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Unformatted text preview: Morby, Grant – Homework 30 – Due: Apr 19 2006, noon – Inst: Drummond 1 This printout should have 9 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points The equation of motion of a simple harmonic oscillator is d 2 x dt 2 = 9 x, where x is displacement and t is time. What is the period of oscillation? 1. T = 2 π 3 correct 2. T = 2 π 9 3. T = 6 π 4. T = 9 2 π 5. T = 3 2 π Explanation: For a simple harmonic oscillator, the equa tion of motion can be written as d 2 x dt 2 = ω 2 x , where ω is the angular frequency, so the period of oscillation is T = 2 π ω = 2 π √ 9 = 2 π 3 . 002 (part 1 of 1) 10 points An oscillator is described by 1 2 3 4 5 6 7 8 9 10 11 12 1 2 x ( t ) t ( sec ) What is the angular frequency ω (in rad/sec ) ? 1. 3.1 2. 2.0 3. 1.6 correct 4. 4.0 Explanation: T = 4 sec ω = 2 π T = 2 π 4 = π 2 = 1 . 6 003 (part 1 of 1) 10 points Given: The oscillation of a massspring sys tem x = x m cos( ω t + φ ) , where x m is a positive number. Attime t = 0, themassisattheequilibrium point x = 0 and it is moving with positive velocity v . What is the phase angle φ ? 1. φ = 3 2 π correct 2. φ = 3 4 π 3. φ = 7 4 π 4. φ = 1 4 π 5. φ = 5 4 π 6. φ = 1 2 π 7. φ = π 8. φ = 2 π 9. φ = 0 Explanation: Since the initial position is x = 0, we know that cos( φ ) = 0, so φ = 1 2 π, 3 2 π, 5 2 π,... . The velocity can be found as v = dx dt = x m ω sin( ω t + φ ) . Morby, Grant – Homework 30 – Due: Apr 19 2006, noon – Inst: Drummond 2 At t = 0 we have v = x m ω sin( φ ) > 0, so the solution is φ = 3 2 π .π ....
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This note was uploaded on 03/02/2009 for the course PHY 58235 taught by Professor Kleinman during the Spring '09 term at University of Texas at Austin.
 Spring '09
 KLEINMAN
 Physics, Work

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