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Superposition waves

  1. Apr 1, 2004 #1
    hi, i was able to get all of my homework problems except this one, i just ccan not seem to figure it out.

    Two identical sinusoidal waves with wavelengths of 6.00 m travel in the same direction at a speed of 2.00 m/s. The second wave originates from the same point as the first, but at a later time. Determine the minimum possible time interval between the starting moments of the two waves if the amplitude of the resultant wave is the same as that of each of the two initial waves.

    any help with this problem would be great.

  2. jcsd
  3. Apr 4, 2004 #2
    If the sum of the waves is to have the same amplitude as each individual wave, then its coefficient must be 1. That means that it is a shifted sine wave. For convenience, let the first wave be positioned such that it's mathematical representation is sin(x). Then the second wave is simply shifted in phase.

    We essentially have two variables running around. We have the phase of the second original wave and we have the phase of the resultant wave. Let a = the phase of the second original wave and let r be the phase of the resultant wave.

    Let us make an expression for the sum of the two original waves.

    sin(x) + sin(x + a)

    using some trig formulas

    sin(x) + [sin(x)cos(a) + cos(x)sin(a)]
    [1 + cos(a)]sin(x) + cos(x)sin(a)

    now, in order for this wave to have the same amplitude, it must be able to be represented as a shifted sine wave of coefficient 1, i.e.

    sin(x + r)


    sin(x)cos(r) + cos(x)sin(r)

    We want this to be equal to the sum of the two original waves, so

    sin(x)cos(r) + cos(x)sin(r) = [1 + cos(a)]sin(x) + cos(x)sin(a)

    this equation is only true if the coefficients on both sides of the equation of sin(x) and cos(x) are equal, so this yields the system of equations

    cos(r) = 1 + cos(a)
    sin(r) = sin(a)

    which must be solved in order to get the answer.

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