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  1. #1
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    Maths help

    Hi,

    Please can you help me with this.

    An oscillating mechanism has a maximum displacement of 3.2 m and a frequency of 50 Hz. At time t = 0 the displacement is 150 cm. Express the displacement in the general form A sin (ωt ± φ).

    Newton’s law of cooling is given by θ = θo e-kt , where the excess of temperature at zero time is θo oC and at time t seconds is θ oC . Given that θo = 15 oC and
    k = - 0.02, find the time when the rate of change of temperature is 1 oC.


    Thanks for your help.
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  2. #2
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    Quote Originally Posted by mathsisfun
    Hi,
    Newton’s law of cooling is given by θ = θo e-kt ,
    If by that you mean,
    \theta=\theta_0e^{-kt}
    Then you are wrong.
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ThePerfectHacker
    If by that you mean,
    \theta=\theta_0e^{-kt}
    Then you are wrong.
    Actually, it is correct if the temperature of the surrounding environment is 0 C.

    Regardless, let \theta = \theta _0 e^{-kt}

    Thus \frac{d \theta}{dt}=\theta _0 (-k) e^{-kt}

    Solving for t we get:
    t = -\frac{1}{k} ln \left ( -\frac{1}{k \theta _0}\frac{d \theta}{dt} \right )

    Thus t = 60.1986 s.

    -Dan
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  4. #4
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    thankyou topsquark.

    For the other one above iam stuck on, I have attached a picture as it didnt come out well.

    Cheers
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by mathsisfun
    thankyou topsquark.

    For the other one above iam stuck on, I have attached a picture as it didnt come out well.

    Cheers
    Sorry it's taken me so long to get to this!

    (Warning! Make sure your calculator is in "radian" mode!)

    Let's do this. Since you have x(t)=150 cm, let's convert the maximum displacement to cm as well. So A = 3.2 m = 320 cm.

    We know that the frequency is f = 50 Hz. We need to convert this to an angular frequency: \omega = 2 \pi f is about 314 rad/s. (This unit is occasionally also given the lable "Hz." Be warned!)

    To find the phase angle ( \phi) we can now plug in some values: (dropping the units for convenience)
    x(t)=A \, sin( \omega t + \phi)

    150 = 320 \, sin( 314*0 + \phi) = 320 \, sin(\phi)

    sin( \phi ) = \frac{150}{320} = 0.46875

    One solution is \phi \approx 0.488 rad.

    So  x(t) = 320 \, sin( 314t+0.488) cm.

    -Dan
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  6. #6
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    many thanks again Dan.

    Cheers
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