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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 back-and-forth 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 counter-clockwise uniform circular motion. Consider the oscillation of a mass-spring 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 x-projection 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