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The position of a certain undamped spring-mass system satisfies the initial value problem $$ u^{\prime \prime}+2 u=0, \quad u(0)=0, \quad u^{\prime}(0)=2 $$ (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 \(u^{\prime}\) versus \(u ;\) that is, plot \(u(t)\) and \(u^{\prime}(t)\) parametrically with \(t\) as the parameter. This plot is known as a phase plot and the \(u u^{\prime}\) -plane is called the phase plane. Observe that a closed curve in the phase plane corresponds to a periodic solution \(u(t) .\) What is the direction of motion on the phase plot as \(t\) increases?

Short Answer

Expert verified
Answer: The solution of the initial value problem is \(u(t) = \sqrt{2}\sin{\sqrt{2}t}\). The position function, \(u(t)\), is sinusoidal with a period of \(2\pi/\sqrt{2}\) and the velocity function, \(u'(t) = -2\cos{\sqrt{2}t}\), is also sinusoidal with the same period. The graphs of these functions show oscillatory behavior versus time.

Step by step solution

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01

Defining the Differential Equation

The given initial value problem is described by the following second-order linear homogeneous differential equation with constant coefficients: $$ u^{\prime\prime} + 2u = 0 $$ Subject to the following initial conditions: $$ u(0) = 0 \\ u'(0) = 2 $$
02

Finding the Complementary Function

To find the complementary function, we assume a solution of the form \(u(t) = e^{rt}\). Substituting this into the differential equation yields: $$ r^2e^{rt} + 2e^{rt} = 0 $$ Dividing through by \(e^{rt}\), we get: $$ r^2 + 2 = 0 $$ Solving for r: $$ r = ±\sqrt{-2} = ±i\sqrt{2} $$ This gives us the complementary function of the form: $$ u_c(t) = C_1\cos{\sqrt{2}t} + C_2\sin{\sqrt{2}t} $$ Where \(C_1\) and \(C_2\) are constants to be determined.
03

Applying Initial Conditions

We have the complementary function: $$ u_c(t) = C_1\cos{\sqrt{2}t} + C_2\sin{\sqrt{2}t} $$ Taking the derivative to find u'(t) : $$ u'(t) = -C_1\sqrt{2}\sin{\sqrt{2}t} + C_2\sqrt{2}\cos{\sqrt{2}t} $$ Now, we apply the initial conditions \(u(0)=0\) and \(u'(0)=2\): $$ u(0) = C_1\cos(0) + C_2\sin(0) = 0 \implies C_1 = 0 \\ u'(0) = -C_1\sqrt{2}\sin(0) + C_2\sqrt{2}\cos(0) = 2 \implies C_2 = \frac{2}{\sqrt{2}} = \sqrt{2} $$ Thus, the solution of the initial value problem is: $$ u(t) = \sqrt{2}\sin{\sqrt{2}t} $$
04

Plotting Graphs

(a) The position function of the spring is given by \(u(t) = \sqrt{2}\sin{\sqrt{2}t}\). This function is sinusoidal with a period of \(2\pi/\sqrt{2}\). Graph this function for the position versus time. (b) The velocity function of the spring is given by \(u'(t) = -2\cos{\sqrt{2}t}\). This function is also sinusoidal with a period of \(2\pi/\sqrt{2}\). Graph this function for the velocity versus time on the same axes as u(t) versus time.
05

Creating the Phase Plot

To create the phase plot, we parametrically plot the position \(u(t) = \sqrt{2}\sin{\sqrt{2}t}\) and velocity \(u'(t)=-2\cos{\sqrt{2}t}\) with respect to time. The phase plot is a closed curve as the position and velocity functions are periodic. As t increases, the motion direction on the phase plot is clockwise.

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Differential Equations
Differential equations are mathematical equations that describe the relationship between a function and its derivatives. They play a crucial role in modeling the behavior of natural systems, such as the motion of a spring-mass system.

