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501_Physics ProblemsTechnical Physics

501_Physics ProblemsTechnical Physics - 503 Chapter 17...

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Chapter 17 503 P17.16 (a) The sound “pressure” is extra tensile stress for one-half of each cycle. When it becomes 0500% 130 10 650 10 10 8 .. . af ej ×= × Pa Pa, the rod will break. Then, Pv s max max = ρω s P v max max . . . == × × = π 8 92 10 5 010 2 500 463 8 3 Nm kg m m s s mm 2 3 bg b g . (b) From ss k x t =− max cos ω v s t sk x t vs = = ωω ωπ max max max sin a f 2 500 145 s m m m s (c) Iv s v v = × 1 2 1 2 1 2 892 10 5010 2 23 2 ρ max max b g kg m m s m s 3 473 10 9 . W m 2 *P17.17 Let Px represent absolute pressure as a function of x . The net force to the right on the chunk of air is +− + PxA Px xA a f . Atmospheric pressure subtracts out, leaving −+ + = ∂∆ ∆∆ x Px A P x xA a f . The mass of the air is mV A x ρρ and its acceleration is 2 2 s t . So Newton’s second law becomes PxA + FIG. P17.17 ∂∆ = F H G I K J = = P x xA A x s t x B s x s t Bs x s t 2 2 2 2 2 2 2 2 Into this wave equation as a trial solution we substitute the wave function
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