homework 30-solutions (1)

# homework 30-solutions (1) - klare(alk736 homework 30...

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klare (alk736) – homework 30 – Turner – (58220) 1 This print-out should have 13 questions. Multiple-choice questions may continue on the next column or page – fnd 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 0 . k m v 0 x =0 x ±ind the phase angle φ . Consider x as the projection oF a counterclockwise uniForm circular motion. 1. φ = 3 2 π correct 2. φ =0 3. φ = π 4. φ = 1 2 π 5. φ = 5 4 π 6. φ = 3 4 π 7. φ =2 π 8. φ = 1 4 π 9. φ = 7 4 π 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 ω ( t + φ ) . At t = 0, x = 0. ±rom 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 6 . 45 kg, spring constant 697 N / m and the initial velocity 2 . 31 m / s. ±ind the amplitude A . Correct answer: 0 . 222216 m. Explanation: Let : m =6 . 45 kg , k = 697 N / m , and v 0 =2 . 31 m / s . v = dx dt = - ω A sin( ω t + φ ) , so the velocity amplitude or the maximum speed is v max = ω A ; i.e. , v 0 = ω A A = v 0 ω = v 0 ± m k = (2 . 31 m / s) ² (6 . 45 kg) (697 N / m) = 0 . 222216 m . 003 (part 3 oF 3) 10.0 points

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klare (alk736) – 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 = 1 2 2 mv 2 0 2. E = mv 2 0 3. E =2 mv 2 0 4. E = 3 4 mv 2 0 5. E = 1 2 mv 2 0 correct 6. E = 3 2 mv 2 0 7. E = 1 4 mv 2 0 8. E = 5 2 mv 2 0 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 0 . 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 . 17 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 70 m / s and the mass of the bullet is 4 . 36 g. 26 N / m 0 . 2 kg 4 . 36 g 70 m / s Find the speed of the block immediately before the collision. Correct answer: 1
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homework 30-solutions (1) - klare(alk736 homework 30...

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