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Math Help - SHM

  1. #1
    Junior Member
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    Jan 2009
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    68

    SHM

    given that x=120 sin (pi/12)t , what is the first time when the particle is (i) 30m to the right of the origin and (ii) 30m to the left of the origin?

    Im thinking

    30=120 sin (pi/12)t

    1/4=sin (pi/12)t

    t=(1/4) / sin (pi/12)

    t=0.96592..... is this correct?

    I don't know how to think about the second part when x=-30. Is it more complicated or do I just solve it as above?

    Thank you
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  2. #2
    Banned
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    Aug 2008
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    Quote Originally Posted by slaypullingcat View Post
    given that x=120 sin (pi/12)t , what is the first time when the particle is (i) 30m to the right of the origin and (ii) 30m to the left of the origin?

    Im thinking

    30=120 sin (pi/12)t

    1/4=sin (pi/12)t

    t=(1/4) / sin (pi/12)

    t=0.96592..... is this correct?

    I don't know how to think about the second part when x=-30. Is it more complicated or do I just solve it as above?

    Thank you
    No, it is wrong.

    30 = 120 \sin \left(\frac{\pi}{12}t\right)

    \frac{1}{4}=\sin \left(\frac{\pi}{12}t\right)

    0.25268 = \frac{\pi}{12}t

    t=0.25268 \times\frac{12}{\pi}

    t=0.9652
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