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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( t + ) , 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 ? 1. = 3 2 correct 2. = 7 4 3. = 0 4. = 5. = 3 4 6. = 1 2 7. = 5 4 8. = 2 9. = 1 4 Explanation: Since the initial position is x = 0, we know that cos( ) = 0, so = 1 2 , 3 2 , 5 2 ,... . The velocity can be found as v = dx dt = x m sin( t + ) . At t = 0 we have v = x m sin( ) > 0, so the solution is = 3 2 . 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 cylinders axis gives = I cm = RF f M R 2 2 = RF f hinojosa (jlh3938) oldmidterm4 02 Turner (58185) 2 R = 2 F f M for the solid cylinder of uniform density. As long as the cylinder rolls without slipping, a x = R, so that M a x = M ( R ) = 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 = 2 x, which leads to simple harmonic motion with angular frequency = 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( t + ) , with amplitude 13 m, angular frequency s 1 , and initial phase 3 radians. Every now and then, the particles kinetic energy and potential energy happen to be equal to each other ( K = U )....
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- Summer '08