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
Unformatted text preview: quiz 04 VENNES, ROSS Due: Dec 5 2007, 10:00 pm 1 Question 1, chap 15, sect 2. part 1 of 1 10 points A simple harmonic oscillator has amplitude . 47 m and period 3 . 2 sec. What is the maxi mum acceleration? 1. . 0458984 m / s 2 2. 1 . 812 m / s 2 correct 3. . 288388 m / s 2 4. . 146875 m / s 2 5. . 905999 m / s 2 6. 5 . 79839 m / s 2 Explanation: For a simple harmonic oscillator, the dis placement is x = A cos( 2 T t + ) So the acceleration is a = d 2 x dt 2 = A ( 2 T ) 2 cos( 2 T t + ) The maximum acceleration is 4 2 A T 2 . Question 2, chap 15, sect 1. part 1 of 1 10 points The body of a 1257 kg car is supported on a frame by four springs. The spring con stant of a single spring is 1 . 54 10 4 N / m. Four people riding in the car have a combined mass of 245 kg. When driven over a pothole in the road, the frame vibrates and for the first few seconds the vibration approximates simple harmonic motion. What is the period of vibration of the car? 1. . 743235 s 2. . 766981 s 3. . 790867 s 4. . 815739 s 5. . 841249 s 6. . 867465 s 7. . 895052 s 8. . 922892 s 9. . 951473 s 10. . 981125 s correct Explanation: Basic Concept: T = 2 radicalbigg m k Given: m c = 1257 kg k = 1 . 54 10 4 N / m m p = 245 kg Solution: m = m p + m c 4 = 245 kg + 1257 kg 4 = 375 . 5 kg T = 2 radicalBigg 375 . 5 kg 15400 N / m = 0 . 981125 s Question 3, chap 18, sect 5. part 1 of 1 10 points A small plastic ball (similar to a pingpong ball) with mass 5 g is attached by a thread to the bottom of a beaker. When the beaker is filled with water so that the ball is totally sub merged, the tension in the thread is 0 . 033 N . The density of water is 1000 kg / m 3 and the acceleration of gravity is 9 . 81 m / s 2 . 5 g quiz 04 VENNES, ROSS Due: Dec 5 2007, 10:00 pm 2 Determine the diameter of the ball. 1. 1 . 94527 cm 2. 2 . 01006 cm 3. 2 . 07235 cm 4. 2 . 14397 cm 5. 2 . 21109 cm 6. 2 . 28685 cm 7. 2 . 35789 cm 8. 2 . 43588 cm 9. 2 . 51847 cm correct 10. 2 . 60556 cm Explanation: V sphere = 4 3 r 3 = 4 3 parenleftbigg d 2 parenrightbigg 3 = 1 6 d 3 . Let : w = 1000 kg / m 3 , g = 9 . 81 m / s 2 , T = 0 . 033 N , and m = 5 g = 0 . 005 kg . Consider the forces acting on the PingPong ball B mg T Applying Archimedes principle, we have B = w f = m f g = w V ball g = 1 6 w g d 3 . Applying summationdisplay F y = 0 to the ball, B mg T = 0 1 6 w g d 3 mg T = 0 . Then we have d = 3 radicalBigg 6 ( T + mg ) w g = 3 radicalBigg 6 [0 . 033 N + (0 . 005 kg) (9 . 81 m / s 2 )] (1000 kg / m 3 ) (9 . 81 m / s 2 ) 100 cm m = 2 . 51847 cm . Question 4, chap 16, sect 2. part 1 of 1 10 points Consider the sinusoidal wave pictured in the figure....
View
Full
Document
This note was uploaded on 04/18/2008 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Turner
 Acceleration

Click to edit the document details