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Unformatted text preview: 6.25) Spring on a T We can express the masses position in cartesian coordinates as x = l cos  r sin y = l sin + r cos where r is the displacement of the mass along the crossbar, and is the angle the arm from the origin to the crossbar (with lenght l ) makes with the horizontal. giving us velocity v x = ( l sin  r cos )  r sin v y = ( l cos  r sin ) + r cos and kinetic energy T = 1 2 m (( l 2 + r 2 ) 2 + 2 l r + r 2 ) We can substitue in = , giving T = 1 2 m (( l 2 + r 2 ) 2 + 2 l r + r 2 ) The only term in the potential energy is from the compression of the spring U = 1 2 kr 2 meaning the Lagrangian is L = 1 2 m (( l 2 + r 2 ) 2 + 2 l r + r 2 ) 1 2 kr 2 Find the equation of motion from L r = mr 2 kr and d dt L r = d dt m ( r + l ) = m r Define the natural frequency of the spring 2 k/m , and we see ( 2 2 ) r = r Evidently, for < , the mass undergoes simple harmonic motion as the arm spins around, with frequency prime = radicalBig 2 2 with r ( t ) = A cos( prime t ) + B sin( prime t ) 1 When > , we instead have solutions of the form r ( t ) = Ae t + Be t where = radicalBig 2 2 At = , the acceleration of the mass along the crossbar is zero it will continue along the crossbar at whatever speed it starts with initially.continue along the crossbar at whatever speed it starts with initially....
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This note was uploaded on 01/22/2012 for the course PHYS 3201 taught by Professor Deheer during the Spring '10 term at Georgia Institute of Technology.
 Spring '10
 deHeer
 mechanics, Mass

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