HW30 - homework 30 PAPAGEORGE, MATT Due: Apr 12 2008, 4:00...

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1 Question 1, chap 15, sect 1. part 1 of 3 10 points A weight suspended from a spring is seen to bob up and down over a distance of 15 cm twice each second. What is its frequency? Correct answer: 2 Hz (tolerance ± 1 %). Explanation: freq = 2 bobs 1 second = 2 Hz . Question 2, chap 15, sect 1. part 2 of 3 10 points What is its period? Correct answer: 0 . 5 s (tolerance ± 1 %). Explanation: period = 1 freq = 1 2 Hz = 0 . 5 s . Question 3, chap 15, sect 1. part 3 of 3 10 points What is its amplitude? Correct answer: 7 . 5 cm (tolerance ± 1 %). Explanation: The amplitude is the distance from the equilibrium position to maximum displace- ment: 1 2 (15 cm) = 7 . 5 cm . Question 4, chap 15, sect 1. part 1 of 1 10 points A particle hangs from a spring and os- cillates with a period of 0 . 37 s. How much will the spring, at rest, shorten from the equilibrium position with the par- ticle hung, if the particle is removed? With Particle Without Particle 1. 0 . 271869 m 2. 0 . 0679673 m 3. 0 . 543738 m 4. 0 . 106763 m 5. 0 . 135935 m 6. 0 . 0339836 m correct Explanation: When the particle hangs and oscillates on the spring, we have T = 2 π r m k , where T is the oscillation period of the spring with particle hung, m is the mass of the particle, and k is the spring constant. So, k = 4 π 2 m T 2 . At equilibrium with the particle hung, the spring is stretched for x = mg k . Put the expression of k into this equation and get x = gT 2 4 π 2 = 0 . 0339836 m Question 5, chap 15, sect 1. part 1 of 1 10 points Given: The oscillation of a mass-spring system x = x m cos( ω t + φ ) , where x m is a positive number. At time t = 0, the mass is at the equilibrium point x = 0 and it is moving with positive velocity v 0 . What is the phase angle
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HW30 - homework 30 PAPAGEORGE, MATT Due: Apr 12 2008, 4:00...

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