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Trigonometry

Given the equation for simple harmonic method d=1/12 sin(60 pi t) answer the following questions:

b)Find the frequency
i got 60pi
c)Calculate the value of d when t=3


d)Determine the least positive value of t for which d=0.

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Ryan

  1. b) correct, the period would be 2π/60π = 1/30
    c) when t = 3
    d = (1/12)sin (180π) = 0

    d) sin(60πt) = 0 when
    t = 0 , 1/30, 1/15 ...

    so the first positive value is 1/30

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    Reiny

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