A mass $m$ is attached to both a spring (with given spring constant $k$ ) and a dashpot (with given damping constant $c$ ). The mass is set in motion with initial position $x_0$ and initial velocity $v_0$. Find the position function $x(t)$ and determine whether the motion is overdamped, critically damped, or underdamped. If it is underdamped, write the position function in the form $x(t)=C_1{e}^{-pt}\cos ({\omega }_{1}t-{\alpha }_1)$. Also, find the undamped position function $u(t)=C_0\cos ({\omega }_{0t}-{\alpha }_0)$ that would result if the mass on the spring were set in motion with the same initial position and velocity, but with the dashpot disconnected (so $c=0$ ). Finally, construct a figure that illustrates the effect of damping by comparing the graphs of $x(t)$ and $u(t)$.
$$m=\frac{3}{4},c=6,k=9,x_0=6,v_0=0$$
$x(t)=◻$, which means the system is
(Use integers or decimals for any numbers in the expression. Round to four decimal places as needed. Type any angle measures in radians. Use angle measures greater than or equal to 0 and less than or equal to $2\pi$.)

Step 1 ：The equation of motion for a damped harmonic oscillator is given by $m\frac{d^{2}x}{d{t}^2}+c\frac{dx}{dt}+kx=0$. Substituting the given values, we get $\frac{3}{4}\frac{d^{2}x}{d{t}^2}+6\frac{dx}{dt}+9x=0$.

Step 2 ：This is a second order homogeneous differential equation. The general solution is of the form $x(t)=e^{rt}$, where $r$ is a root of the characteristic equation $m{r}^2+cr+k=0$. Substituting the given values, we get $\frac{3}{4}{r}^2+6r+9=0$.

Step 3 ：Solving this quadratic equation, we get $r=\frac{-6\pm \sqrt{6^2-4\ast \frac{3}{4}\ast 9}}{2\ast \frac{3}{4}}=-4\pm 2i$.

Step 4 ：Since the roots are complex, the system is underdamped. The general solution is of the form $x(t)=e^{-pt}(C_1\cos ({\omega }_{1}t)+C_2\sin ({\omega }_{1}t))$, where $p=4$ and ${\omega }_1=2$.

Step 5 ：Using the initial conditions $x(0)=x_0=6$ and $v(0)=v_0=0$, we can solve for $C_1$ and $C_2$. We get $C_1=6$ and $C_2=0$.

Step 6 ：So, the position function is $x(t)=6e^{-4t}\cos (2t)$.

Step 7 ：If the dashpot is disconnected (i.e., $c=0$), the equation of motion becomes $m\frac{d^{2}x}{d{t}^2}+kx=0$. The general solution is of the form $u(t)=C_0\cos ({\omega }_{0}t-{\alpha }_0)$, where ${\omega }_0=\sqrt{\frac{k}{m}}=2\sqrt{3}$.

Step 8 ：Using the initial conditions $u(0)=x_0=6$ and $u^{\prime }(0)=v_0=0$, we can solve for $C_0$ and ${\alpha }_0$. We get $C_0=6$ and ${\alpha }_0=0$.

Step 9 ：So, the undamped position function is $u(t)=6\cos (2\sqrt{3}t)$.

Step 10 ：The graphs of $x(t)$ and $u(t)$ would show that the damped oscillator $x(t)$ decreases in amplitude over time, while the undamped oscillator $u(t)$ maintains a constant amplitude.