Harmonic oscillation, spring not attached to center of mass

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SUMMARY

The discussion focuses on the dynamics of a spring attached to a bar that is constrained to move in two dimensions, with the center of mass (COM) located at the center of the rod. The equation of motion for the system is initially given as x = A sin(√(k/m)t + φ₀), where k is the spring constant and m is the mass. However, the position of the COM affects the motion, necessitating a reevaluation of the equation. The use of Lagrange's equations is recommended for a comprehensive analysis of the system's behavior.

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  • Understanding of harmonic oscillation principles
  • Familiarity with Lagrange's equations
  • Knowledge of spring dynamics and coefficients
  • Ability to construct free body diagrams
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This discussion is beneficial for physics students, mechanical engineers, and anyone interested in the dynamics of oscillatory systems and the application of Lagrangian mechanics.

Antti
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A bar is guided so that one end moves on a vertical line and the other on a horizontal line. A spring is attached to the upper end according to the figure. Any friction is neglected.

http://web.comhem.se/~u48800174/springbar.jpg

I want to find out how x varies in time. If the center of mass would have been at x and k is the spring coefficient and m the mass, then

[tex]x = A sin (\sqrt{k/m}t + \phi_{0})[/tex]

But now the center of mass is at the center of the rod (with length d) instead. More specifically:

[tex]\overline{CM} = (x/2, -\sqrt{d^2 - x^2}/2[/tex]

How does this affect the equation of motion? Is it perhaps still the same? My intuition tells me something would change when the spring isn't attached at the COM but i don't know what and why. Thankful for help!
 
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This problem is easiest with Lagrange's equations.
If you don't know them, write a good free body diagram.
 

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