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15homework12

15homework12 - Husain Zeena Homework 12 Due 4:00 am Inst H...

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Husain, Zeena – Homework 12 – Due: Nov 18 2003, 4:00 am – Inst: H L Berk 1 This print-out should have 19 questions, check that it is complete. Multiple-choice questions may continue on the next column or page: find all choices before making your selection. The due time is Central time. 001 (part 1 of 2) 10 points The time interval indicated on this diagram is G . S 0 G t S Which formula corresponds best to the di- agram? 1. S ( t ) = S 0 sin 2 t 3 G 2. S ( t ) = S 0 sin 3 t 2 G 3. S ( t ) = S 0 sin 2 t 3 π G 4. S ( t ) = S 0 sin 2 π t G 5. S ( t ) = S 0 sin t 3 π G 6. S ( t ) = S 0 sin 3 π t G correct 7. S ( t ) = S 0 sin 2 π t 3 G 8. S ( t ) = S 0 sin 3 π t 2 G 9. S ( t ) = S 0 sin t 2 π G 10. S ( t ) = S 0 sin 3 t 2 π G Explanation: The equation of a wave with zero displace- ment at the origin is given by S ( t ) = S 0 sin( ω t ) = S 0 sin 2 π T t , since ω 2 π T . From the figure, we see 1 3 G = 1 2 T which means T = 2 3 G . Since ω 2 π T = 3 π G , we have S ( t ) = S 0 sin 3 π G t . 002 (part 2 of 2) 10 points This wave has period T . S 0 T t S Which formula corresponds best to the di- agram? 1. S ( t ) = S 0 sin 2 π t T - π 2 2. S ( t ) = S 0 sin 2 π t T + π 2 correct 3. S ( t ) = S 0 sin π t T - π 2 4. S ( t ) = S 0 sin π t T + π 2 5. S ( t ) = S 0 sin π 2 - π t T 6. S ( t ) = S 0 sin π t T 7. S ( t ) = S 0 sin 2 π t T 8. S ( t ) = S 0 sin - π 2 - 2 π t T 9. S ( t ) = S 0 sin - 2 π t T 10. S ( t ) = S 0 sin - π t T Explanation: First note that most of the incorrect solu- tions can be eliminated immediately by check- ing the value of the displacement at t = 0. The equation for a wave with period T and maximum displacement at the origin is S ( t ) = S 0 cos 2 π T t .

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Husain, Zeena – Homework 12 – Due: Nov 18 2003, 4:00 am – Inst: H L Berk 2 Since cos( ω t ) = sin ω t + π 2 · we find S ( t ) = S 0 sin 2 π T t + π 2 . 003 (part 1 of 5) 10 points The wave function for a linearly polarized wave on a taut string is y ( x, t ) = A sin( ω t - k x + φ ) , where A = 0 . 34 m, ω = 9 . 1 s - 1 , k = 6 . 2 m - 1 , φ = 0 . 41, t is in seconds and x and y are in meters. What is the speed of the wave? Correct answer: 1 . 46774 m / s. Explanation: Basic Concepts: v = λ T k = 2 π λ f = ω 2 π Solution: For the propagation speed of a wave we have v = ω k = (9 . 1 s - 1 ) (6 . 2 m - 1 ) = 1 . 46774 m / s . 004 (part 2 of 5) 10 points What is the vertical displacement of the string at x 0 = 0 . 07 m, t 0 = 3 s? Correct answer: 0 . 285793 m. Explanation: We can obtain the result if we simply sub- stitute the values for x and t into y ( x, t ) y ( x 0 , t 0 ) = A sin[ ω t - k x + φ ] = (0 . 34 m) sin h (9 . 1 s - 1 ) (3 s) - (6 . 2 m - 1 ) (0 . 07 m) + (0 . 41) i = 0 . 285793 m . 005 (part 3 of 5) 10 points What is the wavelength? Correct answer: 1 . 01342 m. Explanation: For the wave vector we have k = 2 π λ . Thus λ = 2 π k = 2 π (6 . 2 m - 1 ) = 1 . 01342 m . 006 (part 4 of 5) 10 points What is the frequency of the wave?
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