P18_100 - 100. (a) Since the speed of sound is lower in air...

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100. (a) Since the speed of sound is lower in air than in water, the speed of sound in the air-water mixture is lower than in pure water (see Table 18-1). Frequency is proportional to the speed of sound (see Eq. 18-39 and Eq. 18-41), so the decrease in speed is “heard” due to the accompanying decrease in frequency. (b) This follows from Eq. 18-3 and Eq. 18-2 (with ∆’s replaced by derivatives). Thus, 1 v 2 = ρ B = ρ V ¯ ¯ ¯ dp dV ¯ ¯ ¯ = ρ V ¯ ¯ ¯ ¯ dV dp ¯ ¯ ¯ ¯ . (c) Returning to the ∆ notation, and letting the absolute values be “understood,” we write ∆ V = V w +∆ V a as indicated in the problem. Subject to the approximations mentioned in the problem, our equation becomes 1 v 2 = ρ w V w µ V w p + V a p = ρ w V w V w p + ρ w ρ a V a V w µ ρ a V a V a p . In a pure water system or a pure air system, we would have 1 v 2 w = ρ w V w V w p or 1 v 2 a = ρ a V a V a p . Substituting these into the above equation, and using the notation r = V a /V w
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.

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