hw32 - homework 32 Turner (58185) This print-out should...

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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 . (Hint: Consider x as the projection oF a counterclockwise uniForm circular motion.) 1. φ = 2 π 2. φ = 3 2 π correct 3. φ = 5 4 π 4. φ = π 5. φ = 7 4 π 6. φ = 1 2 π 7. φ = 1 4 π 8. φ = 0 9. φ = 3 4 π Explanation: Basic Concepts: x = A cos ω ( t + φ ) ω = r m k 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. ±rom inspection, it should be either C or G . At C , v < 0; while at G , v > 0. So G is the correct choice, or φ = 3 2 π . 002 (part 2 oF 3) 10.0 points Let the mass be m = 5 . 09 kg, spring constant k = 567 N / m and the initial velocity v 0 = 1 . 81 m / s. ±ind the amplitude A . Correct answer: 0 . 171493 m. Explanation: v = dx dt = - ω A sin( ω t + φ ) So the velocity amplitude or the maximum speed is v max = ωA ; i.e. , v 0 = ωA, so A = v 0 ω = v 0 r m k = (1 . 81 m / s) R (5 . 09 kg) (567 N / m) = 0 . 171493 m . 003 (part 3 oF 3) 10.0 points ±ind the total energy oF oscillation at t = T 8 ; i.e. , at one-eighth oF the period. homework 32 – Turner – (58185) 1
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hinojosa (jlh3938) – homework 32 – Turner – (58185) 2 (Hint: Consider what happens to the total energy during oscillatory motion.) 1. E = 2 m v 2 0 2. E = 5 2 mv 2 0 3. E = mv 2 0 4. E = 1 4 mv 2 0 5. E = 3 2 mv 2 0 6. E = 1 2 mv 2 0 correct 7. E = 1 2 2 m v 2 0 8. E = 3 4 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 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 54 m / s and the mass of the bullet is 4 . 29 g. 24 N / m 0 . 2 kg 4 . 29 g 54 m / s Find the speed of the block immediately before the collision. Correct answer: 1 . 20499 m / s. Explanation: Let : M = 0 . 2 kg , A = 0 . 11 m , and k = 24 N / m . Since the block moves in simple harmonic motion, we know that x = A sin( ω 0 t + ϕ ) , and since the only force acting on it is the spring force F s = - k x , the equation of mo- tion is F s = M a = M d 2 x dt 2 = - k x, so the angular velocity of this initial motion is ω 0 = r k M = R 24 N / m 0 . 2 kg = 10 . 9545 rad / s .
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This note was uploaded on 05/02/2009 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.

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hw32 - homework 32 Turner (58185) This print-out should...

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