Normal modes of Mass and two Spings

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SUMMARY

The discussion focuses on determining the normal modes of a mass M constrained to a two-dimensional surface, attached to two identical springs with spring constant k. The springs are fixed at points (-L, 0) and (L, 0), and the equilibrium condition is established under the constraint ℓ < L. The equations of motion (EOM) derived are ma = -2k(L - ℓ) - 2kx, indicating the forces acting on the mass due to the springs. The user seeks confirmation on the correctness of the EOM and guidance on solving for normal modes using the assumed solution x(t) = Acos(wt).

PREREQUISITES
  • Understanding of Newton's second law (F=ma)
  • Knowledge of Hooke's law for springs (T = -kx)
  • Familiarity with normal mode analysis in mechanical systems
  • Basic proficiency in solving differential equations
NEXT STEPS
  • Study the derivation of equations of motion for coupled oscillators
  • Learn about normal mode frequencies in systems with multiple springs
  • Explore the method of solving second-order differential equations
  • Investigate the effects of damping on normal modes in mechanical systems
USEFUL FOR

Students in physics or engineering, particularly those studying dynamics and oscillatory systems, as well as educators looking to enhance their understanding of coupled oscillators and normal modes.

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Homework Statement



Consider a mass M whose motion is confined to a flat, smooth two-dimensional surface. Label the locations in this surface using the Cartesian coordinates (x, y). The mass is attached to two identical springs, each of length ℓ and spring constant k. One spring has one of its ends fixed to the point (-L, 0) and the other spring has one of its ends fixed to the point (L, 0). Find the normal modes of this system when ℓ < L.

Homework Equations



F=ma , T =-kx

The Attempt at a Solution



Well first I've my goal is to find the EOM of the system then plug in the solution to the EOM to find the normal modes. Since ℓ < L , then the springs are stretched in equilibrium and provide constant tension of T = -2k(L- ℓ). Taking into account when the mass is displaced a distance x from equilibrium also provides additional tension T = -2kx. Therefor my EOM is

ma = -2k(L- ℓ) + (-2kx)

However when I plug in a solution of the form x(t) = Acos(wt) it doesn't seem easy to solve.
Can someone please tell me if I am on the right track and if my EOM is correct?
 
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You also need to consider motion in the vertical direction.
 

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