466_PartUniversity Physics Solution

466_PartUniversity Physics Solution - Mechanical Waves...

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Mechanical Waves 15-3 15.9. IDENTIFY: Evaluate the partial derivatives and see if Eq.(15.12) is satisfied. SET UP: cos( ) sin( ) kx t k kx t x ω += + . cos( ) sin( ) kx t kx t t ωω + . sin( ) cos( ) kx t k kx t x + . sin( ) sin( ) kx t kx t t + . EXECUTE: (a) 2 2 2 cos( ) y Ak kx t x =− + . 2 2 2 cos( ) y Ak x t t + . Eq.(15.12) is satisfied, if / vk = . (b) 2 2 2 sin( ) y Ak kx t x + . 2 2 2 sin( ) y x t t + . Eq.(15.12) is satisfied, if / = . (c) sin( ) y kA kx x . 2 2 2 cos( ) y kA k x x . sin( ) y At t . 2 2 2 cos( ) y t . Eq.(15.12) is not satisfied. (d) cos( ) y y vA k x t t == + . 2 2 2 sin( ) y y aA k x t t + EVALUATE: The functions cos( ) kx t + and sin( ) kx t + differ only in phase. 15.10. IDENTIFY: y v and y a are given by Eqs.(15.9) and (15.10). SET UP: The sign of y v determines the direction of motion of a particle on the string. If 0 y v = and 0 y a the speed of the particle is increasing. If 0 y v , the particle is speeding up if y v and y a have the same sign and slowing down if they have opposite signs. EXECUTE: (a) The graphs are given in Figure 15.10. (b) (i) sin(0) 0 y v ω A and the particle is instantaneously at rest. 22 cos(0) y a ω A ω A and the particle is speeding up. (ii) sin( 4) 2, y v ω A πω A and the particle is moving up. cos( 4) 2 , y a ω A A and the particle is slowing down ( y v and y a have opposite sign). (iii) sin( 2) y v ω A A and the particle is moving up. 2 cos( 2) 0 y a ω A π = −= and the particle is instantaneously not accelerating. (iv) sin(3 4) y v ω A A and the particle is moving up. cos(3 4) y a ω A A = and the particle is speeding up. (v) sin( ) 0 y v ω A π and the particle is instantaneously at rest. cos( ) y a ω A A = and the particle is speeding up. (vi) sin(5 4) 2 y v ω A A and the particle is moving down. cos(5 4) 2 y a ω A A = and the particle is slowing down ( y v and y a have opposite sign). (vii) sin(3 2) y v ω A A and the particle is moving down. 2 cos(3 2) 0 y a ω A π = and the particle is instantaneously not accelerating. (viii) sin(7 4) y v ω A A and the particle is moving down. cos(7 4) 2 y a ω A A and the particle is speeding up ( y v and y a have the same sign). EVALUATE: At 0 t = the wave is represented by Figure 15.10a in the textbook: point (i) in the problem corresponds to the origin, and points (ii)-(viii) correspond to the points in the figure labeled 1-7. Our results agree with what is shown in the figure. Figure 15.10
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15-4 Chapter 15 15.11. IDENTIFY and SET UP: Read A and T from the graph. Apply Eq.(15.4) to determine λ and then use Eq.(15.1) to calculate v . EXECUTE: (a) The maximum y is 4 mm (read from graph). (b) For either x the time for one full cycle is 0.040 s; this is the period. (c) Since 0 y = for 0 x = and 0 t = and since the wave is traveling in the -direction x + then (, ) s i n [ 2(/ / ) ] .
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466_PartUniversity Physics Solution - Mechanical Waves...

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