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Homework Help: Small oscillations+normal modes of a system

  1. Feb 20, 2012 #1

    fluidistic

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    1. The problem statement, all variables and given/known data
    Two identical pendulums of length [itex]l[/itex] hang from a ceiling. Their vertical axis is separated by a distance [itex]l_0[/itex]. They are made by 2 masses m. Between these 2 masses we put a spring of constant k and natural length [itex]l_0[/itex].
    Gravity acts verticaly downward.
    1)Calculate the proper frequencies of oscillations in the approximation of small oscillations.
    2)Determine the normal coordinates of oscillations.
    3)Suppose that at [itex]t=0[/itex] both pendulums are at rest for [itex]\theta _1=0[/itex] and [itex]\theta _2=\theta _0[/itex]; find and solve the equations of motion of the system.
    Note that [itex]\theta _1[/itex] and [itex]\theta _2[/itex] are the angles that each mass makes with the vertical.
    2. Relevant equations
    [itex]L=T-V[/itex].


    3. The attempt at a solution
    I tried to get the Lagrangian of the system which consists of the sum of 3 Lagrangians, namely one for each mass and one for the spring.
    I'm having trouble with the potential energy stored in the spring.
    Let [itex]L_1[/itex] and [itex]L_2[/itex] be the Lagrangians for each mass. I found out that [itex]T_1 =\frac{m}{2}l^2 \dot \theta _1 ^2[/itex] while [itex]T_2 =\frac{m}{2}l^2 \dot \theta _2 ^2[/itex]. Potential energy is simply [itex]V_1 = mgl [\cos (\theta _1 ) -1][/itex] and [itex]V_2 = mgl [\cos (\theta _2 ) -1][/itex]. So I can just make up the 2 Lagrangians for these masses.
    Now the hard part: for the spring... [itex]L_3=V_3[/itex].
    [itex]V_3=\frac{k}{2}(\Delta x ) ^2[/itex] where delta x denotes the elongation/compression of the spring with respect to [itex]l_0[/itex]. So if [itex]\vec r_1[/itex] and [itex]\vec r_2[/itex] denotes the positions of mass 1 and mass 2 respectively, I have that [itex]\Delta x = ||\vec r_1 - \vec r_2|-l _0|[/itex]. Since I will consider [itex](\Delta x ) ^2[/itex] I can safely ignore the external "modulus" ||.
    Anyway here is my work: I take my reference system origin at the position of the mass 1 when nothing moves (stable equilibrium position), I have that [itex]\vec r_1 = l\sin \theta _1 \hat i+l (1-\cos \theta _1) \hat j[/itex] and [itex]\vec r _2 =(l_0+l \sin \theta _2)\hat i + (l-l \cos \theta _2 ) \hat j[/itex].
    Thus [itex]\vec r_2 -\vec r_1 =[l _0-l (\sin \theta _2 + \sin \theta _1 )]\hat i +l (\cos \theta _1 - \cos \theta _2 )\hat j[/itex].
    Now I used some trig identities at http://www.sosmath.com/trig/Trig5/trig5/trig5.html to get that [itex]\vec r_2 -\vec r_1 =\{ l_0-l \left [ 2 \sin \left ( \frac{\theta _1+ \theta _2 }{2} \right ) \cos \left ( \frac{\theta _1- \theta _2 }{2} \right ) \right ] \} \hat i+l \left [ -2 \sin \left ( \frac{\theta _1+ \theta _2 }{2} \right ) \sin \left ( \frac{\theta _1- \theta _2 }{2} \right ) \right ] \hat j[/itex].
    I thus calculated [itex]|\vec r_2 - \vec r_1 |=\sqrt {\left [ l_0-2l \sin \left ( \frac{\theta _1 +\theta _2 }{2} \right ) \cos \left ( \frac{\theta _1 -\theta _2 }{2} \right ) \right ]^2+\left [ 2l \sin \left ( \frac{\theta _1 +\theta _2 }{2} \right ) \sin \left ( \frac{\theta _1 -\theta _2 }{2} \right ) \right ]^2}[/itex]. I simplified this to [itex]\sqrt {l_0 ^2 -4ll_0 \sin \left (\frac{\theta _1 + \theta _2 }{2} \right )\cos \left (\frac{\theta _1 - \theta _2 }{2} \right ) +4 l^2\sin ^2 \left (\frac{\theta _1 + \theta _2 }{2} \right ) }[/itex].
    Despite having checked out 3 times the algebra I cannot find any mistake, yet the result makes no sense. Indeed, for when [itex]\theta _1 = \theta _2[/itex], I should get [itex]|\vec r_2 - \vec r_1 |=l_0[/itex] but I get a condition for this to be true instead of it being true no matter what. I get the condition [itex]l_0 \sin \theta =l[/itex] for all angles theta. Since this isn't necessarily true, something is wrong... but I don't see what.
    I'd appreciate any help and would be glad to know where my mistake lies. Great thanks in advance.
     
  2. jcsd
  3. Feb 20, 2012 #2

    fluidistic

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    I just spotted an algebra mistake for [itex]\vec r_2 -\vec r_1[/itex]. Working on that!
     
  4. Feb 27, 2012 #3

    fluidistic

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    I get [itex]L=\frac{m}{2}l^2(\dot \theta _1 ^2+\dot \theta _2^2)+mgl(2-\cos \theta _1 - \cos \theta _2 )-\frac{k}{2} \{ l_0^2+2ll_0 (\sin \theta _2 - \sin \theta _1)+2l^2[1-\cos (\theta _1-\theta _2 )] \}[/itex].
    So that [itex]V(\theta _1 , \theta _2 )=mgl[\cos \theta _1 + \cos ( \theta _2 ) -2]+\frac{k}{2} \{ l_0^2+2ll_0 (\sin \theta _2 - \sin \theta _1)+2l^2[1-\cos (\theta _1-\theta _2 )] \}[/itex].
    Now I am unsure if I must use the small angles approximation already, or first approximate this potential by a function of the form [itex]\frac{k}{2}x^2[/itex] in which case I do not know how to do. Any help is appreciated.
    Edit: Just spotted another algebra mistake, the potential energy should be even more complicated...


    Edit 2:Considering from start that [itex]\sin \theta \approx \theta[/itex] and [itex]1-\cos \theta \approx \frac{\theta ^2}{2}[/itex] I reach [itex]V\approx \frac{mgl}{2} (\theta _1 ^2 + \theta _2 ^2 )+k[l_0 ^2+ll_0(\theta _2 - \theta _1 )+\frac{l^2(\theta _2-\theta _1 )^2}{2}-l_0 \sqrt {l_0 ^2+2ll_0 (\theta _2-\theta _1 )+l^2(\theta_2 - \theta _1)^2}][/itex]. Could someone confirm this?
     
    Last edited: Feb 27, 2012
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