Determine the overshoot of the system of Problem \(2.25\) if it is critically damped and \(v_{0}=-4 \mathrm{~m} / \mathrm{sec}\).

A critically damped system is a system in which the damping ratio, \(\zeta\), is equal to one. The position function for a critically damped system can be written as: \[x(t) = (C_1 + C_2t)e^{-\omega_nt}\] where \(\omega_n\) is the natural frequency, \(C_1\) and \(C_2\) are constants, and \(t\) represents time.

We are given the initial velocity as \(v_0 = -4 m/s\). To find the constants \(C_1\) and \(C_2\), we'll first differentiate the position function with respect to time to obtain the velocity function: \[v(t) = \frac{dx(t)}{dt} = (C_1(-\omega_nt) + C_2(-\omega_n^2 t^2)e^{-\omega_nt}\] Now we can apply the initial condition \(v(0) = -4 m/s\): \[-4 = C_1(-\omega_n \cdot 0) + C_2(-\omega_n^2 \cdot 0^2)e^{-\omega_n \cdot 0}\] Simplifying the equation, we get: \[-4 = C_2 \cdot 0\] So the constant \(C_2 = 0\), and the position function becomes: \[x(t) = C_1e^{-\omega_nt}\]

We are now ready to find the overshoot of the system. The overshoot is the maximum displacement of the position function, which can be found by maximizing the function: \[x(t) = C_1e^{-\omega_nt}\] We do this by differentiating the position function with respect to time and setting the result equal to zero: \[0 = \frac{dx(t)}{dt} = -\omega_nC_1e^{-\omega_nt}\] Dividing by \(-\omega_nC_1\) on both sides of the equation, we find that \(t = 0\) is the only time when the position function has a maximum. Therefore, the overshoot is the maximum value of the position function, which occurs at \(t = 0:\) \[x_{overshoot} = x(0) = C_1e^{-\omega_n \cdot 0} = C_1\] Since we don't have enough information to determine the value of \(C_1\), the overshoot of the system cannot be found explicitly. However, it is important to note that, in practice, the overshoot could be determined by knowing additional parameters of the system, such as the damping coefficient or the mass of the object subjected to the damping action.