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Unformatted text preview: practice 12 – ALIBHAI, ZAHID – Due: Apr 15 2007, 4:00 am 1 Question 1 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. 1740 rpm 10 . 9 cm Suppose the two extremal positions x max and x min of a piston are 10 . 9 cm from each other. When the crankshaft of the engine rotates at 1740 rpm (revolutions per minute), what is the maximal speed  v  max of the piston? Correct answer: 9 . 93057 m / s (tolerance ± 1 %). Explanation: Although a piston in a car engine is not by itself a harmonic oscillator, the crank con necting it to the rotating crankshaft imposes the simple harmonic motion on the piston x ( t ) = x middle + A × cos( ω t + φ ) (1) with the angular frequency ω equal to the angular velocity of the rotating crankshaft. For the problem at hand, ω = 1740 rpm = 182 . 212 rad / s . (2) The amplitude A of the piston’s motion is fixed by the engine’s design. To find out the amplitude for the piston at hand, consider the extreme ponts of the piston’s motion: Ac cording to Eq. (1), x max = x middle + A, x min = x middle − A, (3) hence the distance between the extremes x max − x min = 2 × A. (4) For the problem at hand, x max − x min = 10 . 9 cm , hence A = 10 . 9 cm 2 = 5 . 45 cm . (5) Now consider the piston’s velocity. Differ entiating Eq. (1) WRT time, we find v ( t ) = dx dt = − A × ω × sin( ω t + φ ) , (6) which ranges between + A × ω and − A × ω . In magnitude, the maximal speed of the piston is  v  max = A × ω (7) = (5 . 45 cm) × (182 . 212 rad / s) = 9 . 93057 m / s . Question 2 part 1 of 2 10 points Hint: Consider x as the projection of a counterclockwise uniform circular motion. Consider the oscillation of a massspring system, where x = A cos( ωt + φ ). At the time t = 0, the mass is at x = 0 and it is moving to the right with a speed v . k m v x = 0 x Find the phase angle φ . 1. φ = π 3 2. φ = 5 π 4 3. φ = 0 4. φ = 3 π 2 correct 5. φ = π 4 practice 12 – ALIBHAI, ZAHID – Due: Apr 15 2007, 4:00 am 2 6. φ = 2 π 7. φ = 3 π 4 8. φ = π 2 9. φ = 7 π 4 10. φ = π Explanation: The SHM can be represented by the xprojection of a uniform cir cular motion x = A cos( ωt + φ ) . v B φ A D C x At t = 0, x = 0, we have cos φ = 0 . From inspection on the sketch, this could be either the point B or the point D . At B , v < . But at D , v > . So the point D is the correct choice. Here φ = 3 π 2 . Question 3 part 2 of 2 10 points Denote the mass by m . Find the total energy of the oscillation at t = T 8 , where T is the period....
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This note was uploaded on 10/13/2009 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.
 Spring '08
 Turner
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