F5ch17 - CHAPTER 17 WAVESI 473 CHAPTER 17 Answer to Checkpoint Questions 1 2 3 4 5 6(a 2(b 3(c 1(a 2 3 1(b 3 then 1 and 2 tie a 0:20 and 0:80 tie

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CHAPTER 17 WAVES—I 473 C HAPTER 17 Answer to Checkpoint Questions . ( a ) ; ( b ) ; ( c ) . ( a ) , , ; ( b ) , then and tie . a . : and : tie, then : , : . ( a ) ; ( b ) ; ( c ) . ( a ) Hz; ( b ) Hz Answer to Questions . d . b, a, c . tie of A and B, then C, D . ( a ) ; ( b ) ; ( c ) . intermediate (closer to fully destructive interference) . ( a ) all tie; ( b ) intermediate (closer to fully constructive interference) . a and d tie, then b and c tie . ( a ) and ; ( b ) , , and ; or , , and . ( a ) ; ( b ) antinode; ( c ) longer; ( d ) lower . ( a ) decrease; ( b ) disappears . ( a ) integer multiples of ; ( b ) node; ( c ) node . d . string A . ( a ) increase, decrease; ( b ) same, increase; ( c ) same, decrease
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474 CHAPTER 17 WAVES—I . decrease E ( a ) The frequency is f = v= = ( m/s) = ( : m) = Hz : ( b ) The period is T = =f = = Hz = : s. E ( a ) k = = : m = : m : ( b ) v = f = ! = ( : m)( rad/s) = : m/s : E ( a ) For visible light f min = c max = : m/s m = : Hz and f max = c min = : m/s m = : Hz : ( b ) For radio waves min = c max = : m/s Hz = : m and max = c min = : m/s : Hz = : m : ( c ) For X rays f min = c max = : m/s : m = : Hz
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y (cm) x (cm) 0 80 160 ( t =0) 2.0 -2.0 CHAPTER 17 WAVES—I 475 and f max = c min = : m/s : m = : Hz : E ( a ) T = ( : s) = : s. ( b ) f = =T = = : s = : Hz : ( c ) v = f = ( : m)( : Hz) = : m/s : E ( a ) y ( x; t ) = y m sin( kx + !t ) = y m sin f x v + t = ( : m) sin ( Hz) x m/s + t = : m sin[ ( : x + t )] ; where x is in meters and t is in seconds. E ( a )
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2.0 -2.0 y (cm) 0 80 160 ( t =0.05s) 2.0 -2.0 y (cm) 0 80 160 ( t =0.10s) x (cm) x (cm) 476 CHAPTER 17 WAVES—I ( b ) ( c ) The wave speed is v = ( x= t ) peak = cm = : s = cm/s : The wave travels in the negative x direction. E You can easily verify these alternative forms by using the following relations: k = = , = vT = v=f , and ! = f = =T = kv . E ( a ) Compare the wave f ( x; t ) = h ( x t ) with the general form f ( x; t ) = f ( x vt ) to obtain v = cm/s : ( b ) It travels in the positive x direction.
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h ( x ) 2 11 13 14 x t =2s h ( t ) 2 t x =10cm 1.2 1.4 1.8 CHAPTER 17 WAVES—I 477 ( c ) E ( a ) Use Eq. - : u max = @y @t max = [ !y m cos( kx !t )] max = fy m : ( b ) a y = @u @t = @ @t [ !y m cos( kx !t )] = ! y m sin( kx !t ) : Thus a y; max = [ ! y m sin( kx !t )] max = f y m : E ( a ) A = : mm ; f = s = = Hz ; v = ( s ) = ( m ) = m/s ; and = v=f = ( m/s) = ( Hz) = : m : ( b ) u max = !A = fA = ( Hz)( : mm) = : m/s : E ( a ) Let the displacements of the wave at ( y; t ) be z ( y; t ). Then z ( y; t ) = z m sin( ky !t ), where z m = : mm, k = cm , and ! = =T = = : s = s . Thus z ( y; t ) = ( : mm) sin[( cm ) y ( s ) t ] :
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4 . 0 1 20 y(cm) x (cm) t =0 -4.0 478 CHAPTER 17 WAVES—I ( b ) u z; max = !z m = ( = : s)( : mm) = mm/s : P ( a ) A = : cm : ( b ) Solve from = = : : = cm : ( c ) Solve f = : : f = : Hz : ( d ) v = f = ( cm)( : Hz) = cm/s : ( e ) The wave propagates in the negative x direction. ( f
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This note was uploaded on 01/22/2012 for the course PHYS 2101 taught by Professor Grouptest during the Fall '07 term at LSU.

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F5ch17 - CHAPTER 17 WAVESI 473 CHAPTER 17 Answer to Checkpoint Questions 1 2 3 4 5 6(a 2(b 3(c 1(a 2 3 1(b 3 then 1 and 2 tie a 0:20 and 0:80 tie

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