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Sound wave questinos my tutor couldnt even figure out

  1. Dec 6, 2004 #1
    sound wave questions my tutor couldnt even figure out

    allright, well, i tried these problems, really thought i knew what i was doing, cause this section has been fairly easy, but i just cant get a couple of them! maybe you guys can help?

    1 A stationary motion detector sends sound waves of 0.300 MHz toward a truck approaching at a speed of 32.0 m/s. The speed of sound in the air is 343 m/s. What is the frequency of the waves reflected back to the detector?
    ______ MHz
    my method was to use the whole "f=fo[(v(+-)vo)/(v(+-)vs)] ordeal, using v=343, fo=.3, vo=0, and vs=32
    so i worked it out like that, got .330868, wrong answer... my tutor's idea was that its not doplar at all, and just to enter the orig f, .3, but that wasnt right either... i have no idea what i did wrong

    2. A plane flies at 1.15 times the speed of sound. Its sonic boom reaches a man on the ground 1.00 min after the plane passes directly overhead. What is the altitude of the plane? Assume the speed of sound to be 330 m/s.
    ______ m
    this one should be easy as well, i did sin(theta)=(1/machnumber) and solving for the angle, then using that, as well as the time (60s) times the speed of sound (330 in this case) to find the hypotenuse of a triangle..... did that, solved for the missing side, got 17217.39m, but thats wrong... any ideas?

    and lastly
    3. A trumpet player on a moving railroad flatcar and a second trumpet player standing alongside the track both play a 440 Hz note. The sound waves heard by a stationary observer between the two players have a beat frequency of 5.0 beats/s. The speed of sound in the air is 343 m/s. What is the flatcar's speed?
    _____ m/s
    for this one, figured id use the whole "fb=f1-f2" to get the freq, so that owuld be 445 Hz. then plug the numbers into the f=fo[(1/(1-(vc/vs))] with f=445Hz, fo=440Hz, vs=343m/s, and solve for vc... got 3.8539, wrong....

    those were all the problems out of about 15 that i couldnt get, and i thought i just didnt grasp the concept, but when my tutor couldnt do them, i figured id ask some other experts for their opinions.... any help would be greatly appreciated
    Last edited: Dec 6, 2004
  2. jcsd
  3. Dec 6, 2004 #2

    Andrew Mason

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    There is a double shift because of the reflection. The original sound is doppler shifted up when received by the approaching truck. When it is reflected back to the detector, it is doppler shifted upward again.

    The sound travels along the hypotenuse, so the time of travel is [itex]\Delta t = L_{hyp}/v_{sound}[/itex]. But that time is not 60 seconds. It is 60 seconds + the time it takes for the plane to fly from the opposite end of the hypotenuse to the point directly overhead ([itex]60 + L_{horiz}/v_{plane}[/itex])

    fb=f1-f2. So the approaching sound wave is doppler shifted to 445 Hz. Use:
    [tex]f = f_0 \frac{v_{sound}}{v_{sound} -v_{source}}[/tex]
    I get a speed of about 4m/sec.

  4. Dec 6, 2004 #3
    well, i got the first 2, a friend helped me out, but the 3rd is still a mystery, you said you got about 4m/sec, which is basically what i got with the 3.85m/s, and i dont think theyd mark it wrong with an exact answer..... so that one is still up for grabs on help. thanks for the help though man!
  5. Dec 6, 2004 #4

    Andrew Mason

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    It doesn't say which direction the car is going so there are two potential answers which are slightly different (the doppler shifted frequency would be 435 Hz) and differ by a - sign. So maybe they are looking for both answers.

  6. Dec 6, 2004 #5
    ha, thats it, they wanted the negative answer, just make it postive..... its negative cause the train is going away! hahaha, thanks man for the help!
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