Question

The position of a certain spring-mass system satisfies the initial value problem $$ u^{\prime \prime}+\frac{1}{4} u^{\prime}+2 u=0, \quad u(0)=0, \quad u^{\prime}(0)=2 $$ (a) Find the solution of this initial value problem. (b) Plot \(u\) versus \(t\) and \(u^{\prime}\) versus \(t\) on the same axes. (c) Plot \(u\) versus \(u\) in the phase plane (see Problem 28 ). Identify several corresponding points on the curves in parts (b) and (c). What is the direction of motion on the phase plot as \(t\) increases?

Short answer

Question: Find the specific solution for the given second-order linear differential equation representing the motion of a spring-mass system and plot graphs for parts (b) and (c): (a) $$u^{\prime \prime}+\frac{1}{4} u^{\prime}+2 u=0$$ with initial conditions \(u(0)=0\) and \(u^{\prime}(0)=2\). (b) Plot the graphs of \(u(t)\) and \(u^{\prime}(t)\) versus \(t\) on the same axes. (c) Plot the relationship between \(u(t)\) and \(u^{\prime}(t)\) in the phase plane. Answer: The specific solution for the given differential equation is: $$u(t) = e^{-\frac{1}{8}t}(\frac{16}{\sqrt{15}}\sin(\frac{\sqrt{15}t}{8}))$$ For parts (b) and (c), plot the following graphs: 1. \(u(t)\) and \(u^{\prime}(t)\) versus \(t\) on the same axes. 2. \(u(t)\) versus \(u^{\prime}(t)\) in the phase plane. The direction of motion on the phase plot as \(t\) increases will be counterclockwise.

Step-by-step solution

Identify the differential equation

The given equation is a second-order linear equation with constant coefficients: $$ u^{\prime \prime}+\frac{1}{4} u^{\prime}+2 u=0 $$ With initial conditions \(u(0)=0\) and \(u^{\prime}(0)=2\).

Find the characteristic equation and its roots

The characteristic equation for the given differential equation is: $$ r^2 + \frac{1}{4}r + 2 = 0 $$ Now we need to find the roots of this equation. In this case, we can use the quadratic formula: $$ r = \frac{-b\pm\sqrt{b^2-4ac}}{2a} $$ With \(a = 1\), \(b = \frac{1}{4}\), and \(c = 2\), we get: $$ r = \frac{-\frac{1}{4}\pm\sqrt{(\frac{1}{4})^2-4(1)(2)}}{2(1)} $$ Which simplifies to: $$ r = \frac{-\frac{1}{4}\pm\sqrt{-\frac{15}{16}}}{2} $$ Since the discriminant is negative, we have complex roots: $$ r_1 = -\frac{1}{8} + \frac{\sqrt{15}i}{8}, \quad r_2 = -\frac{1}{8} - \frac{\sqrt{15}i}{8} $$

Find the general and specific solutions

The general solution for a second-order linear equation with complex roots is given by: $$ u(t) = e^{αt}(c_1\cos(βt)+c_2\sin(βt)) $$ Where \(r = α + iβ\). In our case, \(α = -\frac{1}{8}\) and \(β = \frac{\sqrt{15}}{8}\). Thus, the general solution is: $$ u(t) = e^{-\frac{1}{8}t}(c_1\cos(\frac{\sqrt{15}t}{8})+c_2\sin(\frac{\sqrt{15}t}{8})) $$ Now, we will use the initial conditions to find \(c_1\) and \(c_2\). Using \(u(0)=0\): $$ 0 = e^{0}(c_1\cos(0)+c_2\sin(0)) \Rightarrow c_1 = 0 $$ The first derivative of \(u(t)\) is: $$ u^{\prime}(t) = -\frac{1}{8}e^{-\frac{1}{8}t}(c_2\sin(\frac{\sqrt{15}t}{8})) + e^{-\frac{1}{8}t}(c_2\frac{\sqrt{15}}{8}\cos(\frac{\sqrt{15}t}{8})) $$ Using \(u^{\prime}(0)=2\): $$ 2 = -\frac{1}{8}(0) + (c_2\frac{\sqrt{15}}{8}) \Rightarrow c_2 = \frac{16}{\sqrt{15}} $$ The specific solution is: $$ u(t) = e^{-\frac{1}{8}t}(\frac{16}{\sqrt{15}}\sin(\frac{\sqrt{15}t}{8})) $$

