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Math Help - trignometry question

  1. #1
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    trignometry question

    Can someone please help me with this question..
    I attempted it and got the correct ans for the first 3 parts but for the last part my ans was wrong, and I couldn't figure out why

    One end of a piece of elastic is attatched to a point at the top of a door frame and the other end hangs freely. A small ball is attatched to the free end of the elastic.When the ball is hanging freely it is pulled down a distance and then released, so that the ball oscillated up and down on the elastic.The depth d centimetres of the ball from the top of the door frame after t seconds is given by:

    d= 100+10cos 500t degrees

    Find:
    a) the greatest and least depths of the ball
    b)the time at which the ball first reaches its highest position
    c)the time taken for a complete oscillation
    d)the proportion of the time during a complete oscillation for which the depth of the ball is less than 99 cm.

    I got for
    a) 110 cm and 90cm
    b) 0.36s
    c)0.72s

    d)
    99-100=10cos500t
    500t=95.74
    500t=264.3
    t=0.191s
    t=0.528
    the proportion of time where the depth is less than 99 = 0.528-0.191 =0.338

    But correct ans for this part is 0.00213s

    Any help would be appreciated..
    Last edited by puresoul; February 22nd 2013 at 11:56 PM.
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  2. #2
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    Re: trignometry question

    Hey puresoul.

    For d) I got

    99-100=10cos(500t) which means

    -1/10 = cos(500t) which means

    > acos(-1/10)
    [1] 1.670964 implies that

    500t = 1.670964 (approx) which means t is given by
    > acos(-1/10)/500
    [1] 0.003341927

    Checking the answer gives us:

    > x = acos(-1/10)/500
    > 100 + 10*cos(500*x)
    [1] 99

    which is correct.

    Now you try and obtain the next value of t where this is equal and get the length of the appropriate interval.
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  3. #3
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    Re: trignometry question

    Quote Originally Posted by chiro View Post
    Hey puresoul.

    For d) I got

    99-100=10cos(500t) which means

    -1/10 = cos(500t) which means

    > acos(-1/10)
    [1] 1.670964 implies that

    500t = 1.670964 (approx) which means t is given by
    > acos(-1/10)/500
    [1] 0.003341927

    Checking the answer gives us:

    > x = acos(-1/10)/500
    > 100 + 10*cos(500*x)
    [1] 99

    which is correct.

    Now you try and obtain the next value of t where this is equal and get the length of the appropriate interval.
    they gave us the equation d= 100+10cos 500t degrees
    in degrees not radians so i found the value of acos(-1/10)
    in degrees which is 95.74 degrees or 1.670964radians
    Won't we get a different ans if we used radians??



    Continuing what you did,the other value of t will be..

    pi-1.670964 = 1.4706287

    500t=1.4706287 so t=0.0029413 s

    The interval will be (0.003341927-0.0029413= 0.00040066969 s )


    (((And sorry the correct ans that was given in the book is 0.468 not 0.00213s , I copied the ans of another question >< )))
    Its very near the ans that I got (0.338) but still not the same,
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  4. #4
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    Re: trignometry question

    Using R to validate everything:

    > 100+10*cos(500*(acos(-1/10)/500))
    [1] 99
    > 100+10*cos(500*(2*pi - acos(-1/10)/500))
    [1] 99
    > (2*pi - acos(-1/10))/500
    [1] 0.009224443
    > acos(-1/10)/500
    [1] 0.003341927
    > a = (2*pi - acos(-1/10))/500
    > b = acos(-1/10)/500
    > a-b
    [1] 0.005882516
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  5. #5
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    Re: trignometry question

    Quote Originally Posted by chiro View Post
    Using R to validate everything:

    > 100+10*cos(500*(acos(-1/10)/500))
    [1] 99
    > 100+10*cos(500*(2*pi - acos(-1/10)/500))
    [1] 99
    > (2*pi - acos(-1/10))/500
    [1] 0.009224443
    > acos(-1/10)/500
    [1] 0.003341927
    > a = (2*pi - acos(-1/10))/500
    > b = acos(-1/10)/500
    > a-b
    [1] 0.005882516
    I understand what u ar doing, that's what I did but using degrees..
    I still don't understand why u used radians in this situation...
    And the ans. Is 0.468 not 0.00588
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  6. #6
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    Re: trignometry question

    You need to remember that you are looking for a proportion.

    I gave you the time that it was below, now find the proportion of time with respect to the time it takes for a complete oscillation.
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