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Waves: Standing Waves, Superposition, etc.

  1. Oct 8, 2005 #1
    1.) S and P waves, simultaneously radiated from the hypocenter of an earthquake, are received at a seismographic station 17.3 s apart. Assume the waves have traveled over the same path at speeds of 4.50 km/s and 7.80 km/s. Find the distance from the seismograph to the hypocenter of the quake.

    The text doesn't thoroughly explain how to do these types of questions. So, I can't even explain what I did, since i have no beginning point.

    2.) Two waves in one string are described by

    y1 = 3cos(4x - 1.6t)
    y2 = 4sin(5x - 2t)

    Find superposition.

    The formula I have for the addition of two wave functions requires both functions to be sin and also for the amplitudes to be equal. So, I made y1 = 3sin(4x - 1.6 - pi/2). However, how do i deal with the uncommon bases?

    3.) Two pulses travel on string are described as functions:

    y1 = 5 / ( (3x - 4t)^2 + 2)

    y2 = -5 / ( (3x + 4t - 6t^2) + 2)

    When will the two waves cancel everywhere? At what point to the two pulses always cancel?

    I noticed that that y2 is reflected on the x-axis, and has a phase constant. Also, it is moving to the left, while y1 moves to the right.

    Firstly, what are the conditions under which the waves will cancel everywhere? For two waves to cancel everywhere, would they not have to have equal phase constants (which these do not?

    Also, what are the conditions for the waves to cancel? I propose that the functions be added, and the values be found which would cause y to equal = 0. However, there is both x and t variables!

    Thank you for any help you can provide on this urgen issue.
  2. jcsd
  3. Oct 8, 2005 #2


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    Homework Helper

    Not much of a physics question, but rather a math teaser. You know that distance = velocity x time. Now you don't know distance, but you know they are equal:

    d1 = v1 x t1, v1 = 4.5km/s (1)
    d1 = v2 x t2, v2 = 7.8km/s (2)

    You also know that (t2-t1) = 17.3 (3)
    Three unknown variables (d1, t1, t2), three equations as listed. Solved.

    By definition, I believe the superposition is simply the addition of the two functions y1 + y2. They are of different frequency and will not interfere with each others signal. Or better stated, the field caused by a number of sources is determined by adding the individual fields together.

    For the pulses to cancel everywhere, then you must equate
    y1 + y2 = 0, for all x. Solve for t. I can see the math getting pretty ugly however.

    The second part is the opposite of the first. At what point/points do the two pulses always cancel. The condition here is then.
    y1 + y2 = 0, for all t. Solve for x. Again, may get ugly mathematically ^^;;
  4. Oct 8, 2005 #3
    I think this is #5 is my homework, but you're from somewhere else. Interesting. Anyways, it took me a while to figure it out, here's the hints my teacher gave me.
    x=vt. the two waves have the same distance, but different times and velocities. set the equation to solve for the known difference in time and you should get it from there.

    I hope that wasn't too much info.
  5. Oct 15, 2005 #4
    to: mezarashi

    I tried this, and when I solved for x I ended up with x = 1.

    I ended up with

    x = 48t - 36 / 48t - 36 = 1

    However, how do I solve for t?
    Last edited by a moderator: Oct 15, 2005
  6. Feb 19, 2006 #5
    well that means for any value of t other than t=36/48 sec the above equation will be satisfied
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