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test 4 - Overton Mays – Midterm 4 – Due May 4 2005...

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Unformatted text preview: Overton, Mays – Midterm 4 – Due: May 4 2005, 10:00 pm – Inst: Turner 1 This print-out should have 19 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points The motion of a piston in an auto engine is simple harmonic. The piston travels back and forth between the extreme points 13 cm apart. 2225 rpm 13 cm What is the maximum speed of the piston when the engine is running at 2225 rpm (rev- olution per minute)? Correct answer: 15 . 1451 m / s. Explanation: The maximum speed is v = ω A, where A is the oscillation amplitude and ω is the angular frequency. In this case, A is half the distance between the two extreme points. So A = 6 . 5 cm = 0 . 065 m . The angular frequency of SHM is ω = 2225 rev min µ 2 π rad 1 rev ¶ µ 1 min 60 s ¶ = 233 . 001 rad / s . Combining yields v = ω A = (233 . 001 rad / s)(0 . 065 m) = 15 . 1451 m / s . 002 (part 1 of 1) 10 points Sound Level β in Decibels are defined as β ≡ 10log I I , where I = 1 × 10- 12 W / m 2 . The decibel scale intensity for busy traffic is β 1 = 74 dB. Two people having a loud conversation have a decibel intensity of β 2 = 70 dB. What is the approximate combined sound intensity? Correct answer: 3 . 51189 × 10- 5 W / m 2 . Explanation: Inverting the definition of sound level in decibels gives I = I 10 β/ 10 . Therefore, the combined sound intensity I is I = I 1 + I 2 = I h 10 β 1 / 10 + 10 β 2 / 10 i = (1 × 10- 12 W / m 2 ) h 10 7 . 4 + 10 7 i = (2 . 51189 × 10- 5 W / m 2 ) +(1 × 10- 5 W / m 2 ) = 3 . 51189 × 10- 5 W / m 2 . 003 (part 1 of 1) 10 points Consider a light rod of negligible mass and length L pivoted on a frictionless horizontal bearing at a point O . Attached to the end of the rod is a mass M 1 . Also, a second mass M 2 of equal size is attached to the rod µ 1 4 L from the lower end ¶ , as shown in the figure below. 1 4 L L M 2 M 1 O θ Overton, Mays – Midterm 4 – Due: May 4 2005, 10:00 pm – Inst: Turner 2 The period of this pendulum in the small angle approximation is given by 1. T = 2 π s 65 77 L g 2. T = 2 π s 17 20 L g 3. T = 2 π s 53 63 L g 4. T = 2 π s 13 15 L g 5. T = 2 π s 25 28 L g correct 6. T = 2 π s 85 99 L g Explanation: Basic Concepts: The momentum of iner- tia is I ≡ M d 2 . In this case there are two masses with I M 1 = M L 2 I M 2 = M µ L- 1 4 L ¶ 2 = M µ 3 4 L ¶ 2 = 9 16 M L 2 , we get I = I M 1 + I M 2 = µ 1 + 9 16 ¶ M L 2 = 25 16 M L 2 . (1) Torque: “ τ ≡ r F sin φ ” The relationship between torque and angular acceleration is “ τ = I α 00 . We have F = ( M 1 + M 2 ) g exerted on the center of mass, and the distance between O and the center of mass is r cm ≡ r 1 M + r 2 M 2 M = LM + 3 4 LM 2 M = 7 8 L. (2) And φ =- θ . Then (for both masses), using the small angle approximation, we get τ =- 2 M g r cm sin θ ’ - 2 M g r cm θ....
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test 4 - Overton Mays – Midterm 4 – Due May 4 2005...

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