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# HW11 - homework 11 GAUTHIER JOSEPH Due 1:00 am Question 1...

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homework 11 – GAUTHIER, JOSEPH – Due: Nov 16 2007, 1:00 am 1 Question 1, chap 15, sect 2. part 1 of 1 10 points The equation of motion of a simple har- monic oscillator is d 2 x dt 2 = - 9 x , where x is displacement and t is time. What is the period of oscillation? 1. T = 2 π 3 correct 2. T = 3 2 π 3. T = 6 π 4. T = 9 2 π 5. T = 2 π 9 Explanation: For a simple harmonic oscillator, the equa- tion of motion can be written as d 2 x dt 2 = - ω 2 x , where ω is the angular frequency, so the period of oscillation is T = 2 π ω = 2 π 9 = 2 π 3 . Question 2, chap 15, sect 2. part 1 of 1 10 points A body oscillates with simple harmonic mo- tion along the x -axis. Its displacement varies with time according to the equation A = A 0 sin( ω t + π/ 3) , where ω = π radians per second, t is in sec- onds, and A 0 = 1 . 1 m. What is the phase of the motion at t = 2 . 2 s? Correct answer: 7 . 9587 rad (tolerance ± 1 %). Explanation: Basic Concepts: x = A 0 sin( ω t + φ ) The phase is the angle in the argument of the sine function, and from the problem state- ment we see it is φ = π t + π 3 = bracketleftBig ( π rad / s) (2 . 2 s) + π 3 bracketrightBig = 7 . 9587 rad . Question 3, chap 15, sect 1. part 1 of 3 10 points A 424 g mass is connected to a light spring of force constant 3 N / m and it is free to os- cillate on a horizontal, frictionless track. The mass is displaced 5 cm from the equilibrium point and released form rest. 3 N / m 424 g 5 cm x = 0 x Find the period of the motion. Correct answer: 2 . 36212 s (tolerance ± 1 %). Explanation: This situation corresponds to the special case x ( t ) = A cos ωt , with A = 5 cm = 0 . 05 m . Therefore, the frequency is ω = radicalbigg k m = radicalBigg 3 N / m 0 . 424 kg = 2 . 65998 s 1 . For the period we find T = 2 π ω = 2 . 36212 s .

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homework 11 – GAUTHIER, JOSEPH – Due: Nov 16 2007, 1:00 am 2 Question 4, chap 15, sect 1. part 2 of 3 10 points What is the maximum speed of the mass? Correct answer: 0 . 132999 m / s (tolerance ± 1 %). Explanation: The velocity as a function of time is given by v ( t ) = - ω A sin( ω t ) , so the maximum speed of the mass is equal to v max = ω A = (2 . 65998 s 1 ) (0 . 05 m) = 0 . 132999 m / s . Question 5, chap 15, sect 1. part 3 of 3 10 points What is the maximum acceleration of the mass? Correct answer: 0 . 353774 m / s 2 (tolerance ± 1 %). Explanation: The acceleration as a function of time is given by a ( t ) = - ω 2 A cos( ω t ) , so the maximum acceleration of the mass is equal to a max = ω 2 A = (2 . 65998 s 1 ) 2 (0 . 05 m) = 0 . 353774 m / s 2 . Question 6, chap 15, sect 1. part 1 of 1 10 points A large block with mass 25 kg executes horizontal simple harmonic motion as it slides across a frictionless surface with a frequency 1 . 51 Hz . Block smaller block with mass 7 kg rests on it, as shown in the figure, and the coefficient of static friction between the two is μ s = 0 . 646 . The acceleration of gravity is 9 . 8 m / s 2 . k 25 kg μ s = 0 . 646 7 kg What maximum amplitude of oscillation can the system have if the block is not to slip?
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