Question

Assume that the system described by the equation \(m u^{\prime \prime}+\gamma u^{\prime}+k u=0\) is either critically damped or overdamped. Show that the mass can pass through the equilibrium position at most once, regardless of the initial conditions. Hint: Determine all possible values of \(t\) for which \(u=0\).

Short answer

Answer: No, the mass in a critically damped or overdamped mass-spring system can pass through the equilibrium position at most once, regardless of the initial conditions.

Step-by-step solution

Find the characteristic equation and its roots

To find the characteristic equation, we substitute \(u = e^{rt}\) into the given differential equation: \(mu'' + \gamma u' + ku=0\). This results in the quadratic characteristic equation, \(mr^2 + \gamma r + k = 0\). Let's find the roots of this equation, denoted as \(r_1\) and \(r_2\).

Analyze the critically damped case

A system is critically damped if the discriminant of the characteristic equation is equal to zero, i.e., \(\Delta = \gamma^2 - 4mk = 0\). In this case, there is only one distinct root, \(r_1 = r_2 = -\frac{\gamma}{2m}\). Since we have one repeated root, the general solution to the differential equation is: \(u(t) = (A + Bt)e^{-\frac{\gamma}{2m}t}\). Now, let's find all possible values of \(t\) that make \(u=0\): \((A+Bt)e^{-\frac{\gamma}{2m}t}=0\), which simplifies to \(A + Bt = 0\) after dividing both sides by the exponential term. Thus, the only possible value of \(t\) that makes \(u=0\) is \(t = -\frac{A}{B}\). This shows that the system can pass through the equilibrium position at most once.

Analyze the overdamped case

A system is overdamped if the discriminant of the characteristic equation is greater than zero, i.e., \(\Delta = \gamma^2 - 4mk > 0\). In this case, there are two distinct real roots, \(r_1\) and \(r_2\), given by the quadratic formula: \(r_{1,2} = \frac{-\gamma \pm \sqrt{\gamma^2-4mk}}{2m}\). Since we have two distinct roots, the general solution to the differential equation is: \(u(t) = A e^{r_1 t} + B e^{r_2 t}\). Now, let's find all possible values of \(t\) that make \(u=0\): \(A e^{r_1 t} + B e^{r_2 t} = 0\). Calculus shows that, in this case, either there are no intersections at all, or one intersection with the equilibrium position. In either case, we can see that the mass can pass through the equilibrium position at most once. In conclusion, we have shown that in both the critically damped and overdamped cases, the mass can pass through the equilibrium position at most once, regardless of the initial conditions.

Key concepts

Characteristic Equation

Understanding the characteristic equation is essential in solving second-order homogeneous linear differential equations, like the one given in our exercise. To obtain the characteristic equation, we employ a technique where we assume a solution of the form u = e^{rt}, where r is the root that we wish to find and t is the variable of time.

By substituting this assumed solution into the differential equation m u'' + gamma u' + ku=0, we derive a quadratic equation in terms of r: mr^2 + gamma r + k = 0. This equation is the characteristic equation, and its roots will determine the behavior of the system and are key to forming the solution to the differential equation.

The roots can be complex, real and distinct, or real and repeated, depending on the discriminant gamma^2 - 4mk. In the context of mechanical systems, m represents mass, gamma the damping coefficient, and k the spring constant. Each root of this characteristic equation will dictate how the system responds over time—whether it oscillates, exponentially decays, or returns to equilibrium in a critically damped manner.

Critically Damped System

A critically damped system occurs when the system returns to equilibrium as quickly as possible without oscillating. In terms of the characteristic equation, this behavior is observed when the discriminant gamma^2 - 4mk is exactly zero (gamma^2 = 4mk), resulting in a double root of the characteristic equation.

For the given differential equation, a critically damped system will have the general solution u(t) = (A + Bt)e^{-gamma/2m t}. Here, A and B are constants determined by initial conditions, and the term - gamma/2m is the double root. The presence of the t term multiplied by the exponential function reflects the influence of the double root and is characteristic of critically damped motion.

As the exercise reveals, the mass in a critically damped system will pass through the equilibrium position only once. This is intuitive because, in critical damping, there are no oscillations, and the system quickly returns to, but does not cross, its equilibrium position. This is also evident mathematically, as setting u(t) = 0 yields a single value of t, indicating a one-time intersection with the equilibrium point.

Overdamped System

When a system is overdamped, it returns to its equilibrium position without oscillating and does so more slowly compared to a critically damped system. This occurs if the discriminant of the characteristic equation, gamma^2 - 4mk, is greater than zero (gamma^2 > 4mk), resulting in two distinct real roots.

These roots determine the general solution of the differential equation for an overdamped system: u(t) = A e^{r_1 t} + B e^{r_2 t}, where A and B are constants set by initial conditions, and r_1 and r_2 are the roots of the characteristic equation.

When we explore the overdamped case further by setting u(t) = 0, we find that the mass cannot pass through the equilibrium position more than once. This is because the exponentials decay to zero without crossing the equilibrium point again—reflecting an overdamped system’s slower return to stability. Even in the presence of varying initial conditions, this key aspect remains true: the mass in an overdamped system, much like in a critically damped system, cannot cross the equilibrium position more than once, which is aligned with its definition as a non-oscillatory response.