Problem Set 6 - 6.25) Spring on a T We can express the...

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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 cross-bar, and is the angle the arm from the origin to the cross-bar (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 cross-bar is zero- it will continue along the cross-bar at whatever speed it starts with initially.continue along the cross-bar 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.

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Problem Set 6 - 6.25) Spring on a T We can express the...

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