Question

Buoyancy Problems with Drag Force We discussed modeling the bobbing motion of floating cylindrical objects in Section 3.1. Bobbing motion will not persist indefinitely, however. One reason is the drag resistance a floating object experiences as it moves up and down in the liquid. If we assume a drag force proportional to velocity, an application of Newton's second law of motion leads to the differential equation \(y^{\prime \prime}+\mu y^{\prime}+\omega^{2} y=0\), where \(y(t)\) is the downward displacement of the object from its static equilibrium position, \(\mu\) is a positive constant describing the drag force, and \(\omega^{2}\) is a positive constant determined by the mass densities of liquid and object and the vertical extent of the cylindrical object. (See Figure 3.1.) (a) Obtain the general solution of this differential equation, assuming that \(\mu^{2}<4 \omega^{2}\) (b) Assume that a cylindrical floating object is initially displaced downward a distance \(y_{0}\) and released from rest [so the initial conditions are \(\left.y(0)=y_{0}, y^{\prime}(0)=0\right]\).

Short answer

The specific solution is: \(y(t) = e^{-\frac{\mu}{2}t}\left[y_0\cos(\frac{\sqrt{4\omega^2-\mu^2}}{2}t)+\frac{\mu y_0}{\sqrt{4\omega^2-\mu^2}}\sin(\frac{\sqrt{4\omega^2-\mu^2}}{2}t)\right]\)

Step-by-step solution

Analyze the given differential equation

The given differential equation describing the motion of the cylindrical object is: \(y^{\prime \prime}+\mu y^{\prime}+\omega^{2} y=0\) This is a second-order linear differential equation with constant coefficients.

Solve the associated characteristic equation

The characteristic equation associated with the given differential equation is: \(r^2+\mu r+\omega^2=0\) We are given that \(\mu^2 < 4\omega^2\). Now we need to find the roots of this quadratic equation.

Roots of the characteristic equation under given conditions

Using the quadratic formula, we have: \(r=\frac{-\mu\pm\sqrt{\mu^2-4\omega^2}}{2}\) Since \(\mu^2 < 4\omega^2\), the discriminant, \(\mu^2-4\omega^2\), is negative. So, the roots of the characteristic equation are complex and conjugate. They can be written as: \(r_1=\frac{-\mu}{2}+\textit{i}\frac{\sqrt{4\omega^2-\mu^2}}{2}\) and \(r_2=\frac{-\mu}{2}-\textit{i}\frac{\sqrt{4\omega^2-\mu^2}}{2}\)

Obtain the general solution for the given differential equation

Since the roots of the characteristic equation are complex and conjugate, the general solution of the given differential equation is of the form: \(y(t) = e^{-\frac{\mu}{2}t}\left[A\cos(\frac{\sqrt{4\omega^2-\mu^2}}{2}t)+B\sin(\frac{\sqrt{4\omega^2-\mu^2}}{2}t)\right]\) where, A and B are constants that can be determined using initial conditions.

Determine the specific solution using the initial conditions for part (b)

We are given that the cylindrical floating object is initially displaced downward a distance \(y_0\) and released from rest. So, the initial conditions are: \(y(0)=y_0\), \(y^{\prime}(0)=0\) Applying the initial conditions to the general solution, we have: \(y(0)=e^{-\frac{\mu}{2}(0)}\left[A\cos(\frac{\sqrt{4\omega^2-\mu^2}}{2}(0))+B\sin(\frac{\sqrt{4\omega^2-\mu^2}}{2}(0))\right]=y_0\) Since \(\cos(0)=1\) and \(\sin(0)=0\), we have \(A=y_0\). Now we need to find \(y^{\prime}(t)\) and apply the second initial condition to find \(B\). To find \(y^{\prime}(t)\), differentiate \(y(t)\) with respect to \(t\): \(y'(t) = -\frac{\mu}{2}e^{-\frac{\mu}{2}t}\left[A\cos(\frac{\sqrt{4\omega^2-\mu^2}}{2}t)+B\sin(\frac{\sqrt{4\omega^2-\mu^2}}{2}t)\right] + e^{-\frac{\mu}{2}t}\left[-A\frac{\sqrt{4\omega^2-\mu^2}}{2}\sin(\frac{\sqrt{4\omega^2-\mu^2}}{2}t)+B\frac{\sqrt{4\omega^2-\mu^2}}{2}\cos(\frac{\sqrt{4\omega^2-\mu^2}}{2}t)\right]\) Applying the initial condition \(y^{\prime}(0)=0\), we get: \(0= -\frac{\mu}{2}(y_0)+B\frac{\sqrt{4\omega^2-\mu^2}}{2}\) Solving for \(B\), we have: \(B=\frac{\mu y_0}{\sqrt{4\omega^2-\mu^2}}\)

