Answer:
[tex]x(t) = (5.034\,m)\cdot e^{-0.5\cdot t}\cdot \cos ((12\,\frac{rad}{s} )\cdot t + 0.116\,rad)[/tex]
Explanation:
The position and velocity functions have the following forms:
[tex]x(t) = A \cdot e^{-\frac{c}{2\cdot m}\cdot t }\cdot \cos (\omega'\cdot t + \phi)[/tex]
[tex]v(t) = -\omega' \cdot A \cdot e^{-\frac{c}{2\cdot m}\cdot t }\cdot \sin (\omega'\cdot t + \phi)[/tex]
Where:
[tex]\omega' = \sqrt{\frac{k}{m} - \frac{c^{2}}{4\cdot m^{2}} }[/tex]
First, the angular frequency of oscillation is calculated:
[tex]\omega' = \sqrt{\frac{577\,\frac{N}{m}}{4\,kg}-\frac{(4\,\frac{N\cdot s}{m} )^{2}}{4\cdot (4\,kg)^{2}} }[/tex]
[tex]\omega'\approx 12\,\frac{rad}{s}[/tex]
Later, it is needed to determine if system is underdamped, critically damped or overdamped. Critic value is given by [tex]2\cdot \sqrt{k\cdot m}[/tex]. The value is:
[tex]\alpha = 2\cdot \sqrt{(577\,\frac{N}{m} )\cdot (4\,kg)}[/tex]
[tex]\alpha \approx 96.083[/tex]
As c is lower than α, the system has an underdamped behavior. The initial values for position and velocity are, respectively:
[tex]x(0) = 5\,m[/tex]
[tex]v(0) = 7\,\frac{m}{s}[/tex]
Then,
[tex]x(0) = A\cdot \cos \phi\\v(0) = - \omega'\cdot A \cdot \sin \phi[/tex]
By making some algebraic handling:
[tex]\frac{v(0)}{x(0)} = -\omega' \tan \phi[/tex]
[tex]\tan \phi = - \frac{1}{\omega'}\cdot \frac{v(0)}{x(0)}[/tex]
[tex]\phi = \tan^{-1}\left[- \frac{1}{\omega'}\cdot \frac{v(0)}{x(0)}\right][/tex]
[tex]\phi = \tan^{-1} \left[-\frac{1}{12\,\frac{rad}{s} }\cdot \frac{7\,\frac{m}{s} }{5\,m} \right][/tex]
[tex]\phi \approx 0.116\,rad[/tex]
The amplitude is:
[tex]A = \frac{x(0)}{\cos \phi}[/tex]
[tex]A = \frac{5\,m}{\cos (0.116\,rad)}[/tex]
[tex]A \approx 5.034\,m[/tex]
The position function is:
[tex]x(t) = (5.034\,m)\cdot e^{-0.5\cdot t}\cdot \cos ((12\,\frac{rad}{s} )\cdot t + 0.116\,rad)[/tex]
