HW 14-solutions

HW 14-solutions - kuruvila (lk5992) – HW 14 – opyrchal...

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Unformatted text preview: kuruvila (lk5992) – HW 14 – opyrchal – (11113) 1 This print-out should have 9 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A spring stretches 3 . 3 cm when a 14 g object is hung from it. The object is replaced with a block of mass 34 g which oscillates in simple harmonic motion. The acceleration of gravity is 9 . 8 m / s 2 . Calculate the period of motion. Correct answer: 0 . 568197 s. Explanation: Let : x = 3 . 3 cm = 0 . 033 m , m 1 = 14 g = 0 . 014 kg , and m 2 = 34 g = 0 . 034 kg . The force on the spring is F = k x k = F x = m 1 g x . When the 34 g is placed into simple harmonic motion, T = 2 π radicalbigg m 2 k = 2 π radicalbigg m 2 x m 1 g = 2 π radicalBigg (0 . 034 kg) (0 . 033 m) (0 . 014 kg) (9 . 8 m / s 2 ) = . 568197 s . 002 (part 1 of 3) 10.0 points A block of unknown mass is attached to a spring of spring constant 6 . 8 N / m and under- goes simple harmonic motion with an ampli- tude of 9 . 2 cm. When the mass is halfway between its equilibrium position and the end- point, its speed is measured to be 23 . 5 cm / s. Calculate the mass of the block. Correct answer: 0 . 781646 kg. Explanation: Let : k = 6 . 8 N / m , A = 9 . 2 cm , and v = 23 . 5 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 (6 . 8 N / m) (0 . 092 m) 2 4 (0...
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This note was uploaded on 02/12/2011 for the course PHYS 111 taught by Professor Moro during the Fall '08 term at NJIT.

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HW 14-solutions - kuruvila (lk5992) – HW 14 – opyrchal...

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