Homework 12 Solutions - homework 12 GADHIA, TEJAS Due: Apr...

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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 mass-spring system, where x = A cos( t + ) . 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 . (Hint: Consider x as the projection of a counterclockwise uniform circular motion.) 1. = 1 4 2. = 3 2 correct 3. = 5 4 4. = 5. = 0 6. = 1 2 7. = 3 4 8. = 2 9. = 7 4 Explanation: Basic Concepts: x = A cos ( t + ) = radicalbigg m k 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 ( t + ) . 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 = 3 2 . 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 = A sin( t + ) So the velocity amplitude or the maximum speed is v max = A ; i.e. , v = A, so A = v = 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 one-eighth 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 radicalBigg L g Given: L = 1 . 97 m g = 9 . 80 m / s 2 t = 4 . 00 min Solution: oscillations = t 2 radicalBig L g = t 2 radicalbigg g L = (4 min) 2 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 back-and-forth in its cylin- der according to the rules of simple harmonic motion....
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Homework 12 Solutions - homework 12 GADHIA, TEJAS Due: Apr...

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