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Unformatted text preview: Nguyen, Don – Homework 11 – Due: Nov 9 2004, 4:00 am – Inst: Charles Chiu 1 This printout should have 21 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 2) 10 points The motion of a piston in an auto engine is simple harmonic. The piston travels back and forth over a distance of 15 cm, and the piston has a mass of 1 . 5 kg. 7747 rpm 15 cm What is the maximum speed of the piston when the engine is running at 7747 rpm? Correct answer: 60 . 8448 m / s. Explanation: Let : A = d 2 = 15 cm 2 = 0 . 075 m , and f = 7747 rpm ω = 2 π f = 2 π (7747 rpm) (60 s / min) = 811 . 264 rad / s . From conservation of energy, K max = U max , so 1 2 mv 2 = 1 2 k A 2 . This yields v A = r k m , (1) where A is the maximum displacement. In this case, the displacement is half of the dis tance that the piston travels. From the reference circle, the frequency of SHM equals f = 1 T = 2 π ω = 1 2 π r k m = 1 2 π v A , so v = 2 π f d 2 = ω A = (811 . 264 rad / s)(0 . 075 m) = 60 . 8448 m / s , where ω = 2 π f and A = d 2 . Remember to convert the frequency 7747 rpm to Hz by converting minutes to seconds by dividing by 60. 002 (part 2 of 2) 10 points What is the maximum force acting on the piston when the engine is running at the same rate? Correct answer: 74041 . 8 N. Explanation: Using Eq. 1, we have k = m v 2 A 2 = m ( ω A ) 2 A 2 = mω 2 = (1 . 5 kg)(811 . 264 rad / s) 2 = 987224 m / s 2 . k ~ F k = k A = mω 2 A = (1 . 5 kg)(811 . 264 rad / s) 2 (0 . 075 m) = 74041 . 8 N . 003 (part 1 of 1) 10 points A body oscillates with simple harmonic mo tion along the xaxis. Its displacement varies with time according to the equation, x ( t ) = A sin( ω t + φ ) . Nguyen, Don – Homework 11 – Due: Nov 9 2004, 4:00 am – Inst: Charles Chiu 2 If A = 4 . 7 m, ω = 2 . 7 rad / s, and φ = 1 . 0472 rad, what is the acceleration of the body at t = 3 s? Note: The argument of the sine function is given here in radians rather than degrees. Correct answer: 9 . 38907 m / s 2 . Explanation: x = A sin( ω t + φ ) v = dx dt = ω A cos( ω t + φ ) a = dv dt = ω 2 A sin( ω t + φ ) The basic concepts above are enough to solve the problem. Just use the formula for a ob tained by differentiating x twice: a = ω 2 A sin( ω t + φ ) = 9 . 38907 m / s 2 The phase φ (given in radians) incorporates the initial condition where the body started ( t = 0), meaning it started at x = A sin φ = 4 . 07032 m and it is now at x = A sin( ω t + φ ) = 1 . 28794 m (These two last facts are not needed to solve the problem but clarify the physical picture.) 004 (part 1 of 3) 10 points A block of mass 0 . 19 kg is attached to a spring of spring constant 10 N / m on a fric tionless track. The block moves in simple har monic motion with amplitude 0 . 26 m. While passing through the equilibrium point from left to right, the block is struck by a bullet, which stops inside the block.which stops inside the block....
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 Fall '08
 Turner
 Force, Mass, Work, Pressure measurement

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