533_Physics ProblemsTechnical Physics

533_Physics ProblemsTechnical Physics - 535 Chapter 18...

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Chapter 18 535 P18.28 λ G G v f == 20350 . m af ; AA A L v f 2 LLL f f LL f f GAG G A GG G A −=− F H G I K J =− F H G I K J F H G I K J = 10 3 5 0 1 392 440 00382 .. m m a f Thus, AG = = 0 038 2 0 350 0 038 2 0 312 m m , or the finger should be placed 31 2 cm f r om th e b r id g e. L v ff T A 2 1 2 µ ; dL dT fT A A = 4 ; dL L dT T A A = 1 2 dT T dL L A A = 22 0600 382 384% . . . cm 35.0 cm a f P18.29 In the fundamental mode, the string above the rod has only two nodes, at A and B, with an anti-node halfway between A and B. Thus, θ 2 AB L cos or = 2 L cos . Since the fundamental frequency is f , the wave speed in this segment of string is vf Lf 2 cos . Also, v TT mAB TL m = µθ cos where T is the tension in this part of the string. Thus, 2 Lf TL m cos cos θθ = or 4 2 Lf TL m cos cos = and the mass of string above the rod is: m T Lf = cos 4 2 [Equation 1] L M A B T F M g FIG. P18.29 Now, consider the tension in the string. The light rod would rotate about point P if the string exerted
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