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Math Help - Displacement of a particle

  1. #1
    srh
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    Displacement of a particle

    Ok I'm guessing this is simple harmonic motion?
    I don't understand how these values relate to the displacement at all, can anyone help?


    The displacement, d, in cm, of a particle is given by
    d = 3 sin(40 π t) 4 cos( 40 π t ) t ≥ 0
    Find expressions for the velocity and acceleration of the particle
    Calculate the velocity and acceleration when t = 2
    Determine when the particle is first at rest
    Calculate the displacement and acceleration when the particle is at rest

    (π = pi)


    Thanks!
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    Re: Displacement of a particle

    Find the derivative to get the velocity, and the second derivative to find the acceleration with respect to time.
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    srh
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    Re: Displacement of a particle

    Are you sure?
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    Re: Displacement of a particle

    Quote Originally Posted by srh View Post
    Are you sure?
    Yes.
    v=d'~\&~a=v'=d''
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    srh
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    Re: Displacement of a particle

    Then when is the particle at rest?
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    Re: Displacement of a particle

    d = 3 sin(40 nt) - 4 cos( 40 nt )

    d^{'} = 120ncos(40nt) + 160nsin(40nt)

    Particle at rest when velocity =0.

    0 = 120ncos(40nt) + 160nsin(40nt)

    -120ncos(40nt) = 160nsin(40nt)

    -\frac{4}{3}tan(40nt) = 1

    tan(40nt) = -\frac{3}{4}

    and then solve for t....
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    srh
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    Re: Displacement of a particle

    Thanks for your time but that doesn't help me in the slightest I'm afraid.

    Can this be explained in a more simplistic way?
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    srh
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    Re: Displacement of a particle

    Also the π symbol is supposed to be the pi symbol
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    Re: Displacement of a particle

    Quote Originally Posted by srh View Post
    Also the π symbol is supposed to be the pi symbol
    Use the reply with quote tab. On the tool bar you see ~~\Sigma~~ that is the LaTeX rap.
    [TEX]\pi [/TEX] gives \pi .
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    Re: Displacement of a particle

    The displacement, d, in cm, of a particle is given by
    d = 3 sin(40 π t) 4 cos( 40 π t ) t ≥ 0
    Find expressions for the velocity and acceleration of the particle
    So we know how to get velocity and acceleration i.e.

    velocity       =  120n cos(40nt) + 160n sin(40nt) --> which is the derivative of displacement
    acceleration =  -4800 n^2 sin(40nt) + 6400n^2 cos(40nt) --> which is the derivative of the velocity.

    Calculate the velocity and acceleration when t = 2
    now we substitute t=2 into the above:

    velocity       =  120(\pi) cos(40\pi(2)) + 160\pi sin(40\pi(2)) = 120\pi
    acceleration =  -4800 \pi^2 sin(40\pi(2)) + 6400\pi^2 cos(40\pi(2))  = 6400\pi^2

    Say the part you don't understand
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    srh
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    Re: Displacement of a particle

    Any of it really :S
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    Re: Displacement of a particle

    Quote Originally Posted by srh View Post
    Thanks for your time but that doesn't help me in the slightest I'm afraid.

    Can this be explained in a more simplistic way?
    For this problem to make any sense to you at all, you have to know some Calculus. Are you taking a Calculus course? If not, what course is this for?
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  14. #14
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    Re: Displacement of a particle

    I'm studying mechanical engineering, I've got only a basic grasp of calculus from what I've been able to teach myself online.
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    Re: Displacement of a particle

    These are equations that describe the motion.

    Let's take a simple example

    displacement = t for t\geq0

    For the 1st second: displacement = 1 cm
    For the 2nd second: displacement = 2 cm and so on....

    One can easily that the velocity is 1cm per second.
    So how do we do this mathematically,
    we use the formula velocity = \frac{change\hspace{2mm} in\hspace{2mm} displacement}{change\hspace{2mm} in \hspace{2mm}time} = derivative of displacement
    from now on I will use s to refer to displacement...

    velocity = \frac{ds}{dt} (t)= 1 cm per second.

    Now to calculate the acceleration:
    You the same as above but now you take the derivative of the velocity, so we get

    acceleration = \frac{dv}{dt} (1)= 0.

    You can verify this by seeing that the velocity is constant.

    Second example

    displacement = t^2 for t\geq0

    For the 1st second: displacement = 1 cm
    For the 2nd second: displacement = 4 cm and so on....

    we use the formula velocity = \frac{change \hspace{2mm}in\hspace{2mm} displacement}{change \hspace{2mm}in \hspace{2mm}time} = derivative of displacement
    from now on I will use s to refer to displacement...

    velocity = \frac{ds}{dt} (t^2)= 2t cm per second.

    so if t=1, v= 2(1) = 2; if t=2, v= 2(2) = 4

    Now to calculate the acceleration:

    acceleration = \frac{dv}{dt} (2t)= 2cm per second squared.

    Now apply the same principles to your equations...
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