Transcript
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 ρ ρ ρ dV = = = . dp v2 B V dp V dV (c) Returning to the ∆ notation, and letting the absolute values be “understood,” we write ∆V = ∆Vw + ∆Va as indicated in the problem. Subject to the approximations mentioned in the problem, our equation becomes 1 ρw ∆Vw ∆Va ρw ∆Vw ρw Va ρa ∆Va = + = + . v2 Vw ∆p ∆p Vw ∆p ρa Vw Va ∆p In a pure water system or a pure air system, we would have ρw ∆Vw 1 = 2 vw Vw ∆p
or
ρa ∆Va 1 = . va2 Va ∆p
Substituting these into the above equation, and using the notation r = Va /Vw , we arrive at 1 1 ρw r = 2 + 2 v vw ρa va2
1 =⇒ v = . 2 1/vw + r(ρw /ρa )/va2
(d) Dividing our result in the previous part by vw and using the fact that the wave speed is proportional to the frequency, we find v fshift 1 1 = = = 2 2 vw f vw 1/vw + r(ρw /ρa )/va 1 + r(ρw /ρa )(vw /va )2 which becomes the expression shown in the problem when we plug in ρw = 1000 kg/m3 , ρa = 1.21 kg/m3 , vw = 1482 m/s and va = 343 m/s, and round to three significant figures. (e) The graph of fshift /f versus r is shown below. 1
0.8
frequency_ratio
0.6
0.4
0.2 0
0.001
0.002 volume_ratio
0.003
0.004
(f) From the graph (or more accurately by solving the equation itself) we find r = 5.2×10−4 corresponds to fshift /f = 1/3.