Plot the graphs

To complete parts (b) and (c), the student will need to plot the following graphs: 1. \(u(t)\) and \(u^{\prime}(t)\) versus \(t\) on the same axes. This graph will show the behavior of the position and velocity of the spring-mass system over time. 2. \(u(t)\) versus \(u^{\prime}(t)\) in the phase plane. This graph will show the relationship between the position and velocity of the system. For the first graph, plot the functions: $$ u(t) = e^{-\frac{1}{8}t}(\frac{16}{\sqrt{15}}\sin(\frac{\sqrt{15}t}{8})) $$ $$ u^{\prime}(t) = -\frac{1}{8}e^{-\frac{1}{8}t}(\frac{16}{\sqrt{15}}\sin(\frac{\sqrt{15}t}{8})) + e^{-\frac{1}{8}t}(\frac{16}{\sqrt{15}}\frac{\sqrt{15}}{8}\cos(\frac{\sqrt{15}t}{8})) $$ For the second graph (phase plot), plot: $$ u(t) \text{ vs } u^{\prime}(t) $$ The direction of motion on the phase plot as \(t\) increases will be counterclockwise.

Key concepts

Second-Order Linear Differential Equation

In the realm of differential equations, a second-order linear differential equation holds significant importance, especially in physical systems such as oscillations and circuits. It's written generally as:
\[ u'' + p(t)u' + q(t)u = g(t) \]
where \( u'' \) represents the second derivative of \( u(t) \), the function we're solving for, with respect to time \( t \). The terms \( p(t) \) and \( q(t) \) are coefficient functions, and \( g(t) \) is the nonhomogeneous term.For the exercise at hand, our equation simplifies as \( u'' + \frac{1}{4}u' + 2u = 0 \), a homogeneous equation with constant coefficients, meaning the coefficients of \( u' \) and \( u \) are not functions of \( t \) but are mere constants. To address this type of equation, we look for solutions that can be expressed as exponential functions of time. By doing so, we can capture behaviors that include exponential growth or decay, and when we involve complex numbers, it also captures oscillatory motion, which is crucial for understanding many physical systems, from a simple mass-spring setup to electrical circuits.

Characteristic Equation

When we're solving a second-order differential equation with constant coefficients, we're introduced to a critical tool: the characteristic equation. This equation comes from an assumption that the solution is of the form \( e^{rt} \), where \( r \) is a number we're trying to determine.
To derive the characteristic equation, we substitute \( e^{rt} \) (and its derivatives) into the homogeneous differential equation. For the second-order case, this process leads to a quadratic equation in terms of \( r \):
\[ r^2 + ar + b = 0 \]
where \( a \) and \( b \) correspond to the coefficients of \( u' \) and \( u \) respectively in our initial differential equation.

• Finding the roots of this quadratic gives us the values of \( r \) which are crucial to forming our general solution.
• If the roots are real and distinct, our solution involves real exponential functions.
• If we have a repeated root, a slight modification involves multiplying by \( t \).
• Complex roots, as seen in our exercise, lead to trigonometric functions being included in the solution due to Euler's formula relating exponentials and trigonometric expressions.

The characteristic equation is the skeleton key to unlocking the behavior of second-order linear systems, providing a straightforward pathway from complex algebra to the dynamic behavior of a system.

Complex Roots

Our journey through the characteristic equation sometimes takes an interesting turn when we encounter complex roots. These arise when the discriminant \( b^2 - 4ac \) in the quadratic equation is negative, signaling that our roots are of the form \( \text{real part} \text{±} \text{imaginary part} \text{i} \).
In the context of the initial value problem we've been exploring, complex roots like \( r_1 = -\frac{1}{8} + \frac{\sqrt{15}i}{8} \) and \( r_2 = -\frac{1}{8} - \frac{\sqrt{15}i}{8} \) tell us that our solution is going to be oscillatory due to the imaginary parts, with the sine and cosine functions coming into play.The presence of the real parts (in this case, \( -\frac{1}{8} \)) suggests there is also an exponential decay factor. Together, these aspects of the roots describe a system that oscillates while gradually losing amplitude over time due to the exponential decay. This understanding is significant in many scientific and engineering applications, including the study of mechanical vibrations, electrical circuits, and even in modeling population dynamics in biology.

Phase Plane

The phase plane is a powerful visual tool for understanding the behavior of dynamical systems, especially those described by second-order differential equations. It's a two-dimensional plot with axes representing the system's state variables — typically, one axis shows the system's position \( u(t) \), and the other shows its velocity \( u'(t) \).
By plotting the state variables against each other, we bypass time as an explicit variable and instead observe the system's trajectory in this plane. For our initial value problem, the phase plane plot reveals:

• The shape of the trajectory, which, due to our complex roots, will be a spiral.
• The system's behavior over time, as indicated by the trajectory's orientation.
• Specific points on the trajectory corresponding to particular moments in time.

For the spring-mass system we're studying, the phase plane tells us how the position and velocity are interconnected. As one changes, so does the other, in a dance dictated by the differential equation. It's an elegant portrayal of dynamic motion, capturing the essence of the system's behavior in a single snapshot that might otherwise require multiple graphs to fully comprehend.