HW13-solutions-2

# HW13-solutions-2 - elam (lbe244) – HW13 – markert –...

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Unformatted text preview: elam (lbe244) – HW13 – markert – (56475) 1 This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points An ideal massless spring is fixed to the wall at one end, as shown below. A block of mass M attached to the other end of the spring oscillates with amplitude A on a frictionless, horizontal surface. The maximum speed of the block is v m . m k − A + A v m What is the force constant k of the spring? 1. k = mv 2 m 2 A 2. k = mg A 3. k = mg v m 2 A 4. k = mv 2 m 2 A 2 5. k = mv 2 m A 2 correct Explanation: For the ideal harmonic oscillation of the spring system, the kinetic energy maximum is equal to the potential energy maximum which is also the total energy of the system, so we obtain 1 2 k A 2 = 1 2 mv 2 m k = mv 2 m A 2 . 002 (part 1 of 3) 10.0 points A block of unknown mass is attached to a spring of spring constant 7 . 9 N / m and under- goes simple harmonic motion with an ampli- tude of 4 . 8 cm. When the mass is halfway between its equilibrium position and the end- point, its speed is measured to be 29 . 2 cm / s. Calculate the mass of the block. Correct answer: 0 . 160105 kg. Explanation: Let : k = 7 . 9 N / m , A = 4 . 8 cm , and v = 29 . 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 (7 . 9 N / m) (0 . 048 m) 2 4 (0 . 292 m / s) 2 = . 160105 kg . 003 (part 2 of 3) 10.0 points Find the period of the motion. Correct answer: 0 . 894476 s. Explanation: ω = radicalbigg k m = radicalBigg 7 . 9 N / m . 160105 kg = 7 . 02443 rad / s , so the period is T = 2 π ω = 2 π 7 . 02443 rad / s = . 894476 s . 004 (part 3 of 3) 10.0 points Calculate the maximum acceleration of the block. Correct answer: 2 . 36844 m / s 2 . Explanation: Simple harmonic motion is described by x = A cos ωt, elam (lbe244) – HW13 – markert – (56475) 2 so the acceleration is a = − ω 2 A cos ωt. The maximum of the cosine function is 1, so the maximum acceleration is a max = ω 2 A = (7 . 02443 rad / s) 2 (0 . 048 m) = 2 . 36844 m / s 2 . This happens when the block is at its turning point (maximum displacement). 005 10.0 points The displacement in simple harmonic motion is maximum when the 1. velocity is zero. correct 2. acceleration is zero. 3. linear momentum is a maximum. 4. velocity is a maximum. 5. kinetic energy is a maximum. Explanation: The maximum displacement occurs at the turning points, which are the points where the velocity is zero. 006 10.0 points When an object oscillating in simple harmonic motion is at its maximum displacement from the equilibrium position, which of the follow- ing is true of the values of its speed and the magnitude of the restoring force?...
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## This note was uploaded on 02/10/2011 for the course PHY 302K taught by Professor Kaplunovsky during the Fall '08 term at University of Texas at Austin.

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HW13-solutions-2 - elam (lbe244) – HW13 – markert –...

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