# oldhw31 - This print-out should have 12 questions....

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Unformatted text preview: 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 The force required to stretch a spring varies directly with the amount the spring is stretched. A spring stretches by 18 m when a 90 N weight is hung from it and the weight is at rest (at equilibrium). The 90 N weight is replaced by an unknown weight W so that the spring is stetched to a new equilibrium position, 14 m below the position if no weight were attached. The weight W is then dis- placed from equilibrium and released so that it oscillates. 14 m 18 m W 90 N Calculate the period of motion. The accel- eration of gravity is 9 . 8 m / s 2 . Correct answer: 7 . 50984 N. Explanation: Let : W = W , W = 90 N , x 1 = 14 m , and x 2 = 18 m . F = k x , so W W = x 1 x 2 m 1 g W = x 1 x 2 m 1 = W x 1 g x 2 = (90 N)(14 m) (9 . 8 m / s 2 )(18 m) = 7 . 14286 kg = radicalbigg k m 1 , so the (angular) frequency of os- cillation is T = 2 = 2 radicalbigg m 1 k = 2 radicalBigg 7 . 14286 kg 5 N / m = 7 . 50984 s . 002 10.0 points You are designing a pendulum clock to have a period of 1.0 s. The acceleration of gravity is 9 . 81 m / s 2 . How long should the pendulum be? Correct answer: 0 . 24849 m. Explanation: Basic Concept: T = 2 radicalBigg L g Given: T = 1 . 0 s g = 9 . 81 m / s 2 Solution: parenleftbigg T 2 parenrightbigg 2 = L g L = g parenleftbigg T 2 parenrightbigg 2 = (9 . 81 m...
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## This note was uploaded on 05/02/2009 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.

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oldhw31 - This print-out should have 12 questions....

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