = \frac{2 \pi}{\omega_{x}}
    [/tex]
    right on vela, thank you.
     
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