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Chapter 5: Problem 10

A 16 -lb weight is attached to the lower end of a coil spring that is suspended vertically from a support and for which the spring constant \(k\) is \(10 \mathrm{lb} / \mathrm{ft}\). The weight comes to rest in its equilibrium position and is then pulled down 6 in. below this position and released at \(t=0 .\) At this instant the support of the spring begins a vertical oscillation such that its distance from its initial position is given by \(\frac{1}{2} \sin 2 t\) for \(t \geq 0\). The resistance of the medium in pounds is numerically equal to \(2(d x / d t)\), where \(d x / d t\) is the instantaneous velocity of the moving weight in feet per second. (a) Show that the differential equation for the displacement of the weight from its equilibrium position is $$ \frac{1}{2} \frac{d^{2} x}{d t^{2}}=-10(x-y)-2 \frac{d x}{d t}, \quad \text { where } \quad y=\frac{1}{2} \sin 2 t $$ and hence that this differential equation may be written $$ \frac{d^{2} x}{d t^{2}}+4 \frac{d x}{d t}+20 x=10 \sin 2 t $$ (b) Solve the differential equation obtained in step (a), apply the relevant initial conditions, and thus obtain the displacement \(x\) as a function of time.

Short Answer

Expert verified
The displacement of the weight from its equilibrium position as a function of time is given by: \(x(t) = e^{-2t}(-\frac{1}{4}\cos 4t + \frac{3}{2}\sin 4t) + \frac{1}{4}(\sin 2t - \cos 2t)\)

Step by step solution

01

Write the equation of motion

To find the equation of motion for the weight, first we should find the net force acting on the weight. The main forces acting on the weight are: - The spring force = \(-k \cdot(x - y)\), where \(x\) is the displacement from the equilibrium position, and \(y\) is the displacement of the support. - The resistive force of the medium = \(-2 \cdot (dx/dt)\) Now, we can use Newton's second law, \(F = m\cdot a\), where \(F\) is the net force, \(m\) is the mass, and \(a\) is the acceleration. We need first to convert the weight (16 lb) into mass (m in slugs), using \( m = \frac{W}{g}\) where W is the weight in lb and g is the acceleration due to gravity (32.2 ft/s虏). \( m=\frac{16}{32.2}鈮�0.5\,slugs\) We plug in the forces and the mass: \(0.5\frac{d^2x}{dt^2} = -10(x-y) - 2 \frac{dx}{dt}\)
02

Write the differential equation for the displacement

Now, we will substitute \(y\) with the given equation of the support: \(y = \frac{1}{2} \sin 2t\). The differential equation becomes: \(0.5 \frac{d^2x}{dt^2} = -10(x - \frac{1}{2} \sin 2t) - 2 \frac{dx}{dt}\) Multiplying by 2, we can rewrite this as: \(\frac{d^2x}{dt^2} + 4 \frac{dx}{dt} + 20x = 10 \sin 2t\)
03

