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.

Answer 1

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:

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