hw12 - irshad(si2865 – HW12 – distler –(57205 1 This...

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Unformatted text preview: irshad (si2865) – HW12 – distler – (57205) 1 This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A harmonic wave y = A sin[ k x − ω t − φ ] , where A = 1 m, k has units of m − 1 , ω has units of s − 1 , and φ has units of radians, is plotted in the diagram below. +1 − 1 A (meters) x (meters) 6 12 18 At the time t = 0 Which wave function corresponds best to the diagram? 1. y = A sin bracketleftbiggparenleftbigg 2 π 9 m parenrightbigg x − ω t − parenleftbigg 1 π 3 parenrightbiggbracketrightbigg 2. y = A sin bracketleftbiggparenleftbigg 2 π 15 m parenrightbigg x − ω t − parenleftbigg 5 π 3 parenrightbiggbracketrightbigg 3. y = A sin bracketleftbiggparenleftbigg 2 π 9 m parenrightbigg x − ω t − parenleftbigg 4 π 3 parenrightbiggbracketrightbigg 4. y = A sin bracketleftbiggparenleftbigg 2 π 9 m parenrightbigg x − ω t − parenleftbigg 2 π 3 parenrightbiggbracketrightbigg 5. y = A sin bracketleftbiggparenleftbigg 2 π 9 m parenrightbigg x − ω t − parenleftbigg 5 π 3 parenrightbiggbracketrightbigg cor- rect 6. y = A sin bracketleftbiggparenleftbigg 2 π 3 m parenrightbigg x − ω t − parenleftbigg 5 π 3 parenrightbiggbracketrightbigg 7. y = A sin bracketleftbiggparenleftbigg 2 π 15 m parenrightbigg x − ω t − parenleftbigg 2 π 3 parenrightbiggbracketrightbigg 8. y = A sin bracketleftbiggparenleftbigg 2 π 15 m parenrightbigg x − ω t − parenleftbigg 1 π 3 parenrightbiggbracketrightbigg 9. y = A sin bracketleftbiggparenleftbigg 2 π 3 m parenrightbigg x − ω t − parenleftbigg 4 π 3 parenrightbiggbracketrightbigg 10. y = A sin bracketleftbiggparenleftbigg 2 π 15 m parenrightbigg x − ω t − parenleftbigg 4 π 3 parenrightbiggbracketrightbigg Explanation: From the diagram of the wave function the wave-length λ = 9 m (6 horizontal scale divi- sions of 1 . 5 m each, see diagram below). The given wave function (sine function with t = 0) y = A sin( k x − φ ) = A sin bracketleftbiggparenleftbigg 2 π λ parenrightbigg x − φ bracketrightbigg = A sin bracketleftbiggparenleftbigg 2 π 9 m parenrightbigg x − φ bracketrightbigg (dark curve in diagram below) is shifted 5 divisions to the right (negative phase shift) of a no-phase-shift sine function y = sin bracketleftbiggparenleftbigg 2 π 9 m parenrightbigg x bracketrightbigg (gray curve in diagram below), therefore φ = 5 π 3 = 5 π 3 radians . Checking the wave function y at x = 0, we have agreement with the diagram below y = sin parenleftbigg − 5 π 3 parenrightbigg = 0 . 866025 m . +1 − 1 A (meters) x (meters) 9 18 At the time t = 0 φ λ = 9 Therefore, the wave function is y = A sin bracketleftbiggparenleftbigg 2 π 9 m parenrightbigg x − ω t − parenleftbigg 5 π 3 parenrightbiggbracketrightbigg . irshad (si2865) – HW12 – distler – (57205) 2 002 10.0 points A uniform disk of radius 2 . 2 m and mass 8 .....
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This note was uploaded on 07/13/2011 for the course CHEM 301 taught by Professor Wandelt during the Spring '08 term at University of Texas.

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hw12 - irshad(si2865 – HW12 – distler –(57205 1 This...

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