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hw16 - PHYS-1200 PHYSICS II HOMEWORK SOLUTIONS CLASS 16...

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PHYS-1200 PHYSICS II SPRING 2005 HOMEWORK SOLUTIONS CLASS 16. 16.30, 16.42, 16.84, 17.41, 17.111; Q17.09, Q17.10 16.30. According to equation 17-39 on page 384 of the textbook, the amplitude of the resultant wave is given by: y m ’ = 2 y m cos ½ φ . If y m ’ = 1.5 y m , we have, 1.5 y m = 2 y m cos ½ φ , or cos ½ φ = 0.75. Then, ½ φ = 41.4°, and φ = 82.8° In radians, φ = 82.8°× π / 180° φ = 1.45 rad Since one wavelength corresponds to a phase shift of 2 π , this difference, in wavelengths is, φ = 1.45/2 π φ = 0.231 wavelengths 16.42. The string will be flat twice each period, so T = 2(0.50 s) = 1.0 s. Then, λ = vT = (10 cm/s)(1.0 s) λ = 10 cm = 0.10 m 16.84. Since the two waves are given as cosines, they can be combined with the identity, cos α + cos β = 2 cos½( α + β )cos½( α β ) Then the formula for the standing wave is, y = y 1 + y 2 = [2 y m cos kx ] cos ϖ t y = [2(0.050) cos π x ] cos 4 π t = [(0.10) cos π x ] cos 4 π t a) For a node the amplitude must be zero. That occurs when cos π x = 0. The smallest positive value of x that gives that result is x = ½, so x = 0.50 m b) The velocity of a particle of the string at x is given by, dt dy u = = [(0.10) cos π x
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