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Wave and Interference (can you help me?)

  1. Mar 23, 2007 #1
    I am in a terrible headache doing these problems...can someone please help me? I will post my thoughts and the calculations that I have done...

    1) A 50-cm-long wire with a mass of 1.0 g and a tension of 440 N passes across the open top of a vertical tube partially filled with water. The wire, which is fixed at both ends, is bowed at the center so as to vibrate at its fundamental frequency and generate a sound wave. The water level in the tube is slowly lowered until the sound wave from the wire sets up a standing wave in the tube. It is then lowered another 36.0 cm until the next standing wave is detected. Use this information to determine the speed of sound in air.

    I got that the linear density = 1x10^-3/0.5 = 2x10^-3 kg/m
    And so the speed on the string is sqrt(tension/linear density) = 469.04m/s
    Now how can I proceed? I looked through all examples in the textbook, but I still can't figure out what to do...

    2) Two radio antennas are 100 m apart along a north-south line. They broadcast identical radio waves at a frequency of 3.0 MHz. Your job is to monitor the signal strength with a handheld receiver. To get to your first measuring point, you walk 800 m east from the midpoint between the antennas, then 600 m north. What is the phase difference between the waves at this point?
    I found that the path difference is 79.964 m, and lambda = speed/frequency = 3x10^8/3x10^6 = 100m
    Now what can I do? Can someone give me a hint?

    Any help is appreciated! Thanks!
  2. jcsd
  3. Mar 23, 2007 #2
    1) Can someone kindly show me how I can get the fundmental frequency and wavelength? (in which I can plug into the equation v=f(lambda) ?

    I still don't get it.....help......:frown:
  4. Mar 24, 2007 #3
    you should find an eqn, for the string fundamental frequency:

    f0=1/2L*sqrt(T/mu) Now for the speed of sound, I believe using the info from the open ended pipe, where lambda1 and lambda2 are related both by delta X of 36 cm and by a frequency ratio: hint look up resonant modes for a pipe closed on one end, you can solve for c, speed of sound. Hope that helps with first problem.

    for the second, the phase difference is the residual path length difference (the modolo) after dividing by the wavelength, normalized to degrees or radians. In other words if the difference were 3 and a half lambda, you're only worried about the 1/2.
  5. Mar 25, 2007 #4
    1) How can I find the fundamental frequency and the wavelength?
    If I have both, I can plug them into the formula v=(frquency)(wavelength) which gives me the answer. I don't quite understand the situation pictorially...I am not sure what's happening and I don't know how to use the "36.0 cm" data...

    2) How can I know the initial phase difference (specifying the initial conditions?)

    Also, there is a related question: "If you now begin to walk farther north, does the signal strength increase, decrease, or stay the same?"
    The correct answer is "increase", but I don't understand why...can somebody explain this?

    Thanks a lot!
  6. Mar 25, 2007 #5
    I'll come back to 1 in a second, the path length difference I get is:
    sqrt{(800^2+650^2)-(800^2+550^2)} Is this how you calculated it?

    From there, you need to figure out lambda for 3Mhz radiowave.(which you have) Then look for that remainder.

    For the first problem, I'm checking my answer, but believe the answer can be arrived at using an expression such as nC/4L=(n+2)C/4(L+.36) where c = speed of sound. Work on the other one in the meantime.

    Post script/edit: looking at the standing waves a closed/open pipe will support, shows a difference in 1/2 lambda between successive resonance modes, ie delta X=1/2 lambda.

    The frequency will be the same in all such instances, that of the guitar string, which you basically have all but computed above. Than its simply applying the c=F*lambda relation. As to your comment walking North must reduce the phase difference, in other words adding a small delta y to the 650 and 550 terms before squaring and subtracting results in a smaller phase difference. Remember at 180 degrees phase diff, the two cancel, at 0 they reinforce completely so the closer to zero the greater the amplitude.
    Last edited: Mar 25, 2007
  7. Mar 25, 2007 #6
    2) Yes, I got 1.2 rad for the first part, but as you move North, is this phase difference going to increase?
    When it increases to 2 rad, it's maximum constructive interference. But what happens in between?
  8. Mar 25, 2007 #7
    well lets see if we can figure that out assuming x is constant at 800 and y's are 550 and 650,
    the path length=1030-970=60/100 *2pi radians,

    walk north another 50 yards, its 63/100*2*pi radians.

    In other words since the first value was greater than .5 lambda(past 180 degrees phase difference and less than 360), walking north will yield a stronger signal, at least til sqrt(800^2+(y+100)^2)-sqrt(800^2+y^2)=100.
    Last edited: Mar 26, 2007
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