Old Homework 35 - portillo (gdp347) oldhomework 35 Turner...

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portillo (gdp347) – oldhomework 35 – Turner – (58185) 1 This print-out should have 11 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 (part 1 oF 2) 10.0 points The motion oF a piston in an auto engine is simple harmonic. The piston travels back and Forth over a distance oF 26 cm, and the piston has a mass oF 2 . 1 kg. 7817 rpm 26 cm What is the maximum speed oF the piston when the engine is running at 7817 rpm? Correct answer: 106 . 417 m / s. Explanation: Let : A = d 2 = 26 cm 2 = 0 . 13 m , and f = 7817 rpm ω = 2 π f = 2 π (7817 rpm) (60 s / min) = 818 . 594 rad / s . ±rom 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. ±rom 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 = (818 . 594 rad / s) (0 . 13 m) = 106 . 417 m / s , where ω = 2 π f and A = d 2 . Remember to convert the Frequency 7817 rpm to Hz by converting minutes to seconds by dividing by 60 s. 002 (part 2 oF 2) 10.0 points What is the maximum Force acting on the piston when the engine is running at the same rate? Correct answer: 1 . 82936 × 10 5 N. Explanation: Using Eq. 1, we have k = m v 2 A 2 = m ( ω A ) 2 A 2 = 2 = (2 . 1 kg) (818 . 594 rad / s) 2 = 1 . 4072 × 10 6 m / s 2 . b V F b = k A = 2 A = (2 . 1 kg) (818 . 594 rad / s) 2 (0 . 13 m) = 1 . 82936 × 10 5 N . 003 10.0 points The mass oF the deuterium molecule D 2 is twice that oF the hydrogen molecule H 2 . IF the vibrational Frequency oF H 2 is 1 . 36 × 10 14 Hz, what is the vibrational Frequency oF D 2 , assuming that the “spring constant”
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portillo (gdp347) – oldhomework 35 – Turner – (58185) 2 of attracting forces is the same for the two species? Correct answer: 9 . 61665 × 10 13 Hz. Explanation: Let : M D = 2 M H . The angular frequencies depend only on spring constant and mass: ω = r k M r 1 M ω D = R k M D ω H = R k M H The spring constants k are the same, so ω D ω H = R M H M D = R M H 2 M H = r 1 2 = 1 2 .
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This note was uploaded on 01/25/2012 for the course CHEM 302 taught by Professor Mccord during the Spring '10 term at University of Texas at Austin.

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Old Homework 35 - portillo (gdp347) oldhomework 35 Turner...

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