Solve the differential equation with initial conditions

The given differential equation is a nonhomogeneous linear second-order ordinary differential equation with constant coefficients. First, we find the complementary function \(x_c(t)\) by solving the homogeneous equation corresponding to the given equation: \(\frac{d^2x}{dt^2} + 4 \frac{dx}{dt} + 20x = 0\) The characteristic equation is \(r^2+4r+20=0\). Solving for r, we find that the roots are complex: \(r = -2 \pm 4i\). Thus, the complementary function is: \(x_c(t) = e^{-2t}(A \cos 4t + B \sin 4t)\) Next, we find the particular solution, \(x_p(t)\), by considering a trial solution: \(x_p(t) = C \sin 2t + D \cos 2t\) Differentiating \(x_p(t)\) once and then twice: \(\frac{dx_p}{dt} = 2C \cos 2t - 2D \sin 2t\) \(\frac{d^2x_p}{dt^2} = -4C \sin 2t - 4D \cos 2t\) Now, we substitute \(x_p(t)\), \(\frac{dx_p}{dt}\), and \(\frac{d^2x_p}{dt^2}\) into the given differential equation: \(-4C \sin 2t - 4D \cos 2t + 8C \cos 2t + 8D \sin 2t + 20(C \sin 2t + D \cos 2t) = 10 \sin 2t\) By equating the coefficients of the sine and cosine terms, we get: \((-4 + 20)C + 8D = 10\) \(8C + (-4 + 20)D = 0\) Solving this system of linear equations, we find that \(C = 1/4\) and \(D = -1/4\). So, the particular solution is: \(x_p(t) = \frac{1}{4}(\sin 2t - \cos 2t)\) The general solution is the sum of the complementary function and the particular solution: \(x(t) = e^{-2t}(A \cos 4t + B \sin 4t) + \frac{1}{4}(\sin 2t - \cos 2t)\) Now, we use the initial conditions: - The weight is initially pulled down 6 in (0.5 ft) below the equilibrium position: \(x(0)=-0.5\). - The weight is initially at rest: \(\frac{dx}{dt}(0) = 0\). Applying the initial conditions, we get: \(-0.5 = A + \frac{-1}{4}\) and \(0 = B - 2 + \frac{-2}{4}\) Solving, we find that \(A = -\frac{1}{4}\) and \(B = \frac{3}{2}\). The displacement x(t) as a function of time is: \(x(t) = e^{-2t}(-\frac{1}{4}\cos 4t + \frac{3}{2}\sin 4t) + \frac{1}{4}(\sin 2t - \cos 2t)\)

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Key Concepts

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

Second-Order Differential Equations
Second-order differential equations are equations involving a function and its derivatives, where the highest derivative is the second. These equations model various physical phenomena, such as the motion of mechanical systems, circuits, and even some natural processes. In our exercise, we deal with a spring-mass system described by a second-order differential equation. The general form of such an equation is given by:\[\frac{d^2x}{dt^2} + a \frac{dx}{dt} + bx = f(t)\] where \(a\) and \(b\) are constants, \(f(t)\) is a given function, and \(x\) is the function that describes the system's behavior over time.

Second-order differential equations can represent systems with acceleration, such as a mass on a spring where the force depends on position, velocity, and external influences.
Nonhomogeneous Equations
A nonhomogeneous differential equation, like the one in our exercise, includes a non-zero function on the right side, representing external forces or inputs:\[\frac{d^2x}{dt^2} + 4 \frac{dx}{dt} + 20x = 10 \sin 2t\]This external input can represent forces that disrupt the system's natural behavior. In the spring-mass system, the term \(10 \sin 2t\) reflects an external oscillating force due to the vertical movement of the spring's support.

To solve such an equation, we find both the complementary function, which solves the associated homogeneous equation (setting the right side to zero), and a particular solution that fits the nonhomogeneous part. Both solutions are combined to understand the system's behavior under the influence of external factors.
Initial Value Problems
Initial value problems require solving a differential equation with given conditions at the start, such as position and velocity. These initial conditions uniquely determine which of the infinite possible solutions of a differential equation fits the physical scenario.

For our system, the initial conditions are:
  • The initial displacement from equilibrium: \(x(0) = -0.5\) ft
  • The initial velocity: \(\frac{dx}{dt}(0) = 0\)
These conditions describe where the mass starts and how fast it is moving, ensuring that our solution reflects the actual behavior from the beginning of observation. First, find the general solution, then substitute these conditions to identify the specific constants that fit these criteria.
Spring-Mass System
The spring-mass system models how a mass attached to a spring moves when subjected to forces. For our exercise, this involves the interplay of several forces: the spring force returning the mass toward equilibrium, resistive forces slowing it down, and external forces such as the oscillating support.

