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Unformatted text preview: hinojosa (jlh3938) – oldmidterm4 02 – Turner – (58185) 1 This print-out should have 27 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points The oscillation of a mass-spring system x = x m cos( ω t + φ ) , where x m is a positive number. At time t = 0, the mass is at the equilibrium point x = 0 and it is moving with positive velocity v . What is the phase angle φ ? 1. φ = 3 2 π correct 2. φ = 7 4 π 3. φ = 0 4. φ = π 5. φ = 3 4 π 6. φ = 1 2 π 7. φ = 5 4 π 8. φ = 2 π 9. φ = 1 4 π Explanation: Since the initial position is x = 0, we know that cos( φ ) = 0, so φ = 1 2 π, 3 2 π, 5 2 π,... . The velocity can be found as v = dx dt = − x m ω sin( ω t + φ ) . At t = 0 we have v = − x m ω sin( φ ) > 0, so the solution is φ = 3 2 π . 002 10.0 points A solid cylinder of mass 4 kg and radius 35 cm is yoked to a spring as shown in the figure below. To be precise, the axle of the cylinder is attached to a horizontal spring of force constant 2658 N / m. The cylinder rolls back and forth on a horizontal base with- out slipping. For simplicity, assume that the spring, the axle and the yoke which connects them have negligible masses compared to the cylinder itself. 2658 N / m 4 kg What is the angular frequency of the cylin- der rolling back and forth around the equlib- rium position? Correct answer: 21 . 0476 s − 1 . Explanation: Let : M = 4 kg , R = 35 cm , and k = 2658 N / m . Consider the forces and the torques on the rolling cylinder: F s Mg N F f where F s = − k x is the spring force and F f is the static friction force at the bottom of the cylinder. The directions of the forces correspond to x < 0; the spring is compressed and pushes to the right, so F s > 0. Since I cm = M R 2 2 , the torque equation around the cylinder’s axis gives τ = I cm α = RF f M R 2 2 α = RF f hinojosa (jlh3938) – oldmidterm4 02 – Turner – (58185) 2 Rα = 2 F f M for the solid cylinder of uniform density. As long as the cylinder rolls without slipping, a x = Rα, so that M a x = M ( Rα ) = M 2 F f M = 2 F f . (2) The horizontal force equation gives us F net x = M a x = F s − F f 2 F f = F x − F f F f = 1 3 F x and M a x = F s − F f = 2 3 F s , so a x = − 2 k 3 M x. The equation of motion is harmonic, so a x ≡ d 2 x dt 2 = − ω 2 x, which leads to simple harmonic motion with angular frequency ω = radicalbigg 2 k 3 M = radicalBigg 2 (2658 N / m) 3 (4 kg) = 21 . 0476 s − 1 . 003 10.0 points A particle oscillates harmonically x = A sin( ωt + φ ) , with amplitude 13 m, angular frequency π s − 1 , and initial phase π 3 radians. Every now and then, the particle’s kinetic energy and potential energy happen to be equal to each other ( K = U )....
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This note was uploaded on 07/07/2010 for the course PHY 302 taught by Professor Staff during the Summer '08 term at University of Texas.
- Summer '08