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491_PartUniversity Physics Solution

# 491_PartUniversity Physics Solution - 15-28 Chapter 15(c...

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15-28 Chapter 15 (c) Denote the tension as 0 ( ) , F x F ax = + where 0 392 N F = and 7.70 N/m. a = Then the speed of transverse waves as a function of x is 0 ( ) dx v F ax dt μ = = + and the time t needed for a wave to reach the surface is found from 0 . μ dx t dt dx dx dt F ax = = = + Let the length of the cable be L , so ( ) 0 0 0 0 0 0 2 2 L L dx t F ax F aL F a a F ax μ μ μ = = + = + + . 2 1.10 kg m ( 392 N (7.70 N m)(100 m) 392 N) 3.98 s. 7.70 N m t = + = E VALUATE : If the weight of the cable and the buoyant force on the cable are neglected, then the tension would have the constant value calculated in part (a). Then 392 N 18.9 m/s 1.10 kg/m F v μ = = = and 5.92 s L t v = = . The weight of the cable increases the tension along the cable and the time is reduced from this value. 15.83. I DENTIFY : The tension in the rope will vary with radius r . S ET U P : The tension at a distance r from the center must supply the force to keep the mass of the rope that is further out than r accelerating inward. The mass of this piece in , L r m L and its center of mass moves in a circle of radius 2 L r + . E XECUTE : 2 2 2 2 ( ) ( ). 2 2 L r L r m T r m L r L L ω ω + = = The speed of propagation as a function of distance is 2 2 ( ) ( ) , 2 dr T r TL v r L r dt m ω μ = = = = where 0 dr dt > has been chosen for a wave traveling from the center to the edge. Separating variables and integrating, the time t is 2 2 0 2 . L dr t dt ω L r = = The integral may be found in a table, or in Appendix B. The integral is done explicitly by letting 2 2 sin , cos , cos , r L θ dr L θ d θ L r L θ = = = so that 2 2 2 arcsin , and arcsin(1) . 2 dr r t L L r π θ ω ω = = = = E VALUATE : An equivalent method for obtaining ( ) T r is to consider the net force on a piece of the rope with length dr and mass . dm dr m L = The tension must vary in such a way that 2 2 ( ) ( ) , or ( ) . dT T r T r dr ω rdm m ω L rdr dr + = − = − This is integrated to obtained 2 2 ( ) ( 2 ) , T r m ω L r C = − + where C is a constant of integration. The tension must vanish at , r L = from which 2 ( 2) C m ω L = and the previous result is obtained. 15.84. I DENTIFY : Carry out the calculation specified in part (a). S ET U P : SW SW cos sin sin cos y y kA kx t, A kx t x t ω ω ω = = − . 1 2 sin cos sin2 θ θ θ = . E XECUTE : The instantaneous power is 2 2 SW SW 1 (sin cos )(sin cos ) sin(2 )sin(2 ). 4 P FA k kx kx t t FA k kx t ω ω ω ω ω = = (b) The average value of P is proportional to the average value of sin(2 ), t ω and the average of the sine function is zero; av 0. P = (c) The graphs are given in Figure 15.84. The waveform is the solid line, and the power is the dashed line. At time 0 t = , 0 y = and 0 P = and the graphs coincide. (d) When the standing wave is at its maximum displacement at all points, all of the energy is potential, and is concentrated at the places where the slope is steepest (the nodes). When the standing wave has zero displacement, all of the energy is kinetic, concentrated where the particles are moving the fastest (the antinodes). Thus, the energy must be transferred from the nodes to the antinodes, and back again, twice in each cycle. Note that P is

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491_PartUniversity Physics Solution - 15-28 Chapter 15(c...

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