1. During a training exercise a fastattack submarine making a speed of 10 knots

Question

1.      During a training exercise a fastattack submarine making a speed of 10 knots approaches a missilesubmarine which is stationary (0 knot speed). The fast attacksubmarine then transmits a sonar pulse of 7.0 kHz. Uponhearing this pulse the missile submarine opens a torpedo tube doorand increases speed to 10 knots in the direction of the fast attacksubmarine. Upon hearing the torpedo tube door go open thefast attack submarine reverses course and increases speed to 20knots. After this maneuver the fast attack submarinetransmits a second pulse at the same frequency. At whatfrequencies does the missile submarine receive the first and secondsonar pulses? Assume the speed of sound in water is 1500m/sand that 1 knot is 2.0 km/h.

Explanation / Answer

There seems to be multiple postings of this problem. I alwayshave to look up Doppler questions because they are confusing. The general formula is f = ((v ± vr)/(v ±vs))*f0; where v is the velocity of thesignal in the medium, vr is the velocity of the receiverWRT the medium, vs is the velocity of the source WRT themedium, and f0 is the emitted frequency of thesource. We now have to decide how to assign the plus-minussigns (if the source and/or receiver are moving towards each other,then the frequency should increase; and the frequency decreases ifthey are moving away from each other): 1. f = (v/(v - vs))*f0; source movingtowards the receiver 2. f = (v/(v + vs))*f0; source movingaway from the receiver 3. f = ((v + vr)/v)*f0; receiver movingtowards the source 4. f = ((v - vr)/v)*f0; receiver moving awayfrom the source As a check, we notice that when the source and receiver are movingin the same direction at the same speeds (1,4 or 2,3) then we getno change in the perceived frequency (the velocity ratios reduce toone over one). We also need to convert the knot conversion into something moreuseful: 2.0 km/hr/knot = 2000/3600 m/s/knot = 5/9 m/s/knot. In the first case we can use equation 1: f = ((1500 m/s)/(1500 m/s- 50/9 m/s))*7.0 kHz. In the second case, we must combine equations 2 and 3: f = ((1500m/s + 50/9 m/s)/(1500 m/s + 100/9 m/s)*7.0kHz.

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