1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Small oscillations+normal modes of a system

  1. Feb 20, 2012 #1

    fluidistic

    User Avatar
    Gold Member

    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

    User Avatar
    Gold Member

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

    fluidistic

    User Avatar
    Gold Member

    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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook