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Transverse Waves

  1. Feb 26, 2015 #1
    1. The problem statement, all variables and given/known data
    8. Four wave functions are given below. Rank them in order of the magnitude of the wave
    speeds, from least to greatest.
    I. y(x,t) = 5sin(4x − 20t + 4)
    II. y(x,t) = 5sin(3x −12t + 5)
    III. y(x,t) = 5cos(4x + 24t + 6)
    IV. y(x,t) =14cos(2x − 8t + 3)
    (A) IV, II, I, III
    (B) IV = II, I, III
    (C) III, I, II, IV
    (D) IV, I, II=III
    (E) III, IV, II, I

    2. Relevant equations

    3. The attempt at a solution
    I don't know how to go about this!
  2. jcsd
  3. Feb 26, 2015 #2


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    Have you not learnt any equations related to one dimensional waves?
    You can approach it this way: consider some particular x and t. The equation tells you the value of y at that position and point of time. A short time later, the wave has moved along a bit, so some nearby point has that value of y. Can you see by looking at the equation how to balance a small change in x with a small change in t so that y does not change? What is the ratio of the changes in x and t?
  4. Feb 26, 2015 #3
    Can u explain how the x:t ratio idicates anything about the speed of the wave
  5. Feb 26, 2015 #4


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    I thought that's what I explained.
    Suppose the wave is y=Asin(ax+bt). For some given x, t, consider a nearby position x+dx at time t+dt. If the wave moves distance dx in time dt then y will be the same: y=Asin(ax+bt)=Asin(a(x+dx)+b(t+dt)). If dx and dt are small, that cannot be achieved by moving along a whole number of wavelengths, so it must be that ax+bt=a(x+dx)+b(t+dt). What do you deduce from that?
  6. Feb 28, 2015 #5
    The functions have the following general form if the wave is in the +x direction

    y(x,t) = A sin [k(x − vt) + initial phase]

    where the A is amplitude, k is propagation constant, v is velocity, and t is time.

    In this case, the velocities are 5, 4, - 6, and 4. So

    (B) IV = II, I, III

    is correct.

    Örsan Yüksek
  7. Mar 1, 2015 #6


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    Looks right.
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