hw11 solutions

# hw11 solutions - homework 11 – VENNES ROSS – Due 1:00...

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Unformatted text preview: homework 11 – VENNES, ROSS – 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 9 2. T = 3 2 π 3. T = 9 2 π 4. T = 2 π 3 correct 5. T = 6 π 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 sin( ω t + π/ 3) , where ω = π radians per second, t is in sec- onds, and A = 1 . 4 m. What is the phase of the motion at t = 6 s? Correct answer: 19 . 8968 rad (tolerance ± 1 %). Explanation: Basic Concepts: x = A 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) (6 s) + π 3 bracketrightBig = 19 . 8968 rad . Question 3, chap 15, sect 1. part 1 of 3 10 points A 954 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 6 cm from the equilibrium point and released form rest. 3 N / m 954 g 6 cm x = 0 x Find the period of the motion. Correct answer: 3 . 54318 s (tolerance ± 1 %). Explanation: This situation corresponds to the special case x ( t ) = A cos ωt, with A = 6 cm = 0 . 06 m . Therefore, the frequency is ω = radicalbigg k m = radicalBigg 3 N / m . 954 kg = 1 . 77332 s − 1 . For the period we find T = 2 π ω = 3 . 54318 s . Question 4, chap 15, sect 1. part 2 of 3 10 points homework 11 – VENNES, ROSS – Due: Nov 16 2007, 1:00 am 2 What is the maximum speed of the mass? Correct answer: 0 . 106399 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 = (1 . 77332 s − 1 ) (0 . 06 m) = 0 . 106399 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 . 188679 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 = (1 . 77332 s − 1 ) 2 (0 . 06 m) = 0 . 188679 m / s 2 . Question 6, chap 15, sect 1. part 1 of 1 10 points A large block with mass 20 kg executes horizontal simple harmonic motion as it slides across a frictionless surface with a frequency 1 . 18 Hz . Block smaller block with mass 8 kg rests on it, as shown in the figure, and the coefficient of static friction between the two is μ s = 0 . 533 ....
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## This homework help was uploaded on 04/18/2008 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.

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hw11 solutions - homework 11 – VENNES ROSS – Due 1:00...

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