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Unformatted text preview: kuehner (rjk679) HW14 Mackie (20211) 1 This printout should have 16 questions. Multiplechoice 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 16 g object is hung from it. The object is replaced with a block of mass 25 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 . 455757 s. Explanation: Let : x = 3 . 3 cm = 0 . 033 m , m 1 = 16 g = 0 . 016 kg , and m 2 = 25 g = 0 . 025 kg . The force on the spring is F = k x k = F x = m 1 g x . When the 25 g is placed into simple harmonic motion, T = 2 radicalbigg m 2 k = 2 radicalbigg m 2 x m 1 g = 2 radicalBigg (0 . 025 kg) (0 . 033 m) (0 . 016 kg) (9 . 8 m / s 2 ) = . 455757 s . 002 10.0 points Having landed on a newly discovered planet, an astronaut sets up a simple pendulum of length 1 . 57 m and finds that it makes 302 complete oscillations in 593 s. The amplitude of the oscillations is very small compared to the pendulums length. What is the gravitational acceleration on the surface of this planet? Correct answer: 16 . 0755 m / s 2 . Explanation: Basic Concept A simple pendulum oscil lating with small amplitude has period T = 2 radicalBigg g where is the pendulums length and g is the gravitational acceleration at the pendulums location. The pendulum in question made n = 302 complete oscillations in time t = 593 s, which implies the period of T = t n . Thus t n = 2 radicalBigg g t 2 n = radicalBigg g t 2 4 n 2 2 = g g = 4 n 2 2 t 2 Thus the planets surface gravity is g = 4 n 2 2 t 2 = 16 . 0755 m / s 2 . 003 10.0 points You need to know the height of a tower, but darkness obscures the ceiling. You note that a pendulum extending from the ceiling almost touches the floor and that its period is 26 s. The acceleration of gravity is 9 . 81 m / s 2 . How tall is the tower? Correct answer: 167 . 979 m. Explanation: Basic Concept: T = 2 radicalBigg L g Given: T = 26 s g = 9 . 81 m / s 2 kuehner (rjk679) HW14 Mackie (20211) 2 Solution: parenleftbigg T 2 parenrightbigg 2 = L g L = g parenleftbigg T 2 parenrightbigg 2 = (9 . 81 m / s 2 ) parenleftbigg 26 s 2...
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 Spring '09
 MIHALISIN
 Physics

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