Homework 18 Solutions

# Homework 18 Solutions - lim(kl9356 – Homework 18 –...

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

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: lim (kl9356) – Homework 18 – Weathers – (17104) 1 This print-out should have 12 questions. Multiple-choice 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

## This note was uploaded on 09/07/2010 for the course PHYS 1710 taught by Professor Weathers during the Fall '08 term at North Texas.

### Page1 / 7

Homework 18 Solutions - lim(kl9356 – Homework 18 –...

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

View Full Document
Ask a homework question - tutors are online