This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: lim (kl9356) – Homework 18 – Weathers – (17104) 1 This printout should have 12 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A particle executes simple harmonic motion with an amplitude of 4 . 5 cm. At what positive displacement from the midpoint of its motion does its speed equal one half of its maximum speed? Correct answer: 3 . 89711 cm. Explanation: The potential energy of a simple harmonic oscillator at displacement x from the equilib rium point is U osc = 1 2 k x 2 = 1 2 mω 2 x 2 , since k = mω 2 . When the particle is at maximum displacement A , the energy is all potential: U = 1 2 mω 2 A 2 . At other points x , the energy might be both kinetic (speed v ) and potential K + U = 1 2 mv 2 + 1 2 mω 2 x 2 . Conservation of energy gives 1 2 mv 2 + 1 2 mω 2 x 2 = 1 2 mω 2 A 2 , or v 2 + ω 2 x 2 = ω 2 A 2 . The speed v is v = − Aω sin( ω t ) , and the sine is never more than 1, meaning v max = ω A. We are asked for the displacement at half this speed, v = ωA 2 , so conservation of energy is now parenleftbigg ω A 2 parenrightbigg 2 + ω 2 x 2 = ω 2 A 2 , or 1 4 A 2 + x 2 = A 2 , from which we see x = ± √ 3 2 A = ± √ 3 2 (4 . 5 cm) = ± 3 . 89711 cm . 002 (part 1 of 3) 10.0 points A block of unknown mass is attached to a spring of spring constant 8 . 9 N / m and under goes simple harmonic motion with an ampli tude of 10 . 4 cm. When the mass is halfway between its equilibrium position and the end point, its speed is measured to be 33 . 2 cm / s. Calculate the mass of the block. Correct answer: 0 . 655001 kg. Explanation: Let : k = 8 . 9 N / m , A = 10 . 4 cm , and v = 33 . 2 cm / s . If the maximum displacement (amplitude) is A , the halfway displacement is A 2 . By energy conservation, K i + U i = F f + U f 0 + 1 2 k A 2 = 1 2 mv 2 + 1 2 k parenleftbigg A 2 parenrightbigg 2 k A 2 = mv 2 + 1 4 k A 2 m = 3 k A 2 4 v 2 = 3 (8 . 9 N / m) (0 . 104 m) 2 4 (0 . 332 m / s) 2 = . 655001 kg . 003 (part 2 of 3) 10.0 points lim (kl9356) – Homework 18 – Weathers – (17104) 2 Find the period of the motion. Correct answer: 1 . 70453 s. Explanation: ω = radicalbigg k m = radicalBigg 8 . 9 N / m . 655001 kg = 3 . 68616 rad / s , so the period is T = 2 π ω = 2 π 3 . 68616 rad / s = 1 . 70453 s . 004 (part 3 of 3) 10.0 points Calculate the maximum acceleration of the block. Correct answer: 1 . 41313 m / s 2 . Explanation: Simple harmonic motion is described by x = A cos ωt, so the acceleration is a = − ω 2 A cos ωt....
View
Full Document
 Fall '08
 Weathers
 Magnetism, Simple Harmonic Motion, Work, lim, Correct Answer

Click to edit the document details