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

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6.25) Spring on a T We can express the masses position in cartesian coordinates as x = l cos θ - r sin θ y = l sin θ + r cos θ where r is the displacement of the mass along the cross-bar, and θ 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 θ - r cos θ ) ˙ θ - ˙ r sin θ v y = ( l cos θ - r sin θ ) ˙ θ + ˙ r cos θ and kinetic energy T = 1 2 m (( l 2 + r 2 ) ˙ θ 2 + 2 l ˙ r ˙ θ + ˙ r 2 ) We can substitue in ˙ θ = ω , giving T = 1 2 m (( l 2 + r 2 ) ω 2 + 2 l ˙ + ˙ 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 ) ˙ θ 2 + 2 l ˙ + ˙ r 2 ) - 1 2 kr 2 Find the equation of motion from ∂L ∂r = mrω 2 - kr and d dt ∂L ˙ r = d dt m ( ˙ r + l ˙ ω ) = m ¨ r Define the natural frequency of the spring ω 2 0 - k/m , and we see ( ω 2 - ω 2 0 ) r = ¨ r Evidently, for ω < ω 0 , the mass undergoes simple harmonic motion as the arm spins around, with frequency ω prime = radicalBig ω 2 0 - ω 2 with r ( t ) = A cos( ω prime t ) + B sin( ω prime t ) 1
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When ω > ω 0 , we instead have solutions of the form r ( t ) = Ae γt + Be - γt where γ = radicalBig ω 2 - ω 2 0 At ω = ω 0 , 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.
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