Write the final specific solution

Combining the values of \(A\) and \(B\), the specific solution of the differential equation for the given initial conditions is: \(y(t) = e^{-\frac{\mu}{2}t}\left[y_0\cos(\frac{\sqrt{4\omega^2-\mu^2}}{2}t)+\frac{\mu y_0}{\sqrt{4\omega^2-\mu^2}}\sin(\frac{\sqrt{4\omega^2-\mu^2}}{2}t)\right]\) This is the specific solution that describes the motion of the cylindrical floating object under the given conditions.

Key concepts

Second-Order Linear Differential Equation

Understanding the basics of a second-order linear differential equation is crucial to solving many physical problems, including those involving mechanical vibrations like the bobbing motion of a floating object. These equations have the general form \( y'' + p(t)y' + q(t)y = g(t) \) where \( y'' \) denotes the second derivative of \( y \) with respect to time, \( t \) and \( p(t) \) and \( q(t) \) are functions of \( t \) or constants. For our specific exercise, the form is simplified as \( y'' + \mu y' + \omega^2 y = 0 \) with \( \mu \) and \( \omega^2 \) as constants, representing a homogeneous equation with constant coefficients. These types of differential equations are pivotal in modeling dynamic systems and understanding their behavior over time.

In the buoyancy problem given, it represents how the cylindrical object's displacement changes purely under the influence of a damping force and restoring force, ignoring any external forces such as a driving force. The solution to such equations determines the system's response, predicting how the object will move after being displaced and released.

Characteristic Equation

The characteristic equation plays a pivotal role in finding the solution to a second-order linear differential equation with constant coefficients. By substituting \( y = e^{rt} \) into the differential equation, we can derive a polynomial that we call the characteristic equation. In our example, this characteristic equation is \( r^2 + \mu r + \omega^2 = 0 \) and can be solved for \( r \) using the quadratic formula. The roots of the characteristic equation provide critical information about the behavior of the system over time, offering a path to the equation's general solution.

The process of finding the characteristic equation simplifies solving differential equations, transforming a problem of calculus into one of algebra, which is generally easier to handle. The nature of the equation's roots, whether they are real, repeated, or complex, will shape the form of the general solution and, consequently, how we eventually apply initial conditions to determine the system's specific response.

Complex Conjugate Roots

When the characteristic equation's discriminant is negative, which occurs under the condition \( \mu^2 < 4\omega^2 \) in our problem, the roots of the equation will be complex conjugates. These roots take the form \( r_1 = \alpha + i\beta \) and \( r_2 = \alpha - i\beta \), where \( i \) is the imaginary unit. The presence of complex roots indicates an oscillatory motion in the system, which is precisely what we observe with the bobbing of the floating object.

Complex conjugate roots led us to a general solution that combines exponential decay, representing energy loss due to the drag force, and trigonometric functions, representing periodic motion. This reflects how the floating object gradually ceases the bobbing motion until it ultimately becomes at rest, showcasing the intertwined effects of damping and the system's natural tendency to oscillate.

Initial Conditions

Initial conditions are the given values of the function and its derivatives at the start of the observation, which for our exercise are \( y(0) = y_0 \) and \( y'(0) = 0 \). These conditions allow us to pinpoint a specific solution from the infinite family of solutions in the general form. By applying the initial conditions to the general solution, we determine the constants in the expression that uniquely define the system's behavior.

In our bobbing object example, the values of \( y_0 \) and the initial velocity inform us about the starting position and the initial motion state of the object, enabling us to solve for the coefficients in the general solution and accurately describe the object's movement over time. The application of initial conditions is akin to customizing the mathematical model to fit the real-world scenario at hand, affirming the importance of accurately knowing or measuring these conditions for an effective analysis.