Practice Problems 3 Solutions

# Practice Problems 3 Solutions - khounvivongsy(sk27799 –...

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Unformatted text preview: khounvivongsy (sk27799) – Practice Problems 3 Solutions – Weathers – (17104) 1 This print-out should have 19 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points Two harmonic waves are described by y 1 = A sin( k x- ω t ) , y 2 = A sin( k x + ω t ) , where A = 3 . 3 mm, k = 7 . 5 rad / m and ω = 3600 rad / s. What is the amplitude y max ( x ) of the resul- tant wave at x = 1 . 55 m? Correct answer: 5 . 33521 mm. Explanation: The superposition of two harmonic waves travelling in opposite directions y 1 ( x, t ) = A sin( k x- ω t ) , (1) y 2 ( x, t ) = A sin( k x + ω t ) , (2) is the standing wave y ( x, t ) = y 1 ( x, t ) + y 2 ( x, t ) (3) = A bracketleftBig sin( kx- ωt ) + sin( kx + ωt ) bracketrightBig = 2 A sin( kx ) cos( ωt ) (4) where the last equality follows from the trigonometric identity sin( α- β )+sin( α + β ) = 2 sin( α ) cos( β ) . (5) At any particular point x , the standing wave (4) oscillates according to y ( t ) = A ( x ) cos( ω t ) (6) with x –dependent amplitude A ( x ) = 2 A sin( kx ) . (7) To be precise, since this A ( x ) function could be either positive or negative, the true ampli- tude at x = 1 . 55 m is the magnitude y max ( x ) = |A ( x ) | = 2 A | sin( k x ) | = 2 A vextendsingle vextendsingle vextendsingle sin bracketleftBig (7 . 5 rad / m) (1 . 55 m) bracketrightBigvextendsingle vextendsingle vextendsingle = 2 (3 . 3 mm) | sin(11 . 625 rad) | = 5 . 33521 mm . 002 10.0 points Two harmonic waves are described by y 1 = A sin(3 x- 5 t ) , y 2 = A sin(3 x- 5 t- 4) . A = 16 m, with x and t being given in SI units and the phase angles in radians. What is the displacement of the sum of these two harmonic waves, y 1 + y 2 , at x = 1 m, t = 1 s? Correct answer:- 10 . 0781 m. Explanation: First evaluate y 1 and y 2 at x = 1 m and t = 1 s. Then add the two results by superpo- sition. y 1 = (16 m) sin(3- 5) =- 14 . 5488 m , y 2 = (16 m) sin(3- 5- 4) = 4 . 47065 m y 1 (1 m , 1 s) + y 2 (1 m , 1 s) =- 10 . 0781 m . Alternative Solution: Using the basic trigonometric relation sin θ 1 + sin θ 2 = 2 sin θ 1 + θ 2 2 cos θ 1- θ 2 2 , we have the resultant wave y = y 1 + y 2 = 2 A sin(3 x- 5 t- 2) cos 2 . Therefore the displacement of the resultant wave at x = t = 1 is A ′ = 2 A sin(3- 5- 2) cos2 =- 10 . 0781 m . 003 10.0 points Two identical harmonic waves with wave- lengths of 1 . 8 m travel in the same direction khounvivongsy (sk27799) – Practice Problems 3 Solutions – Weathers – (17104) 2 at a speed of 3 . 9 m / s. The second wave origi- nates from the same point as the first, but at a later time. Determine the minimum possible time in- terval between the starting moments of the two waves if the amplitude of the resultant wave is the same as that of the two initial waves....
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## This note was uploaded on 05/09/2010 for the course PHYS 1710 taught by Professor Weathers during the Spring '09 term at University of North Carolina Wilmington.

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Practice Problems 3 Solutions - khounvivongsy(sk27799 –...

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