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Unformatted text preview: homework 22 – FIERRO, JEFFREY – Due: Apr 15 2008, 11:00 pm 1 Question 1, chap 15, sect 4. part 1 of 3 10 points A certain pendulum clock that works per fectly on Earth is taken to the moon, where g = 1 . 63 m / s 2 . The acceleration of gravity is 9 . 81 m / s 2 . The clock is started at 12:00:00 A.M. and runs for 16 h. a) What will be the reading for the hours? Correct answer: 6 h (tolerance ± 1 %). Explanation: Basic Concept: T = 2 π radicalBigg L g Given: g moon = 1 . 63 m / s 2 Δ t = 16 h g Earth = 9 . 81 m / s 2 Solution: T Earth = 2 π radicalBigg L g Earth T moon = 2 π radicalBigg L g moon T moon T Earth = radicalbigg g Earth g moon T moon T Earth = radicalBigg 9 . 81 m / s 2 1 . 63 m / s 2 = 2 . 45324 Thus the period of one oscillation of the pen dulum on the moon takes 2 . 45324 times longer than one oscillation on Earth, or Deltat moon = Deltat Earth 2 . 45324 = 16 h 2 . 45324 = 6 . 52198 h = 6 h + 0 . 521977 h The hours will be 6. Question 2, chap 15, sect 4. part 2 of 3 10 points b) What will be the reading for the min utes? Correct answer: 31 min (tolerance ± 1 %). Explanation: . 521977 h = (0 . 521977 h) parenleftbigg 60 min 1 hr parenrightbigg = 31 min + 0 . 318636 min The minutes will be 31 min. Question 3, chap 15, sect 4. part 3 of 3 10 points c) What will be the reading for the seconds? Correct answer: 19 . 1182 sec (tolerance ± 1 %). Explanation: . 318636 min = (0 . 318636 min) parenleftbigg 60 sec 1 min parenrightbigg = 19 . 1182 s The seconds will be 19 . 1182 s. Question 4, chap 15, sect 5. part 1 of 3 0 points A damped massspring system oscillates at 267 Hz . The oscillation of the amplitude has a time constant of 2 . 5 s . At t = 0 the amplitude of oscillation is 5 . 2 cm and the energy of the oscillating system is 16 J . What is the amplitude of oscillation at t = 8 . 4 s? Correct answer: 0 . 180623 cm (tolerance ± 1 %). Explanation: homework 22 – FIERRO, JEFFREY – Due: Apr 15 2008, 11:00 pm 2 Let : f = 267 Hz , t = 8 . 4 s , τ amp = 2 . 5 s and A = 5 . 2 cm . The equation describing damped harmonic motion is − k x − b dx dt = m d 2 x dt 2 , or x = A exp parenleftbigg − b 2 m t parenrightbigg cos( ω t + φ ) Let : τ = 2 m b = 8 . 4 s , time constant x = A e − t τ cos( ω t + π ) , where ω = radicalbigg 2 τ − 1 τ 2 , and A = A e − t τ The amplitude of the oscillations at t = 8 . 4 s is A = A e − t/τ amp A (8 . 4 s) = (5 . 2 cm) e ( − 8 . 4 s) / (2 . 5 s) = . 180623 cm . Question 5, chap 15, sect 5. part 2 of 3 0 points Hint: The oscillator’s energy is propor tional to the angular velocity squared and the amplitude squared, consequently the time constant for the energy may be different from the time constant for the amplitude....
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This note was uploaded on 05/11/2008 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas at Austin.
 Spring '08
 Turner
 Acceleration, Work

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