Test4 - quiz 04 VENNES, ROSS Due: Dec 5 2007, 10:00 pm 1...

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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 ping-pong 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 Ping-Pong 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....
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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.

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Test4 - quiz 04 VENNES, ROSS Due: Dec 5 2007, 10:00 pm 1...

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