hw38 - This print-out should have 13 questions...

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Unformatted text preview: This print-out should have 13 questions. Multiple-choice questions may continue on the next column or page – find 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 Find the speed at which waves propagate on the string. 1. 10 m / s correct 2. 5 m / s 3. 20 m / s 4. 2 . 5 m / s 5. . 4 m / s Explanation: Let : f = 5 Hz and λ = 2 m . The wavelength is λ = 2 m, so the wave speed is | vectorv | = f λ = (5 Hz)(2 m) = 10 m/s . 002 (part 2 of 2) 10.0 points Find the fundamental frequency of vibration of the string. 1. 1 Hz 2. 7 . 5 Hz 3. 10 Hz 4. 5 Hz 5. 2 . 5 Hz correct 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 4 m = 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 first wire. When the two wires are stretched, and the tension in the second wire is also twice the tension in the first wire, the fundamental frequency of the first wire is 610 Hz. What is the fundamental frequency of the second wire? Correct answer: 215 . 668 Hz. Explanation: Let : f 1 = 610 Hz . The second wire has twice the radius and hence four times the cross sectional area (= π r 2 ) of the first 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 first: μ 2 = 4 μ 1 . The speed of transverse waves in a string or a wire is v = radicalBigg T μ , and since the second wire has twice the tension of the first wire, v 2 = radicalBigg T 2 μ 2 = radicalBigg 2 T 1 4 μ 1 = v 1 radicalbigg 1 2 . homework 38 – Turner – (58185) 1 hinojosa (jlh3938) – homework 38 – Turner – (58185) 2 The fundamental frequency of standing waves in the wire is given by the condition L = λ 2 , so f = v λ = v 2 L . For the two wires, v 2 = v 1 √ 2 and L 2 = 2 L 1 , so f 2 = f 1 2 √ 2 = 610 Hz 2 √ 2 = 215 . 668 Hz . 004 10.0 points A nylon guitar string vibrates in a standing wave pattern shown below. . 9 m What is the wavelength of the wave? Correct answer: 0 . 6 m. Explanation: Let : l = 0 . 9 m . The wavelength is the length of two loops: λ = 2 l 3 = 2 (0 . 9 m) 3 = . 6 m . 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 . What is the wavelength of its third har- monic? 1. λ 3 = L 2. λ 3 = 2 L 3 correct 3. λ 3 = L 3 4. λ 3 = d 3 5. λ 3 = L 2 6. λ 3 = 3 √ d L 7. λ 3 = 3 L 2 8. λ 3 = 6 L 9. λ 3 = L 6 10. λ 3 = 3 L Explanation: Basic Concepts For a string fixed at both ends, the normal modes have a wavelength λ n = 2 L n ....
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hw38 - This print-out should have 13 questions...

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