# old31 - hernandez(ejh742 oldhomework 31 Turner(58120 This...

This preview shows pages 1–3. Sign up to view the full content.

hernandez (ejh742) – oldhomework 31 – Turner – (58120) 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 9 . 1 cm, and the piston has a mass of 2 . 4 kg. 4991 rpm 9 . 1 cm What is the maximum speed of the piston when the engine is running at 4991 rpm? Correct answer: 23 . 7809 m / s. Explanation: Let : A = d 2 = 9 . 1 cm 2 = 0 . 0455 m , and f = 4991 rpm ω = 2 π f = 2 π (4991 rpm) (60 s / min) = 522 . 656 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 = (522 . 656 rad / s) (0 . 0455 m) = 23 . 7809 m / s , where ω = 2 π f and A = d 2 . Remember to convert the frequency 4991 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: 29830 . 1 N. Explanation: Using Eq. 1, we have k = m v 2 A 2 = m ( ω A ) 2 A 2 = m ω 2 = (2 . 4 kg) (522 . 656 rad / s) 2 = 6 . 55607 × 10 5 m / s 2 . bardbl vector F bardbl = k A = m ω 2 A = (2 . 4 kg) (522 . 656 rad / s) 2 (0 . 0455 m) = 29830 . 1 N . 003 10.0 points The mass of the deuterium molecule D 2 is twice that of the hydrogen molecule H 2 .

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
hernandez (ejh742) – oldhomework 31 – Turner – (58120) 2 If the vibrational frequency of H 2 is 1 . 21 × 10 14 Hz, what is the vibrational frequency of D 2 , assuming that the “spring constant” of attracting forces is the same for the two species? Correct answer: 8 . 55599 × 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 .
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern