homework 33-solutions - klare (alk736) homework 33 Turner...

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klare (alk736) – homework 33 – Turner – (58220) 1 This print-out should have 13 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 (part 1 oF 2) 10.0 points A standing wave oF Frequency 5 hertz is set up on a string 2 meters long with nodes at both ends and in the center, as shown. 2 meters ±ind the speed at which waves propagate on the string. 1. 2 . 5m / s 2. 20 m / s 3. 0 . 4m / s 4. 10 m / s correct 5. / s Explanation: Let : f = 5 Hz and λ = 2 m . The wavelength is λ = 2 m, so the wave speed is | ± v | = f λ = (5 Hz)(2 m) = 10 m/s . 002 (part 2 oF 2) 10.0 points ±ind the Fundamental Frequency oF vibration oF the string. 1. 5 Hz 2. 7 . 5 Hz 3. 2 . 5 Hz correct 4. 1 Hz 5. 10 Hz Explanation: 2 meters The Fundamental wave has only two nodes at the ends, so its wavelength is λ = 4 m and the Fundamental Frequency is f = v λ = 10 m / s = 2.5 Hz . 003 10.0 points Two wires are made oF the same material but the second wire has twice the diameter and twice the length oF the frst wire. When the two wires are stretched, and the tension in the second wire is also twice the tension in the frst wire, the Fundamental Frequency oF the frst wire is 270 Hz. What is the Fundamental Frequency oF the second wire? Correct answer: 95 . 4594 Hz. Explanation: Let : f 0 1 = 270 Hz . The second wire has twice the radius and hence Four times the cross sectional area (= π r 2 ) oF the frst wire. Since the two wires are made From the same material, the linear density μ = π d 2 4 ρ oF the second wire is Four times that oF the frst: μ 2 =4 μ 1 . The speed oF transverse waves in a string or a wire is v = ± T μ , and since the second wire has twice the tension oF the frst wire, v 2 = ± T 2 μ 2
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klare (alk736) – homework 33 – Turner – (58220) 2 = ± 2 T 1 4 μ 1 = v 1 ² 1 2 . The fundamental frequency of standing waves in the wire is given by the condition L = λ 2 , so f 0 = v λ = v 2 L . For the two wires, v 2 = v 1 2 and L 2 =2 L 1 , so f 0 2 = f 0 1 2 2 = 270 Hz 2 2 = 95 . 4594 Hz . 004 10.0 points A nylon guitar string vibrates in a standing wave pattern shown below. 0 . 6m What is the wavelength of the wave? Correct answer: 0 . 4 m. Explanation: Let : l =0 . . The wavelength is the length of two loops: λ = 2 l 3 = 2 (0 . 6 m) 3 = 0 . 4m . 005 (part 1 of 2) 10.0 points Consider a vibrating piano string. The string is under a tension T , has a length L and diameter d .
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homework 33-solutions - klare (alk736) homework 33 Turner...

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