newmember Posted April 16, 2006 Share Posted April 16, 2006 Hi see the attached picture... 2 coupled masses, each suspended from spring in gravitational field... also entire construction can vibrate only vertically... I need to write lagrangian for this system in the following form: [math] L = \frac{1}{2}\sum_{i=1,j=1}^{2}m_{ij}\dot{x}_i\dot{x}_j-\frac{1}{2}\sum_{i=1,j=1}^{2}k_{ij}x_ix_j [/math] [math]x_i[/math] and [math]x_j[/math] are displacements of masses from equalibrium positions and springs are identical the kinetic part of L is easy one but i stack with potential part... i get to this expression(i always confused about right signs): [math]m_1g(l+x_1)+m_2g(l+x_1+l+x_2) + \frac{1}{2}m_1{x_1}^2+\frac{1}{2}m_2{x_2}^2[/math] - positive direction taken along g direction - l is length of unstreched spring every potential term in L must in [math]k_{ij}x_ix_j[/math] form but i don't see a way how do i get it from my expression for potential energy because there are g force terms containing only [math]x_i[/math] or [math]x_j[/math] thanks Link to comment Share on other sites More sharing options...

nicobudini Posted April 23, 2006 Share Posted April 23, 2006 Ok, I tried to calculate de Lagrangian and I arrived to this expression: L = T + m1·g·x1 + m2·g·x2 +1/2·k·[2·x1^2 + x2^2 - 2·x1·x2] where T is the same kinetic term you wrote, and the other terms are minus the potential energy (-U). I calculate this supposing that the origin of coordinates is located at the roof and is positive downwards. I called x1, and x2 the positions of the masses from the origin. For the first mass you have: U1 = - m1·g·x1 - 1/2·k(x1 - l)^2. And for the second mass: U2 = - m2·g·x2 - 1/2·k·(x2 - x1 - l)^2. Expanding these terms and just keeping terms of the form xi·xj you arrive to the expression I wrote above. Observe that there is a term that couples both masses. I am sorry, i don't know how to insert a formula as an image. Hope this helps. Greetings. Nicolas Link to comment Share on other sites More sharing options...

nicobudini Posted April 25, 2006 Share Posted April 25, 2006 Corrected version (using LaTeX) of my previous post . [math] L=T+m_1gx_1+m_2gx_2+\frac{k}{2}(2x_1^2+x_2^2-2x_1x_2) [/math] where [math]T[/math] is the same kinetic term you wrote, and the other terms are minus the potential energy [math](-U)[/math]. I calculate this supposing that the origin of coordinates is located at the roof and is positive downwards. I called [math]x_1[/math], and [math]x_2[/math] the positions of the masses from the origin. For the first mass you have: [math]U_1=-m_1gx_1-\frac{k}{2}(x_1-l)^2[/math]. And for the second mass: [math]U_2=-m_2gx_2-\frac{k}{2}(x_2-x_1-l)^2[/math]. Expanding these terms and just keeping terms of the form [math]x_ix_j[/math] you arrive to the expression I wrote above. Observe that there is a term that couples both masses. Greetings. Nicolas Link to comment Share on other sites More sharing options...

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