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**Unformatted text preview: **sanchez (ds28677) – homework 30 – Turner – (58220) 1 This print-out should have 13 questions. Multiple-choice 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 mass-spring system where x = A cos( ω t + φ ) . 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 φ . Consider x as the projection of a counterclockwise uniform circular motion. 1. φ = 2 π 2. φ = π 3. φ = 0 4. φ = 1 4 π 5. φ = 5 4 π 6. φ = 1 2 π 7. φ = 3 4 π 8. φ = 3 2 π correct 9. φ = 7 4 π Explanation: A B C D E F G H φ ϖ x The SHM can be represented by the x- projection of a uniform circular motion: x = A cos ω ( t + φ ) . 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 2 . 37 kg, spring constant 898 N / m and the initial velocity 2 . 36 m / s. Find the amplitude A . Correct answer: 0 . 121241 m. Explanation: Let : m = 2 . 37 kg , k = 898 N / m , and v = 2 . 36 m / s . v = dx dt =- ω A sin( ω t + φ ) , so the velocity amplitude or the maximum speed is v max = ωA ; i.e. , v = ωA A = v ω = v radicalbigg m k = (2 . 36 m / s) radicalBigg (2 . 37 kg) (898 N / m) = . 121241 m . 003 (part 3 of 3) 10.0 points sanchez (ds28677) – homework 30 – Turner – (58220) 2 Find the total energy of oscillation at t = T 8 ; i.e. , at one-eighth of the period. Consider what happens to the total energy during os- cillatory motion. 1. E = 5 2 mv 2 2. E = 3 2 mv 2 3. E = 3 4 mv 2 4. E = 2 m v 2 5. E = 1 2 mv 2 correct 6. E = 1 4 mv 2 7. E = mv 2 8. E = 1 2 √ 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 mv 2 . 004 (part 1 of 3) 10.0 points A block of mass 0 . 3 kg is attached to a spring of spring constant 24 N / m on a fric- tionless track. The block moves in simple har- monic motion with amplitude 0 . 11 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 49 m / s and the mass of the bullet is 1 . 97 g. 24 N / m . 3 kg 1 . 97 g 49 m / s Find the speed of the block immediately before the collision. Correct answer: 0 . 98387 m / s. Explanation: Let : M = 0 . 3 kg , A = 0 . 11 m , and k = 24 N / m ....

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