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Verify by direct substitution that the wave function for a standing wave given in Equation 17.1,
is a solution of the general linear wave equation, Equation 16.27:
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Chapter 17 Solutions
PHYSICS:F/SCI.+ENGRS.,V.1
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- An E&M wave has an e-field given by E(x,t) - -(4.950V/m)k cos(ky + (1.38 x 101 rud/)t) Which of the following is correct for B(x, t) and the direction of propagation? (c - 3x10 m/s) B(7, t) - -(0.165 NA)i cos ((4.6 x 10° rad/m)y + (1.38 x 1015 rad/)e),y B(x, t) - (33.0 nT)j cos (2.3 x 10*rad/m)y + (2.78 x 101 rad/s)e). - d/m)y + (1.38 x 10 rad 1. I. rad/).-y B(x, t) = (11.5 aT)t cos ((4.6 x 10 rad/mly + (1.38 x 10 rad (1.38 x 101 rad, /,). -y. B(x, t) - (16.5 nT)t cos ((4.6 x 10* rad II. IV. IV None of the abovearrow_forwardQuestion 19 A string hanging from a shelf supports aarrow_forwardConsider the one-dimensional wave equation Pu u 0<<1, t20 u(0, t) = u(1, t) = 0 u(x, 0) = 3 sin(27x) du (x, 0) = x – x². Use CTCS scheme with Ar = 0.1, At = 0,05 to estimate the value of the displacement u(0.9,0.15).arrow_forward
- Example 14-8 depicts the following scenario. Two people relaxing on a deck listen to a songbird sing. One person, only 1.66 m from the bird, hears the sound with an intensity of 2.86×10−6 W/m^2. A bird-watcher is hoping to add the white-throated sparrow to her "life list" of species. How far could she be from the bird described in example 14-8 and still be able to hear it? Assume no reflections or absorption of the sparrow's sound.arrow_forwardGiven the wave functions y1 (x, t) = A sin (kx − ωt) and y2 (x, t) = A sin (kx − ωt + ϕ) with ϕ ≠ π/2 , show that y1 (x, t) + y2 (x, t) is a solution to the linear wave equation with a wave velocity of v = √(ω/k).arrow_forwardA traveling wave is described by y = 10 sin (βz – ωt). Sketch the wave at t = 0 and at t = t1, when ithas advanced λ /8, if the velocity is 3 X 108 m/s and the angular frequency ω = 1 X 106 rad/s. Repeat for ω = 2 X 106 rad/s and the same t1.arrow_forward
- At high frequencies, the value of the propagation constant (y) is approximately :* Y = (R + j@ L ) ( G + j w C ) Y = a Option 4 Option 3 Y = a +jß Y = j ß Option 1 Option 2arrow_forwardO F= 9f A traveling wave on a taut string with a tension force T is given by the wave function: y(x,t) = 0.1sin(2tx-300t), where x and y are in meters and t is in %3D seconds. The linear mass density of the string is u = 100 g/m, and the string is 5 m-long. The total energy on the string is O 90 J O 225 J O 450 J 45 J O 22.5 J -rhy thewevearrow_forwardLet us assume we are using SI units (kg, m, s). Consider the harmonic travelling wave 17 y(x, t) = = 12 cos(3x - 12πt). Assume the above harmonic wave is a solution of the motion of a string with tension T = 1 Newton. What is the string's density?p= kg/m What is its kinetic energy density? What is its kinetic energy over one wavelength? Ek What is its potential energy density? What is its potential energy over one wavelength? Ey = J/m J/marrow_forward
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