Small oscillations of rhombus figure
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Four massless rods of length L are hinged together at their ends to form a rhombus. a particle of mass M is attached at each joint. the opposite corners of the rhombus are joined by springs, each with a spring constant k. In the equilibrium (square) configuration, the springs are unstretched. The motion is confined to a plane, and the particles move only along the diagonals of the rhombus. Introduce suitable generalized coordinates and find the Lagrangian of the system.
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Solution Summary
Describes to form Lagrangian corresponding to an oscillating system. This can be extended to find Euler-Lagrangian equation of motion.
Solution Preview
Consider, Diagonals of the square/rhombus along X and Y axes respectively, and centre at the origin.
In equilibrium, two particles on X-axis, and two on y axis, each at distance of L/sqrt(2) from the origin.
In general condition, particles on X-axis are at x and -x position and y-axis particles and y and -y.
Hence,
x^2 + y^2 = L^2 -- (1) [Length of each rod]
differentiate with respect to t:
2.x.dx/dt + 2.y.dy/dt = ...
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- BEng, Allahabad University, India
- MSc , Pune University, India
- PhD (IP), Pune University, India
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