A simple yet important type of differential equation is the second-order linear homogeneous differential equation with constant coefficients. In the context of the exercise, the equation is \( u'' + 2u = 0 \), representing the movement of an undamped spring-mass system.

To solve such an equation, we look for a complementary function, which typically takes the form of an exponential function. Through this approach, constants and the structure of the solution are derived, leading to a function that describes the system's position over time. The method requires the application of initial conditions to find specific values for the constants, thereby arriving at the exact solution that satisfies the problem's requirements.
Deciphering the Phase Plot
A phase plot is a graphical representation that showcases the relationship between a system's state variables, often position and velocity. It is especially useful in visualizing the behavior of dynamical systems.

In the exercise at hand, the phase plot is created by parametrically plotting the position function \(u(t)\) and its derivative \(u'(t)\) over time. The result is a closed curve that reflects the periodicity of the system. A closed curve in the phase plot confirms that a function, in this case \(u(t)\), is periodic.

The Direction of Motion

The direction of the motion on the phase plot, which is clockwise in our example, gives insights into the energy conservation and the system's stability. Observing this plot helps in understanding how the system evolves over time without delving into the specifics of temporal evolution.
The Sinusoidal Function and its Properties
Sinusoidal functions are a cornerstone of mathematics when modeling oscillations and waves. These functions, including the sine and cosine functions, describe smooth periodic oscillations.

In the given spring-mass system described by \(u(t) = \sqrt{2}\sin(\sqrt{2}t)\), the solution represents a sinusoidal function that models the periodic displacement of the spring. Such functions have a set of important properties, including amplitude, period, and frequency, which dictate the behavior of the wave.

Sinusoidal functions are particularly important because they can be used to represent complex oscillatory behavior through simple, uniform equations. By understanding and analyzing the solution in the form of a sinusoidal function, students can predict how the spring system will act over time—rising and falling in a consistent, repeatable pattern.

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Most popular questions from this chapter

Use the method of reduction of order to find a second solution of the given differential equation. \(x^{2} y^{\prime \prime}+x y^{\prime}+\left(x^{2}-0.25\right) y=0, \quad x>0 ; \quad y_{1}(x)=x^{-1 / 2} \sin x\)

By combining the results of Problems 24 through \(26,\) show that the solution of the initial value problem $$ L[y]=\left(a D^{2}+b D+c\right) y=g(t), \quad y\left(t_{0}\right)=0, \quad y^{\prime}\left(t_{0}\right)=0 $$ where \(a, b,\) and \(c\) are constants, has the form $$ y=\phi(t)=\int_{t_{0}}^{t} K(t-s) g(s) d s $$ The function \(K\) depends only on the solutions \(y_{1}\) and \(y_{2}\) of the corresponding homogeneous equation and is independent of the nonhomogeneous term. Once \(K\) is determined, all nonhomogeneous problems involving the same differential operator \(L\) are reduced to the evaluation of an integral. Note also that although \(K\) depends on both \(t\) and \(s,\) only the combination \(t-s\) appears, so \(K\) is actually a function of a single variable. Thinking of \(g(t)\) as the input to the problem and \(\phi(t)\) as the output, it follows from Eq. (i) that the output depends on the input over the entire interval from the initial point \(t_{0}\) to the current value \(t .\) The integral in Eq. (i) is called the convolution of \(K\) and \(g,\) and \(K\) is referred to as the kernel.

Use the method of Problem 32 to solve the given differential $$ y^{\prime \prime}+2 y^{\prime}+y=2 e^{-t} \quad(\text { see Problem } 6) $$

(a) Determine a suitable form for \(Y(t)\) if the method of undetermined coefficients is to be used. (b) Use a computer algebra system to find a particular solution of the given equation. $$ y^{\prime \prime}+4 y=t^{2} \sin 2 t+(6 t+7) \cos 2 t $$

Use the method of Problem 33 to find a second independent solution of the given equation. \(t^{2} y^{\prime \prime}+3 t y^{\prime}+y=0, \quad t>0 ; \quad y_{1}(t)=t^{-1}\)

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