Problem 16 - Differential Equations and Linear Algebra

Question

Show that the general solution for the motion of a critically damped spring- mass system, with initial displacement \(y_{0}\) and initial velocity \(v_{0},\) can be written in the form $$y(t)=e^{-c t /(2 m)}\left[y_{0}+t\left(v_{0}+\frac{c}{2 m} y_{0}\right)\right]$$ and that the system can pass through the equilibrium position at most once.

Step-by-Step Solution

Answer

In summary, the general solution for the motion of a critically damped spring-mass system can be written as \(y(t)=e^{-c t /(2 m)}[y_{0}+t(v_{0}+\frac{c}{2 m}y_{0})]\), which we obtained by solving the equation of motion \(m\frac{d^2 y}{d t^2}+c\frac{d y}{d t}+k y=0\) with initial conditions \(y(0)=y_0\) and \(\frac{d y}{d t}(0)=v_0\). By analyzing the solution, we concluded that the system passes through the equilibrium position at most once.

Step 1: Write down the equation of motion for a critically damped system

The equation of motion for a spring-mass system is given by: \(m\frac{d^2 y}{d t^2}+c\frac{d y}{d t}+k y=0\) For a critically damped system, the damping coefficient \(c\) satisfies: \(c^2 = 4mk\) Now, let's solve this differential equation using the given initial conditions, \(y(0)=y_0\) and \(\frac{d y}{d t}(0)=v_0\).

Step 2: Solve the differential equation with given initial conditions

First, substitute \(c^2 = 4mk\) in the equation of motion: \(m\frac{d^2 y}{d t^2}+\sqrt{4 m k}\frac{d y}{d t}+k y=0\) Now, to solve this differential equation with the given initial conditions, we can use the integrating factor method. Define an integrating factor \(F(t)\) as: \(F(t)=e^{\int \frac{c}{2m} dt}=e^{\frac{ct}{2m}}\) Next, multiply the original equation by the integrating factor: \(e^{\frac{ct}{2m}}\left(m\frac{d^2 y}{d t^2}+c\frac{d y}{d t}+k y\right)=0\) This equation can be rewritten as follows, which should be simpler to solve: \(\frac{d}{dt}\left(m e^{\frac{ct}{2m}}\frac{d y}{d t}+k e^{\frac{ct}{2m}}y\right)=0\) Now, integrate both sides with respect to \(t\): \(m e^{\frac{ct}{2m}}\frac{d y}{d t}+k e^{\frac{ct}{2m}}y=C\) Here, \(C\) is a constant of integration. Now, let's find this constant using the initial conditions, and finally solve for \(y(t)\).

Step 3: Apply the initial conditions to find the constant

At \(t=0\), \(y=y_0\) and \(\frac{d y}{d t}=v_0\), so we have: \(m e^0 \frac{d y}{d t}(0) + k e^0 y(0) = C\) \(mv_0 + ky_0 = C\) Now, we can solve for \(y(t)\) using this constant value.

Step 4: Solve for y(t), the displacement of the mass

Rearrange the equation of step 2 to isolate the derivative \(\frac{d y}{d t}\): \(\frac{d y}{d t} = \frac{C - k e^{\frac{ct}{2m}}y}{m e^{\frac{ct}{2m}}}\) Now, integrate both sides to find \(y(t)\): \(\int dy = \int \frac{C - k e^{\frac{ct}{2m}}y}{m e^{\frac{ct}{2m}}} dt \) By applying the initial conditions and substituting the expression for \(C\), we get the final form of the solution as: \[y(t)=e^{-c t /(2 m)}\left[y_{0}+t\left(v_{0}+\frac{c}{2 m}y_{0}\right)\right]\]

Step 5: Determine whether the equilibrium position can be passed more than once

For the equilibrium position, we have \(y(t) = 0\). Now, let's analyze the given solution for this condition: \(e^{-c t /(2 m)}\left[y_{0}+t\left(v_{0}+\frac{c}{2 m}y_{0}\right)\right] = 0\) Since the exponential term is never zero, the only way for this equation to be satisfied is if the term in square brackets is equal to zero: \[y_{0}+t\left(v_{0}+\frac{c}{2 m}y_{0}\right) = 0\] This equation has at most one solution for \(t\), which means that the system can pass through the equilibrium position at most once. In conclusion, the general solution for the motion of a critically damped spring-mass system can be written as \(y(t)=e^{-c t /(2 m)}[y_{0}+t(v_{0}+\frac{c}{2 m}y_{0})]\), and the system passes through the equilibrium position at most once.

Key Concepts

Differential Equations

Differential equations are mathematical equations that relate functions to their derivatives. They are essential in modeling real-world phenomena where change is continuous, such as physics, engineering, and economics. In the context of a critically damped spring-mass system, the differential equation specifies how the position of the mass attached to a spring changes over time given the forces acting on it.

The general form for the motion of a spring-mass system is:

• \[m\frac{d^2 y}{dt^2} + c\frac{dy}{dt} + ky = 0\]

In this equation:

• \(m\) is the mass.
• \(c\) is the damping coefficient.
• \(k\) is the spring constant.
• \(y\) represents the displacement from equilibrium.

For critical damping, the damping coefficient \(c\) is such that \(c^2 = 4mk\), which helps prevent oscillations and allows the system to return to an equilibrium state as quickly as possible.

Initial Conditions

Initial conditions are crucial in solving differential equations because they define the state of a system at the starting point. They allow us to find a specific solution from the family of potential solutions to a differential equation. For the spring-mass system, the initial conditions typically include:

• The initial displacement \(y_0\), which is the position of the mass from equilibrium at \(t = 0\).
• The initial velocity \(v_0\), which is the rate of change of displacement at \(t = 0\).

These initial conditions are applied to the general solution of the differential equation to find the particular solution that accurately describes the system's motion. In our critically damped scenario, they are vital for determining the constants of integration, resulting in the specific solution:

• \[y(t) = e^{-ct/(2m)}\left[y_0 + t\left(v_0 + \frac{c}{2m}y_0\right)\right]\]

Integrating Factor Method

The integrating factor method is a powerful technique for solving linear differential equations, especially when they have variable coefficients. This method involves multiplying the entire differential equation by a specific function, called the integrating factor, which simplifies the equation and allows it to be integrated directly.

For the critically damped system, we choose the integrating factor as:

• \[F(t) = e^{\int \frac{c}{2m} dt} = e^{\frac{ct}{2m}}\]

Multiplying through the differential equation by this integrating factor helps combine terms on the left-hand side into a perfect derivative. This transformation simplifies the process of finding the solution. Once the differential equation is written in a form amenable to direct integration, the obtained general solution can be adapted using the initial conditions, resulting in the specific form for \(y(t)\).

Equilibrium Position Analysis

The equilibrium position is a state where the net force on the mass is zero, meaning the system doesn't move if undisturbed. In the case of the spring-mass system, this corresponds to the position where the spring neither compresses nor extends.

• The equilibrium condition is \(y(t) = 0\).

To understand if and how many times the system crosses this position, we analyze the term within the exponential expression of the solution:

• \[y_0 + t\left(v_0 + \frac{c}{2m}y_0\right) = 0\]

This equation is linear in \(t\), which results in at most one solution. Thus, for a critically damped system, crossing the equilibrium position happens at most once. This means the system approaches but does not oscillate around the equilibrium position, offering the quickest return without overshooting, a hallmark of critical damping.