hw11 solutions - homework 11 VENNES, ROSS Due: Nov 16 2007,...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 9 2. T = 3 2 3. T = 9 2 4. T = 2 3 correct 5. T = 6 Explanation: For a simple harmonic oscillator, the equa- tion of motion can be written as d 2 x dt 2 =- 2 x , where is the angular frequency, so the period of oscillation is T = 2 = 2 9 = 2 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( t + / 3) , where = 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( t + ) The phase is the angle in the argument of the sine function, and from the problem state- ment we see it is = t + 3 = bracketleftBig ( rad / s) (6 s) + 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 t, with A = 6 cm = 0 . 06 m . Therefore, the frequency is = radicalbigg k m = radicalBigg 3 N / m . 954 kg = 1 . 77332 s 1 . For the period we find T = 2 = 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 ) =- A sin( t ) , so the maximum speed of the mass is equal to v max = 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 ) =- 2 A cos( t ) , so the maximum acceleration of the mass is equal to a max = 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 s = 0 . 533 ....
View Full Document

Page1 / 8

hw11 solutions - homework 11 VENNES, ROSS Due: Nov 16 2007,...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online