PHY 303K - Homework 22

PHY 303K - Homework 22 - homework 22 – FIERRO JEFFREY –...

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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 mass-spring 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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PHY 303K - Homework 22 - homework 22 – FIERRO JEFFREY –...

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