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Unformatted text preview: homework 30 BAUTISTA, ALDO Due: Apr 14 2006, 4:00 am 1 Version number encoded for clicker entry: V1:1, V2:4, V3:2, V4:4, V5:3. Question 1 Part 1 of 3. 10 points. Consider the oscillation of a massspring system, where x = A cos( t + ) . At the time t = 0, the mass m is at x = 0 (the equilibrium point) and it is moving with a positive velocity v . k m v x = 0 x Find the phase angle . (Hint: Consider x as the projection of a counterclockwise uniform circular motion.) 1. = 7 4 2. = 3. = 3 4 4. = 5 4 5. = 1 2 6. = 0 7. = 1 4 8. = 2 9. = 3 2 correct Explanation: Basic Concepts: x = A cos ( t + ) = r m k A B C D E F G H x The SHM can be represented by the x projection of a uniform circular motion: x = A cos ( t + ) . At t = 0 , x = 0. From inspection, it should be either C or G . At C , v < 0; while at G , v > 0. So G is the correct choice, or = 3 2 . Question 2 Part 2 of 3. 10 points. Let the mass be m = 8 . 59 kg, spring con stant k = 566 N / m and the initial velocity v = 3 . 25 m / s. Find the amplitude A . Correct answer: 0 . 400379 m (tolerance 1 %). Explanation: v = d x dt = A sin( t + ) So the velocity amplitude or the maximum speed is v max = A ; i.e. , v = A , so A = v = v r m k = (3 . 25 m / s) s (8 . 59 kg) (566 N / m) = 0 . 400379 m . Question 3 Part 3 of 3. 10 points. homework 30 BAUTISTA, ALDO Due: Apr 14 2006, 4:00 am 2 Find the total energy of oscillation at t = T 8 ; i.e. , at oneeighth of the period. (Hint: Consider what happens to the total energy during oscillatory motion.) 1. E = 5 2 m v 2 2. E = 1 4 m v 2 3. E = 3 4 m v 2 4. E = 1 2 2 m v 2 5. E = 1 2 m v 2 correct 6. E = m v 2 7. E = 3 2 m v 2 8. E = 2 m v 2 Explanation: Since the spring force is a conservative force the total energy is conserved. One may equate the energy at t = 0 where x = 0. So E = K + U = K max = 1 2 m v 2 . Question 4 Part 1 of 1. 10 points. A mass, m 1 = 12 . 1 kg, is in equilibrium while connected to a light spring of constant k = 112 N / m that is fastened to a wall (see a). A second mass, m 2 = 7 . 8 kg, is slowly pushed up against mass m 1 , compressing the spring by the amount A i = 0 . 32 m (see b). The system is then released, causing both masses to start moving to the right on the frictionless surface. When m 1 is at the equi librium point, m 2 loses contact with m 1 (see c) and moves to the right with speed v max ....
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 Spring '08
 Turner
 Mass, Work

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