midterm04 - Version 097/ABCAB – midterm 04 – Turner –...

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Unformatted text preview: Version 097/ABCAB – midterm 04 – Turner – (58120) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points An experiment requires sound intensity 1 . 44 W / m 2 at the distance of 6 . 15 m in any direction from a speaker. What total sound power is required? 1. 78.1094 2. 321.952 3. 549.375 4. 26.9233 5. 87.6437 6. 684.42 7. 443.126 8. 510.938 9. 272.446 10. 101.263 Correct answer: 684 . 42 W. Explanation: Let : I = 1 . 44 W / m 2 and r = 6 . 15 m . I = P 4 π r 2 P = 4 π r 2 I = 4 π (6 . 15 m) 2 (1 . 44 W / m 2 ) = 684 . 42 W . 002 10.0 points A car with bad shock absorbers bounces up and down with a period of 2 . 21 s after hitting a bump. The car has a mass of 1630 kg and is supported by four springs of equal force constant k . Determine a value for k . 1. 4368.22 2. 3694.7 3. 23440.6 4. 3293.84 5. 2476.5 6. 4309.79 7. 9344.0 8. 37182.0 9. 1323.75 10. 1718.0 Correct answer: 3293 . 84 N / m. Explanation: Let : T = 2 . 21 s and m = 1630 kg . The spring constant is k = F x . Assuming that each spring supports an equal portion of the car’s mass; i.e. , m 4 , the angular frequency of each spring is ω = radicalbigg k M = radicalbigg 4 k m and the period is T = 2 π ω T = 2 π radicalbigg m 4 k T 2 = 4 π 2 m 4 k k = π 2 m T 2 = π 2 (1630 kg) (2 . 21 s) 2 = 3293 . 84 N / m . 003 10.0 points A large block with mass 13 kg executes horizontal simple harmonic motion as it slides across a frictionless surface with a frequency 2 . 31 Hz . A smaller block with mass 6 kg rests on it, as shown in the figure, and the coefficient of static friction between the two is μ s = 0 . 569 . k 13 kg μ s = 0 . 569 6 kg Version 097/ABCAB – midterm 04 – Turner – (58120) 2 What maximum amplitude of oscillation can the system have if the top block is not to slip? The acceleration of gravity is 9 . 8 m / s 2 . 1. 1.68851 2. 10.6265 3. 11.2912 4. 2.87605 5. 14.3968 6. 8.64958 7. 12.442 8. 2.45612 9. 2.647 10. 15.691 Correct answer: 2 . 647 cm. Explanation: Let : m b = 13 kg , m t = 6 kg , f = 2 . 31 Hz , μ s = 0 . 569 , and g = 9 . 8 m / s 2 . f s mg N For simple harmonic motion, we have x = A cos( ω t + φ ) and a = − Aω 2 cos( ω t + φ ) , where ω = 2 π f = 2 π (2 . 31 Hz) = 14 . 5142 rad / s and since the extreme values of the cosine are ± 1, a max = Aω 2 . The force of friction is f s ≤ μN . Horizontally, the large block only feels the force of friction, so f s = ma and the maximum frictional force is f s = μ s N = μ s mg since the setup is horizontal. (If it were not, the normal force would also depend on the angle of inclination.) From Newton’s 2 nd law summationdisplay vector F = mvectora μ s mg = mA max ω 2 , so A max = μ s g ω 2 = (0 . 569)(9 . 8 m / s 2 ) (14 . 5142 rad / s) 2 · 100 cm 1 m = 2 . 647 cm ....
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This note was uploaded on 04/07/2011 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.

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midterm04 - Version 097/ABCAB – midterm 04 – Turner –...

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