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Unformatted text preview: hernandez (ejh742) homework 30 Turner (58120) 1 This printout should have 13 questions. Multiplechoice questions may continue on the next column or page find all choices before answering. 001 (part 1 of 3) 10.0 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 . Consider x as the projection of a counterclockwise uniform circular motion. 1. = 5 4 2. = 0 3. = 2 4. = 1 2 5. = 7 4 6. = 1 4 7. = 8. = 3 2 correct 9. = 3 4 Explanation: 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. 002 (part 2 of 3) 10.0 points Let the mass be 7 . 58 kg, spring constant 938 N / m and the initial velocity 3 . 81 m / s. Find the amplitude A . Correct answer: 0 . 342498 m. Explanation: Let : m = 7 . 58 kg , k = 938 N / m , and v = 3 . 81 m / s . v = dx dt = A sin( t + ) , so the velocity amplitude or the maximum speed is v max = A ; i.e. , v = A A = v = v radicalbigg m k = (3 . 81 m / s) radicalBigg (7 . 58 kg) (938 N / m) = . 342498 m . 003 (part 3 of 3) 10.0 points hernandez (ejh742) homework 30 Turner (58120) 2 Find the total energy of oscillation at t = T 8 ; i.e. , at oneeighth of the period. Consider what happens to the total energy during os cillatory motion. 1. E = 3 4 mv 2 2. E = 1 2 mv 2 correct 3. E = 1 2 2 m v 2 4. E = mv 2 5. E = 2 m v 2 6. E = 1 4 mv 2 7. E = 5 2 mv 2 8. E = 3 2 mv 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 mv 2 . 004 (part 1 of 3) 10.0 points A block of mass 0 . 2 kg is attached to a spring of spring constant 26 N / m on a fric tionless track. The block moves in simple har monic motion with amplitude 0 . 15 m. While passing through the equilibrium point from left to right, the block is struck by a bullet, which stops inside the block. The velocity of the bullet immediately be fore it strikes the block is 52 m / s and the mass of the bullet is 4 . 27 g. 26 N / m . 2 kg 4 . 27 g 52 m / s Find the speed of the block immediately before the collision. Correct answer: 1 . 71026 m / s. Explanation: Let : M = 0 . 2 kg , A = 0 . 15 m , and k = 26 N / m ....
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This note was uploaded on 06/14/2011 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Turner
 Physics, Mass, Work

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