# hw16 - PHYS-1200 PHYSICS II SPRING 2006 HOMEWORK SOLUTIONS...

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Unformatted text preview: PHYS-1200 PHYSICS II SPRING 2006 HOMEWORK SOLUTIONS CLASS 16. 16.42, 16.84, 17.41, 17.111; Q17.09, Q17.10 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 ][– 4 π sin 4 π t ] This will be zero at t = 0, ¼, ½ etc., so, the first will occur at: t = 0 c) The second time that u will be zero is: t = 0.25 s d) The third time that u will be zero is: t = 0.50 s 17.41. The well acts the same as a closed pipe, so its fundamental (lowest) frequency will be,...
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hw16 - PHYS-1200 PHYSICS II SPRING 2006 HOMEWORK SOLUTIONS...

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