Spring mass spring mass spring - 2 balls oscillations

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SUMMARY

The discussion focuses on deriving the equations of motion for a system of two oscillating masses connected by springs. The participants assume equal masses and spring constants, leading to the formulation of two second-order differential equations. The equations are derived using Newton's second law, resulting in expressions for the accelerations of both masses. The natural frequencies of the system are determined to be √(k/m) and √(3k/m), indicating the oscillatory behavior of the masses.

PREREQUISITES
  • Understanding of Newton's second law (F = ma)
  • Familiarity with differential equations
  • Knowledge of harmonic motion and natural frequencies
  • Basic concepts of spring mechanics (Hooke's Law)
NEXT STEPS
  • Study the derivation of second-order differential equations in mechanical systems
  • Explore the concept of coupled oscillators and their solutions
  • Learn about the implications of natural frequencies in oscillatory systems
  • Investigate the effects of varying mass and spring constants on system behavior
USEFUL FOR

Physics students, mechanical engineers, and anyone interested in the dynamics of oscillating systems will benefit from this discussion.

mmatras
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http://fatcat.ftj.agh.edu.pl/~i7matras/hej.jpg

Both masses are in point.

I need to count displacement x(t) but I don't know how to write derivative equations? Could someone help? Or at least give me a tip?
 
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Are we to include gravity or can we think of this as horizontal? Can we assume that the two masses are equal? Can we assume that the two springs have the same spring constant and natural length?

Assuming all of that, let [itex]x_1[/itex] be the distance from the lower edge to the center of mass of the first ball, [itex]x_2[/itex] be the distance from the lower edge to the center of mass of the second ball and let m be the mass of each ball. Let L be the distance from the lower edge to the upper, let k be the spring constant of each spring and let l be their natural lengths.

"F= ma" of course, so [itex]m d^2x_1/dt^2[/itex] is equal to the total force on the first mass. There is a force from the spring below it equal to the spring contant times its extension: [itex]-k(x_1- l)[/itex]. There is a force from the spring above it, equal to the spring constant times its extension: [itex]k(x_2- x_1- l)[/itex] (positive because if the spring is stretched, it will pull the mass upward).
[tex]m\frac{d^2x_1}{dt^2}= -k(x_1-l)+ k(x_2- x_1- l)= -2kx_1+ kx_2[/tex]

[itex]md^2x_2/dt^2[/itex] is equal to the total force on the second mass. There is a force from the spring below it of [itex]-k(x_2-x_1- l)[/itex], equal and opposite to the force of that spring on the lower mass. There is a force from the spring above it: [itex]k(L- x_2- l[/itex].
[tex]m\frac{d^2 x_2}{dt^2}= -k(x_2-x_1-l)+ k(L- x_2- l)= -2kx_2+ kx_1+ kL[/itex].<br /> <br /> Those are your two equations. When you solve them, you should find that you have <b>two</b> "natural frequencies", [itex]\sqrt{k/m}[/itex] and [itex]\sqrt{3k/m}[/itex].-[/tex]
 

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