501_Physics ProblemsTechnical Physics

501_Physics ProblemsTechnical Physics - 503 Chapter 17...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

Ask a homework question - tutors are online