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Unformatted text preview: obennoskey (blo242) – hw13 – Demkov – (59910) 1 This printout should have 49 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points What requires a physical medium in which to travel? 1. sound correct 2. Both sound and light 3. light 4. Neither sound nor light Explanation: Sound requires a physical medium in which to travel; light does not. 002 10.0 points A piano string of mass per unit length . 00225 kg / m is under a tension of 2160 N. Find the speed with which a wave travels on this string. Correct answer: 979 . 796 m / s. Explanation: Let : μ = 0 . 00225 kg / m and T = 2160 N . The wave speed for a string of mass per unit length μ is v = radicalBigg T μ = radicalBigg 2160 N . 00225 kg / m = 979 . 796 m / s . keywords: 003 (part 1 of 5) 10.0 points A sinusoidal wave train is described by y = (0 . 25 m) sin bracketleftbig (0 . 3 m − 1 ) x − (40 s − 1 ) t bracketrightbig , where x and y are in meters and t is in seconds. Determine for this wave the amplitude. Correct answer: 0 . 25 m. Explanation: For a sinusoidal wave described by y = A sin( k x + ω t ) , where its amplitude is A = 0 . 25 m, angular wave number k = 0 . 3 m − 1 , angular frequency ω = 40 s − 1 , wavelength λ = 2 π k = 20 . 944 m and wave speed v = ω k = 133 . 333 m / s. 004 (part 2 of 5) 10.0 points Determine for this wave the angular fre quency. Correct answer: 40 s − 1 . Explanation: See Part 1. 005 (part 3 of 5) 10.0 points Determine for this wave the wave number. Correct answer: 0 . 3 m − 1 . Explanation: See Part 1. 006 (part 4 of 5) 10.0 points Determine for this wave the wavelength. Correct answer: 20 . 944 m. Explanation: λ = 2 π k = 2 π (0 . 3 m − 1 ) = 20 . 944 m . 007 (part 5 of 5) 10.0 points Determine for this wave the wave speed. Correct answer: 133 . 333 m / s. Explanation: v = ω k = (40 s − 1 ) (0 . 3 m − 1 ) = 133 . 333 m / s . obennoskey (blo242) – hw13 – Demkov – (59910) 2 008 10.0 points When a pebble falls into a pond, it produces ripples on the (twodimensional) water sur face. The ripples spread in all directions pro ducing a circular wavefront of increasing ra dius r . The amplitude of spreading ripples changes with the radius r . Assume: The energy spreading the water waves does not dissipated into the water or lost in any other way (no frictional losses). Determine which of the following formulae correctly describes the way the ripple ampli tude A depends on the radius r of the circular wavefront. 1. A ∝ r 2 2. A ∝ e − r/L for some constant L 3. A ∝ log r L for some constant L 4. A ∝ r 5. A ∝ 1 r 6. A = const (same A for all r ) 7. A ∝ 1 r √ r 8. A ∝ √ r 9. A ∝ 1 r 2 10. A ∝ 1 √ r correct Explanation: In two dimensions, the wave intensity is the wave power per unit length of the wave front; for a circular wavefront of radius r , I = P/ (2 πr ). If the wave’s energy is not lost in any way, than the total wave power...
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This note was uploaded on 12/04/2008 for the course PHY PHY301 taught by Professor Demovk during the Spring '08 term at University of Texas at Dallas, Richardson.
 Spring '08
 DEMOVK

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