midterm04

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

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

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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 ....
View Full Document

## 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.

### Page1 / 13

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online