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  1. #1
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    Further Maths

    An oscillating mechanism has a maximum displacement of 3.2 meters and a frequency of 50HZ. At time T=0 the displacement is 150cm. express the displacement in the general form A sin (Wt +or-theta).
    (W= omega)

    Can anyone help me out please?
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  2. #2
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    Quote Originally Posted by batman121 View Post
    An oscillating mechanism has a maximum displacement of 3.2 meters and a frequency of 50HZ. At time T=0 the displacement is 150cm. express the displacement in the general form A sin (Wt +or-theta).
    (W= omega)

    Can anyone help me out please?
    x = A \sin (\omega t + \theta).

    "maximum displacement of 3.2 meters": A = 3.2.

    "frequency of 50HZ: Period = \frac{2 \pi}{\omega} \Rightarrow f = \frac{\omega}{2 \pi} = 50 \Rightarrow \omega = 100 \pi.

    "At time T=0 the displacement is 150cm": Solve 1.5 = 3.2 \sin (\theta) for \theta. A decimal approximation will be required.
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  3. #3
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    1.5 = 3.2 \sin \theta
    so 1.5/sin \theta = 3.2

    sin \theta = 3.2/1.5
    sin \theta = 0.46875
    sin^-1  (0.46875) = \theta = 27.953

    That looks about right...
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  4. #4
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    Quote Originally Posted by batman121 View Post
    1.5 = 3.2 \sin \theta
    so 1.5/sin \theta = 3.2

    sin \theta = 3.2/1.5 Mr F says: 1.5/3.2

    sin \theta = 0.46875
    sin^-1  (0.46875) = \theta = 27.953

    That looks about right...
    The problem with your answer is that it's in degrees. It needs to be in radians.
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  5. #5
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    27.953 to radians = 27.953/57.2957= 0.48787r
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