Key aspects of a spring-mass system include:
  • Mass (\(m\)): Affects how quickly the system responds to forces.
  • Spring constant (\(k\)): Describes the spring's stiffness, determining how strong the restoring force is. Here, \(k = 10\ lb/ft\).
  • Damping: Resistive forces modeled by terms like \(-2 \frac{dx}{dt}\). This reflects how the velocity affects the motion, slowing down the mass's movement.
  • External Forces: In this case, the \(\frac{1}{2} \sin 2t\), which reflects the motion of the spring's support.
These aspects govern complex motions that can be simplified and anticipated through differential equations, showing the interconnectedness of mathematics and physical reality.

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Most popular questions from this chapter

A simple pendulum is composed of a mass \(m\) (the bob) at the end of a straight wire of negligible mass and length \(L\). It is suspended from a fixed point \(S\) (its point of support) and is free to vibrate in a vertical plane (sec Figure 5.4). Let SP denote the straight wire; and let \(\theta\) denote the angle that \(S P\) makes with the vertical \(S P_{0}\) at time \(t\), positive when measured counterclock wise. We neglect air resistance and assume that only two forces act on the mass \(m: F_{1}\), the tension in the wire; and \(F_{2}\), the force due to gravity, which acts vertically downward and is of magnitude mg. We write \(F_{2}=F_{r}+F_{N}\), where \(F_{r}\) is the component of \(F_{2}\) along the tangent to the path of \(m\) and \(F_{N}\) is the component of \(F_{2}\) normal to \(F_{r}\). Then \(F_{N}=-F_{1}\) and \(F_{\mathrm{r}}=-m g \sin \theta\), and so the net force acting on \(m\) is \(F_{1}+F_{2}=F_{1}+F_{r}+F_{N}=-m g \sin \theta\), along the arc \(P_{0} P\). Letting s denote the length of the arc \(P_{0} P\), the acceleration along this arc is \(d^{2} s / d t^{2} .\) Hence applying Newton's second law, we have \(m d^{2} s / d t^{2}=-m g \sin \theta\) But since \(s=1 \theta\), this reduces to the differential equation $$ m l \frac{d^{2} \theta}{d t^{2}}=-m g \sin \theta \quad \text { or } \quad \frac{d^{2} \theta}{d t^{2}}+\frac{g}{l} \sin \theta=0 . $$ (a) The equation $$ \frac{d^{2} \theta}{d t^{2}}+\frac{g}{l} \sin \theta=0 $$ is a nonlinear second-order differential equation. Now recall that $$ \sin \theta=\theta-\frac{\theta^{3}}{3 !}+\frac{\theta^{5}}{5 !}-\cdots \cdot $$ Hence if \(\theta\) is sufficiently small, we may replace \(\sin \theta\) by \(\theta\) and consider the approximate linear equation $$ \frac{d^{2} \theta}{d t^{2}}+\frac{g}{l} \theta=0 $$ Assume that \(\theta=\theta_{0}\) and \(d \theta / d t=0\) when \(t=0\). Obtain the solution of this approximate equation that satisfies these initial conditions and find the amplitude and period of the resulting solution. Observe that this period is independent of the initial displacement. (b) Now return to the nonlinear equation $$ \frac{d^{2} \theta}{d t^{2}}+\frac{g}{l} \sin \theta=0 $$ Multiply through by \(2 d \theta / d t\), integrate, and apply the initial condition \(\theta=\theta_{\text {o. }}\) \(d \theta / d t=0 .\) Then separate variables in the resulting equation to obtain $$ \frac{d \theta}{\sqrt{\cos \theta-\cos \theta_{0}}}=\pm \sqrt{\frac{2 g}{l}} d t . $$ From this equation determine the angular velocity \(d \theta / d t\) as a function of \(\theta\). Note that the left member cannot be integrated in terms of elementary functions to obtain the exact solution \(\theta(t)\) of the nonlinear differential equation.

