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Unformatted text preview: homework 10 – KIM, JI – Due: Nov 7 2007, 4:00 am 1 Question 1, chap 15, sect 2. part 1 of 1 10 points A particle oscillates up and down in simple harmonic motion. Its height y as a function of time t is shown in the diagram. 1 2 3 4 5 5 5 y (cm) t (s) At what time t in the period shown does the particle achieve its maximum positive ac celeration? 1. t = 3 s 2. t = 4 s 3. t = 2 s 4. t = 1 s correct 5. None of these, because the acceleration is constant Explanation: This oscillation is described by y ( t ) = − sin π t 2 . The acceleration is given by the second deriva tive d 2 y dt 2 = parenleftBig π 2 parenrightBig 2 sin π t 2 . The maximum acceleration will occur, then, when t = 1 s . From a noncalculus perspective, the veloc ity is negative just before t = 1 s since the particle is slowing down. At t = 1 s , the par ticle is momentarily at rest and v = 0. Just after t = 1 s, the velocity is positive since the particle is speeding up. Remembering that a = Δ v Δ t , acceleration is a positive maximum because the velocity is changing from a negative to a positive value. Question 2, chap 15, sect 3. part 1 of 1 10 points A(n) 0 . 107 kg mass is attached to a spring and undergoes simple harmonic motion with a period of 0 . 67 s. The total energy of the system is 3 J. Find the amplitude of the motion. Correct answer: 0 . 798507 m (tolerance ± 1 %). Explanation: Basic Concepts: From T = 2 π ( radicalbigg m k ) , we have k = 4 π 2 m T 2 = 4 π 2 (0 . 107 kg) (0 . 67 s) 2 = 9 . 41009 N / m . The total energy E is, E = 1 2 k A 2 , where A is the amplitude of motion. Thus A = radicalbigg 2 E k = radicalBigg 2 × (3 J) 9 . 41009 N / m = 0 . 798507 m . Alternative solution: E = ( 1 2 ) m ( ω A ) 2 , from T given, first find ω . Then based on the known quantities: E , m , and ω , solve for A . Question 3, chap 15, sect 3. part 1 of 1 10 points A particle executes simple harmonic motion with an amplitude of 4 . 64 cm. homework 10 – KIM, JI – Due: Nov 7 2007, 4:00 am 2 At what positive displacement from the midpoint of its motion does its speed equal one half of its maximum speed? Correct answer: 4 . 01836 cm (tolerance ± 1 %). 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....
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 Fall '08
 Turner
 Physics, Harmonic Series, Simple Harmonic Motion, Work, Wavelength

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