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Old Homework 35

# Old Homework 35 - portillo(gdp347 oldhomework 35...

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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 – find 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 . From conservation of energy, K max = U max , so 1 2 m v 2 = 1 2 k A 2 . This yields v A = radicalbigg 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 radicalbigg 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 = m ω 2 = (2 . 1 kg) (818 . 594 rad / s) 2 = 1 . 4072 × 10 6 m / s 2 . bardbl vector F bardbl = k A = m ω 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 = radicalbigg k M radicalbigg 1 M ω D = radicalBigg k M D ω H = radicalBigg k M H The spring constants k are the same, so ω D ω H = radicalBigg M H M D = radicalBigg M H 2 M H = radicalbigg 1 2 = 1 2 .
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