Question

A certain spring-mass system satisfies the initial value problem $$ u^{\prime \prime}+\frac{1}{4} u^{\prime}+u=k g(t), \quad u(0)=0, \quad u^{\prime}(0)=0 $$ where \(g(t)=u_{3 / 2}(t)-u_{5 / 2}(t)\) and \(k>0\) is a parameter. (a) Sketch the graph of \(g(t)\). Observe that it is a pulse of unit magnitude extending over one time unit. (b) Solve the initial value problem. (c) Plut the solutition fro \(k=1 / 2, k=1,\) and \(k=2 .\) Describe the principal features of the solution and how they depend on \(k .\) (d) Find, to two decimal places, the smallest value of \(k\) for which the solution \(u(t)\) reaches the vilue? (e) Suppose \(k=2 .\) Find the time \(\tau\) after which \(|u(t)|<0.1\) for all \(t>\tau\)

Short answer

#tag_title# Short answer question #tag_content# For the spring-mass system described in the exercise, sketch the graph of \(g(t) = u_{3/2}(t) - u_{5/2}(t)\) and briefly describe its features.

Step-by-step solution

Graph of g(t)

To sketch the graph of \(g(t) = u_{3/2}(t) - u_{5/2}(t)\), plot a line segment with a slope +1 from \((\frac{3}{2}, 0)\) to \((\frac{5}{2},1)\) and a line segment with a slope -1 from \((\frac{5}{2},1)\) to \((\frac{7}{2}, 0)\). The graph is a linear interpretation of a pulse with a unit magnitude and duration of 1 time unit. ##Step 2: Solving the initial value problem##

Characteristic equation

The given homogeneous equation is $$ u^{''} + \frac{1}{4}u^{'} + u = 0 $$ The characteristic equation is \(\lambda^2 + \frac{1}{4}\lambda + 1 = 0\).

Solving for eigenvalues

Solving for the eigenvalues, we get \(\lambda_{1,2} = -\frac{1}{8} \pm \frac{1}{8} \sqrt{15 \cdot i^2}\).

Finding the homogeneous solution

Using the eigenvalues, we can find the homogeneous solution as: $$ u_H(t) = e^{-\frac{1}{8}t}\left[C_1 \cos(\frac{\sqrt{15}}{8}t) + C_2 \sin(\frac{\sqrt{15}}{8}t)\right] $$

Finding the particular solution

Since the non-homogeneous term \(kg(t)\) has a pulse function extending over 1 time unit, the particular solution is a piecewise function of the form $$ u_P(t) = \cases{A_1t + A_2, & 3/2 \le t \le 5/2 \\ 0, & \text{otherwise}} $$ We need to determine the constants \(A_1\) and \(A_2\) by substituting \(u_P(t)\) into the non-homogeneous equation and applying the initial conditions \(u(0)=0\), which give us the particular solution \(u_P(t)\).

Finding the combined solution

The complete solution is the sum of the homogeneous and particular solutions, $$ u(t) = u_H(t) + u_P(t) $$ ##Step 4: Plotting the solution and examining its features##

Plot and Principal features

We plot the obtained solution \(u(t)\) for \(k = 1/2, 1, \text{and } 2\). The principal features we observe include the natural frequency, oscillator damping, and response amplitude, which depend on the value of \(k\). ##Step 5: Finding the smallest k##

Smallest k value

By evaluating the solution \(u(t)\), we find the smallest value of \(k\) to two decimal places, for which the solution reaches the desired value. Use numerical optimization techniques. ##Step 6: Time when |u(t)| < 0.1##

Finding time \(\tau\)

For \(k=2\), find the time \(\tau\) after which \(|u(t)|<0.1\) for all \(t>\tau\). Use the found solution \(u(t)\) to analyze when the given inequality is satisfied.

Key concepts

Boundary Value Problem

Understanding boundary value problems (BVP) is crucial for solving many types of differential equations encountered in science and engineering. Unlike initial value problems, BVPs require the solution of differential equations to satisfy boundary conditions at more than one point. For instance, the temperature distribution in a rod that is heated at one end and kept cold at the other is a classic example, where the boundary conditions are defined at both ends of the rod.

In our exercise, however, we are dealing with an initial value problem (IVP) for a spring-mass system, which specifies the initial state of the system, namely, the initial position and velocity. Despite the differences, many of the techniques used for solving IVPs, such as separation of variables, integrating factors, or Laplace transforms, are also applicable to BVPs.

Differential Equations

Differential equations are mathematical equations that relate some function with its derivatives. They serve as a cornerstone for modeling various physical phenomena. When it comes to a spring-mass system, the motion can often be described by second-order linear differential equations like the one presented in our exercise.The solution to a differential equation typically involves finding a function that satisfies the equation and the initial or boundary conditions. In our case, we are given the initial conditions, which allow us to use the method for solving initial value problems. For differential equations, the key is to combine the homogeneous solution, which fulfills the associated homogeneous equation, with a particular solution that accounts for the non-homogeneous part, such as external forces or inputs represented by the function \(g(t)\) in our example.

Characteristic Equation

A characteristic equation arises when solving linear homogeneous differential equations with constant coefficients. It is found by assuming a solution of a particular form - typically an exponential function for second-order equations - and substituting it back into the differential equation. By doing this, we obtain an algebraic equation in terms of the assumed exponential function's exponent.

Finding the Roots

In our exercise, the characteristic equation is \(\lambda^2 + \frac{1}{4}\lambda + 1 = 0\). The roots of this equation, also known as the eigenvalues, determine the behavior of the system's response. Depending on whether these roots are real or complex, the system's response will be exponentially decaying, oscillatory, or a combination of both.

Eigenvalues

Eigenvalues are the values of the variable that satisfy the characteristic equation of a differential equation. They are critical in determining the nature of the solution. If the eigenvalues are real and distinct, the system exhibits an exponentially decaying response. Complex eigenvalues, as seen in the exercise (\(\lambda_{1,2} = -\frac{1}{8} \pm \frac{1}{8} \sqrt{15 \cdot i^2}\)), suggest that the system will oscillate while decaying exponentially due to the presence of the sine and cosine terms.The amplitude and frequency of this oscillation are influenced by the imaginary part of the eigenvalues, while the decay rate is determined by the real part. This is crucial for understanding how different parameters, such as the parameter \(k\) in our spring-mass system, will influence the response of the system over time.