A 12-lb weight is attached to the lower end of a coil spring suspended from the ceiling. The weight comes to rest in its equilibrium position thereby stretching the spring 6 in. Beginning at \(t=0\) an external force given by \(F(t)=2 \cos \omega t\) is applied to the system. (a) If the damping force in pounds is numerically equal to \(3(d x / d t)\), where \(d x / d t\) is the instantaneous velocity in feet per second, determine the resonance frequency of the resulting motion and find the displacement as a function of the time when the forcing function is in resonance with the system. (b) Assuming there is no damping, determine the value of \(\omega\) that gives rise to undamped resonance and find the displacement as a function of the time in this case.

A coil spring having spring constant \(20 \mathrm{lb} / \mathrm{ft}\) is suspended from the ceiling. A \(32 \cdot \mathrm{b}\) weight is attached to the lower end of the spring and comes to rest in its equilibrium position. Beginning at \(t=0\) an external force given by \(F(t)=\) \(40 \cos 2 t\) is applied to the system. This force then remains in effect until \(t=\pi\), at which instant it ceases to be applied. For \(t>\pi\), no external forces are present. The medium offers a resistance in pounds numerically equal to \(4(d x / d t)\), where \(d x / d t\) is the instantancous velocity in feet per second. Find the displacement of the weight as a function of the time for all \(t \geq 0 .\)

A 10-lb weight is attached to the lower end of a coil spring suspended from the ceiling, the spring constant being \(20 \mathrm{lb} / \mathrm{ft}\). The weight comes to rest in its equilibrium position. It is then pulled down 6 in. below this position and released at \(t=0\) with an initial velocity of \(1 \mathrm{ft} / \mathrm{sec}\), directed downward. The resistance of the medium in pounds is numerically equal to \(a(d x / d t)\), where \(a>0\) and \(d x / d t\) is the instantaneous velocity in feet per second. (a) Determine the smallest value of the damping coefficient \(a\) for which the motion is not oscillatory. (b) Using the value of \(a\) found in part (a) find the displacement of the weight as a function of the time. (c) Show that the weight attains a single extreme displacement from its equilibrium position at time \(t=\frac{1}{46}\), determine this extreme displacement, and show that the weight then tends monotonically to its equilibrium position as \(t \rightarrow \infty\). (d) Graph the displacement found in step (b).

A circuit has in series an electromotive force given by \(E(t)=E_{0} \sin \omega t \mathrm{~V}\), a resistor of \(R \Omega\), an inductor of \(L H\), and a capacitor of \(C\) farads. (a) Show that the steady-state current is $$ i=\frac{E_{0}}{Z}\left(\frac{R}{Z} \sin \omega t-\frac{X}{Z} \cos \omega t\right) $$ where \(X=L \omega-1 / C \omega\) and \(Z=\sqrt{X^{2}+R^{2}}\). The quantity \(X\) is called the reactance of the circuit and \(Z\) is called the impedance. (b) Using the result of part (a) show that the steady-state current may be written $$ i=\frac{E_{0}}{Z} \sin (\omega t-\phi) $$ where \(\phi\) is determined by the equations $$ \cos \phi=\frac{R}{Z}, \quad \sin \phi=\frac{X}{Z} $$ Thus show that the steady-state current attains its maximum absolute value \(E_{0} / Z\) at times \(t_{n}+\phi / \omega\), where $$ t_{n}=\frac{1}{\omega}\left[\frac{(2 n-1) \pi}{2}\right] \quad(n=1,2,3, \ldots) $$ are the times at which the clectromotive force attains its maximum absolute value \(E_{\mathrm{o}}\) - (c) Show that the amplitude of the steady-state current is a maximum when $$ \omega=\frac{1}{\sqrt{L C}} $$ For this value of \(\omega\) electrical resonance is said to occur. (d) If \(R=20, L=\frac{1}{4}, C=10^{-4}\), and \(E_{0}=100\), find the value of \(\omega\) that gives rise to electrical resonance and determine the amplitude of the steady-state current in this case.
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