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oldhw35 - This print-out should have 11 questions...

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Unformatted text preview: 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 18 cm, and the piston has a mass of 2 . 3 kg. 8734 rpm 18 cm What is the maximum speed of the piston when the engine is running at 8734 rpm? Correct answer: 82 . 316 m / s. Explanation: Let : A = d 2 = 18 cm 2 = 0 . 09 m , and f = 8734 rpm ω = 2 π f = 2 π (8734 rpm) (60 s / min) = 914 . 622 rad / s . From conservation of energy, K max = U max , so 1 2 mv 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 = (914 . 622 rad / s) (0 . 09 m) = 82 . 316 m / s , where ω = 2 π f and A = d 2 . Remember to convert the frequency 8734 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 . 73163 × 10 5 N. Explanation: Using Eq. 1, we have k = m v 2 A 2 = m ( ω A ) 2 A 2 = mω 2 = (2 . 3 kg) (914 . 622 rad / s) 2 = 1 . 92403 × 10 6 m / s 2 . bardbl vector F bardbl = k A = mω 2 A = (2 . 3 kg) (914 . 622 rad / s) 2 (0 . 09 m) = 1 . 73163 × 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 . 33 × 10 14 Hz, what is the vibrational frequency of D 2 , assuming that the “spring constant” oldhomework 35 – Turner – (58185) 1 hinojosa (jlh3938) – oldhomework 35 – Turner – (58185) 2 of attracting forces is the same for the two species? Correct answer: 9 . 40452 × 10 13 Hz....
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