486_PartUniversity Physics Solution

486_PartUniversity Physics Solution - Mechanical Waves...

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Mechanical Waves 15-23 15.70. IDENTIFY: The displacement of the string at any point is SW ( , ) ( sin )sin . yxt A k x t ω = For the fundamental mode 2, L λ = so at the midpoint of the string sin sin(2 )( 2) 1, kx L πλ = = and SW sin . yA t = The transverse velocity is / y vy t =∂ ∂ and the transverse acceleration is / yy av t = ∂∂ . SET UP: Taking derivatives gives SW cos y y vA t t == , with maximum value ,max SW y = , and 2 SW sin y y v aA t t , with maximum value 2 , max SW y = . EXECUTE: 32 3 ,max ,max (8.40 10 m s ) (3.80 m s) 2.21 10 rad s × = × , and then 33 SW ,max (3.80 m s) (2.21 10 rad s) 1.72 10 m y Av × = × . (b) 3 (2 )( 2 ) (0.386 m)(2.21 10 rad s) 272 m s vf L L λω π = = × = . EVALUATE: The maximum transverse velocity and acceleration will have different (smaller) values at other points on the string. 15.71. IDENTIFY: To show this relationship is valid, take the second time derivative. SET UP: sin cos tt t = . cos sin t ωω =− . EXECUTE: (a) 22 SW SW (,) [( sin )sin ] [( sin )cos ] Ak xt x t t [] 2 sw 2 ( sin )sin ( , ) x t y x t t . This equation shows that 2 y ay . This is characteristic of simple harmonic motion; each particle of the string moves in simple harmonic motion. (b) Yes, the traveling wave is also a solution of this equation. When a string carries a traveling wave each point on the string moves in simple harmonic motion. EVALUATE: A standing wave is the superposition of two traveling waves, so it is not surprising that for both types of waves the particles on the string move in SHM. 15.72. IDENTIFY and SET UP: Carry out the analysis specified in the problem. EXECUTE: (a) The wave moving to the left is inverted and reflected; the reflection means that the wave moving to the left is the same function of , x and the inversion means that the function is ( ). f x (b) The wave that is the sum is ( ) ( ) f xfx (an inherently odd function), and for any , (0) ( 0) 0. ff f −−= (c) The wave is reflected but not inverted (see the discussion in part (a) above), so the wave moving to the left in Figure 15.21 in the textbook is ( ). f x +− (d) () ( ) ( ) ( ) (() ) x x dy d df x df x df x df x d x df df fx f x dx dx dx dx dx d x dx dx dx =− −− =+ = + = + = . At 0 x = , the terms are the same and the derivative is zero. EVALUATE: Our results verify the behavior shown in Figures 15.20 and 15.21 in the textbook. 15.73. IDENTIFY: Carry out the derivation as done in the text for Eq.(15.28). The transverse velocity is / y t and the transverse acceleration is / . t =∂ (a) SET UP: For reflection from a free end of a string the reflected wave is not inverted, so 12 ( , ) ( , ), yx t y x t where 1 (, ) c o s ( ) yxt A k x t (traveling to the left) 2 ) c o s ( ) k (traveling to the right) Thus ( , ) [cos( ) cos( )]. A k x t k x t + EXECUTE: Apply the trig identity cos( ) cos cos sin sin ab a b ± = with ak x = and : bt = cos( ) cos cos sin sin kx t kx t kx t += and cos( ) cos cos sin sin . kx t kx t kx t −= + Then ( , ) (2 cos )cos A k x t = (the other two terms cancel) (b) For 0, x = cos 1 kx = and ( , ) 2 cos . A t =
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This note was uploaded on 07/16/2011 for the course PHY 2053 taught by Professor Buchler during the Spring '06 term at University of Florida.

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486_PartUniversity Physics Solution - Mechanical Waves...

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