A mass m=4kg is attached to both a spring with spring constant k=577N/m and a dash-pot with damping constant c=4N⋅s/m . The mass is started

Question

A mass m=4kg is attached to both a spring with spring constant k=577N/m and a dash-pot with damping constant c=4N⋅s/m . The mass is started in motion with initial position x0=5m and initial velocity v0=7m/s . Determine the position function x(t) in meters.

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Philomena 3 years 2021-08-25T03:51:04+00:00 1 Answers 17 views 0

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  1. ngochoa
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    2021-08-25T03:52:12+00:00
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    Answer:

    x(t) = (5.034\,m)\cdot e^{-0.5\cdot t}\cdot \cos ((12\,\frac{rad}{s} )\cdot t + 0.116\,rad)

    Explanation:

    The position and velocity functions have the following forms:

    x(t) = A \cdot e^{-\frac{c}{2\cdot m}\cdot t }\cdot \cos (\omega'\cdot t + \phi)

    v(t) = -\omega' \cdot A \cdot e^{-\frac{c}{2\cdot m}\cdot t }\cdot \sin (\omega'\cdot t + \phi)

    Where:

    \omega' = \sqrt{\frac{k}{m} - \frac{c^{2}}{4\cdot m^{2}} }

    First, the angular frequency of oscillation is calculated:

    \omega' = \sqrt{\frac{577\,\frac{N}{m}}{4\,kg}-\frac{(4\,\frac{N\cdot s}{m} )^{2}}{4\cdot (4\,kg)^{2}} }

    \omega'\approx 12\,\frac{rad}{s}

    Later, it is needed to determine if system is underdamped, critically damped or overdamped. Critic value is given by 2\cdot \sqrt{k\cdot m}. The value is:

    \alpha = 2\cdot \sqrt{(577\,\frac{N}{m} )\cdot (4\,kg)}

    \alpha \approx 96.083

    As c is lower than α, the system has an underdamped behavior. The initial values for position and velocity are, respectively:

    x(0) = 5\,m

    v(0) = 7\,\frac{m}{s}

    Then,

    x(0) = A\cdot \cos \phi\\v(0) = - \omega'\cdot A \cdot \sin \phi

    By making some algebraic handling:

    \frac{v(0)}{x(0)} = -\omega' \tan \phi

    \tan \phi = - \frac{1}{\omega'}\cdot \frac{v(0)}{x(0)}

    \phi = \tan^{-1}\left[- \frac{1}{\omega'}\cdot \frac{v(0)}{x(0)}\right]

    \phi = \tan^{-1} \left[-\frac{1}{12\,\frac{rad}{s} }\cdot \frac{7\,\frac{m}{s} }{5\,m} \right]

    \phi \approx 0.116\,rad

    The amplitude is:

    A = \frac{x(0)}{\cos \phi}

    A = \frac{5\,m}{\cos (0.116\,rad)}

    A \approx 5.034\,m

    The position function is:

    x(t) = (5.034\,m)\cdot e^{-0.5\cdot t}\cdot \cos ((12\,\frac{rad}{s} )\cdot t + 0.116\,rad)

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