# hw27 - alexander(jra2623 – homework 27 – Turner...

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Unformatted text preview: alexander (jra2623) – homework 27 – Turner – (92510) 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 130 N object vibrates with a period of 3.12 s when hanging from a spring. What is the spring constant of the spring? The acceleration of gravity is 9 . 81 m / s 2 . Correct answer: 53 . 7434 N / m. Explanation: Let : F g = 130 N , T = 3 . 12 s , and g = 9 . 81 m / s 2 . T = 2 π radicalbigg m k parenleftbigg T 2 π parenrightbigg 2 = m k k = m parenleftbigg 2 π T parenrightbigg 2 = F g g parenleftbigg 2 π T parenrightbigg 2 = 130 N 9 . 81 m / s 2 parenleftbigg 2 π 3 . 12 s parenrightbigg 2 = 53 . 7434 N / m . 002 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? Magnitude of Speed Restoring Force 1. 1 2 maximum 1 2 maximum 2. Zero Zero 3. Maximum Zero 4. Maximum 1 2 maximum 5. Zero Maximum correct Explanation: The maximum displacement occurs at the turning points (where the velocity or speed is zero). The magnitude of restoring force is given by Hooke’s law F =- k x , where k is the spring constant and x is the displacement. Since x is a maximum, F is a maximum. From a different perspective, the displace- ment from the equilibrium position can be written as y = A sin θ , where θ is the phase of the oscillation. When the object is at its maximum displacement sin θ = 1 θ = π 2 so its speed is...
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hw27 - alexander(jra2623 – homework 27 – Turner...

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