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Unformatted text preview: homework 12 – GADHIA, TEJAS – Due: Apr 18 2007, 4:00 am 1 Question 1 part 1 of 3 10 points Consider the oscillation of a massspring 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 . (Hint: Consider x as the projection of a counterclockwise uniform circular motion.) 1. φ = 1 4 π 2. φ = 3 2 π correct 3. φ = 5 4 π 4. φ = π 5. φ = 0 6. φ = 1 2 π 7. φ = 3 4 π 8. φ = 2 π 9. φ = 7 4 π Explanation: Basic Concepts: x = A cos ω ( t + φ ) ω = radicalbigg 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. From 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 . Question 2 part 2 of 3 10 points Let the mass be m = 1 . 15 kg, spring con stant k = 804 N / m and the initial velocity v = 3 . 28 m / s. Find the amplitude A . Correct answer: 0 . 124049 m (tolerance ± 1 %). 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 = ωA, so A = v ω = v radicalbigg m k = (3 . 28 m / s) radicalBigg (1 . 15 kg) (804 N / m) = 0 . 124049 m . Question 3 part 3 of 3 10 points Find the total energy of oscillation at t = T 8 ; i.e. , at oneeighth of the period. homework 12 – GADHIA, TEJAS – Due: Apr 18 2007, 4:00 am 2 (Hint: Consider what happens to the total energy during oscillatory motion.) 1. E = 5 2 mv 2 2. E = 1 4 mv 2 3. E = 1 2 mv 2 correct 4. E = mv 2 5. E = 1 2 √ 2 m v 2 6. E = 3 4 mv 2 7. E = 3 2 mv 2 8. E = 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 . Question 4 part 1 of 1 10 points A simple 1.97 m long pendulum oscillates. The acceleration of gravity is 9 . 8 m / s 2 . How many complete oscilations does this pendulum make in 4.00 min? Correct answer: 85 . 1944 (tolerance ± 1 %). Explanation: Basic Concepts: oscillations = ∆ t T T = 2 π radicalBigg L g Given: L = 1 . 97 m g = 9 . 80 m / s 2 ∆ t = 4 . 00 min Solution: oscillations = ∆ t 2 π radicalBig L g = ∆ t 2 π · radicalbigg g L = (4 min) 2 π · parenleftbigg 60 s 1 min parenrightbigg · radicalbigg 9 . 8 m / s 2 1 . 97 m = 85 . 1944 The pendulum makes 85 complete oscila tions. Question 5 part 1 of 1 10 points Consider a car engine running at constant speed. That is, the crankshaft of the en gine rotates at constant angular velocity while each piston moves backandforth in its cylin der according to the rules of simple harmonic motion....
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This note was uploaded on 03/09/2011 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.
 Spring '08
 Turner
 Mass, Work

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