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16 December, 10:01

Two particles oscillate in simple harmonic motion along a common straight-line segment of length

a. each particle has a period of 1.5 s, but they differ in phase by pi/6 rad. (a) how far apart are they (in terms of

a. 0.50 s after the lagging particle leaves one end of the path

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Answers (1)
  1. 16 December, 12:28
    0
    Assuming the two functions are sine functions.

    From the problem:

    T = 1.5

    So

    b = 2π/T

    b = 2π/1.5

    b = 4π/3

    The first sine function has an equation

    y1 = sin 4π/3x

    and since they have a difference in phase of π/6, the other sine function has an equation

    y2 = sin 4π/3x + π/6

    Their difference is simply π/6.

    If we substitute 0.5 s to each equation,

    y1 = 2π/3

    y2 = 5π/6
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