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A 4-lb weight is hung upon the lower end of a coil spring hanging vertically from a fixed support. The weight comes to rest in its equilibrium position, thereby stretching the spring 8 in. The weight is then pulled down a certain distance below this equilibrium position and released at \(t=0 .\) The medium offers a resistance in pounds numerically equal to \(a x^{\prime}\), where \(a>0\) and \(x^{\prime}\) is the instantaneous velocity in feet per second. Show that the motion is oscillatory if \(a<\sqrt{3}\), critically damped if \(a=\sqrt{3}\), and overdamped if \(a>\) \(\sqrt{3}\).

Short Answer

Expert verified
The motion of the given spring system is oscillatory if \(a < \sqrt{96}\), critically damped if \(a = \sqrt{96}\), and overdamped if \(a > \sqrt{96}\).

Step by step solution

01

Identify Given Information

From the given problem, we can extract the following information: - Mass (m): 4 lb - Equilibrium position stretch (∆x): 8 inches = 2/3 ft - Damping resistance: \(a \cdot x'\), where \(x'\) is the instantaneous velocity in ft/s
02

Calculate Spring Constant (k)

We can calculate the spring constant (k) using Hooke's law, which states that the force exerted by a spring is proportional to its displacement from the equilibrium position: \( F = k \Delta{x} \) Since the force acting on the spring is equal to the weight, we have \( 4 = k \cdot \frac{2}{3} \) Solving for k, we get \( k = 6 \)
03

Formulate the Equations of Motion

Now, we can set up the differential equation governing this mass-spring-damper system. The equation of motion can be written as: \( m x''(t) + a x'(t) + k x(t) = 0 \) Plugging in the values we derived, we get: \( 4 x''(t) + a x'(t) + 6 x(t) = 0 \)
04

Analyze the Characteristic Equation

To solve this differential equation, we need to find the roots of the characteristic equation. The characteristic equation can be obtained by converting the above equation to a quadratic equation in terms of λ: \( 4 \lambda^2 + a \lambda + 6 = 0 \) The roots of this quadratic equation will determine the general solution of the given differential equation. Now, for the motion to be oscillatory, the roots must be complex; for critical damping, the roots must be equal; and for overdamping, the roots must be real and distinct. Here, the discriminant \(\Delta\) plays a crucial role in determining the type of roots: \( \Delta = a^2 - 4 \cdot 4 \cdot 6 \)
05

Determine the Conditions for Oscillatory, Critically Damped, and Overdamped Motion

1. For oscillatory motion, \(\Delta < 0\): \( a^2 - 96 < 0 \) \( a^2 < 96 \) Thus, the motion is oscillatory if \( a < \sqrt{96} \). 2. For critically damped motion, \(\Delta = 0\): \( a^2 - 96 = 0 \) \( a^2 = 96 \) Thus, the motion is critically damped if \( a = \sqrt{96} \). 3. For overdamped motion, \(\Delta > 0\): \( a^2 - 96 > 0 \) \( a^2 > 96 \) Thus, the motion is overdamped if \( a > \sqrt{96} \). Therefore, the motion of the given spring system is oscillatory if \(a < \sqrt{96}\), critically damped if \(a = \sqrt{96}\), and overdamped if \(a > \sqrt{96}\).

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Key Concepts

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

Damped Oscillatory Motion
Damped Oscillatory Motion is a phenomenon commonly found in mechanical systems such as the mass-spring-damper setup from our textbook problem. This type of motion is characterized by the repetitive movement around an equilibrium position, but unlike simple harmonic motion, it has a resistance that causes the amplitude of the oscillations to gradually decrease over time.

The resistance is provided by the damping force, which in our example is proportional to the velocity, as represented by the term \( a x'(t) \). The coefficient \( a \) is crucial in determining the nature of the motion. If the damping is not too strong, specifically if \( a \) is less than the square root of 3, the system will still oscillate but with diminishing amplitude. This is because the discriminant of the characteristic equation, \( a^2 - 4mk \), is negative, leading to complex roots that are responsible for oscillatory behavior.
Critical Damping
Critical damping occurs at the precise point where oscillations stop and the system returns to equilibrium in the shortest possible time without oscillating. This happens when the damping coefficient \( a \) is equal to the critical value, which in our textbook example is \( \sqrt{96} \) but was symbolized as \( \sqrt{3} \) for the initial concept explanation.

At the point of critical damping, the discriminant of the characteristic equation is zero, resulting in a repeated real root of the characteristic equation. The solution to the equation of motion in critically damped scenarios comprises terms that are linear combinations of exponential functions of time, featuring no sinusoidal components. The system reaches equilibrium as quickly as possible without overshooting.
Overdamping
Overdamping is a situation where the damping force is so large that the system returns to equilibrium without oscillating, but more slowly compared to critical damping. This occurs when the damping coefficient \( a \) is greater than \( \sqrt{96} \), equivalent to going beyond the critical damping value.

In an overdamped system, the discriminant of the characteristic equation is positive, leading to two distinct real roots. Due to the absence of complex roots, the solution to the motion equation does not involve oscillatory sin and cos functions. Instead, the system's movement towards equilibrium is an exponential decay dominated by the root with the smaller absolute value.
Characteristic Equation
The characteristic equation is a cornerstone for analyzing the behavior of differential equations, particularly in mechanical systems like our spring-damper model. By converting the second-order linear differential equation into a quadratic equation of the form \( 4\lambda^2 + a\lambda + 6 = 0 \), where \( \lambda \) corresponds to complex frequencies, we can analyze the system's motion.

The roots of the characteristic equation are determined by its discriminant, \( \Delta = a^2 - 4mk \). These roots can be real or complex and hold the key to understanding whether the system will exhibit oscillatory motion, critical damping, or overdamping. By identifying the type of roots of the characteristic equation, we can infer the nature of the system's response to displacement from equilibrium.

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Most popular questions from this chapter

A 64-lb weight is attached to the lower end of a coil spring suspended from a fixed beam. The weight comes to rest in its equilibrium position, thereby stretching the spring \(\frac{1}{3}\) foot. The weight is then pulled down 2 feet below its equilibrium position and released at \(t=0 .\) The medium offers a resistance in pounds numerically equal to \(a x^{\prime}\), where \(a>0\) and \(x^{\prime}\) is the instantaneous velocity in feet per second. (a) Find \(a\) if the resulting motion of the weight is critically damped, and show that in this case the displacement of the weight from its equilibrium position decreases monotonically for all \(t>0\).

A circuit has in series an electromotive force given by \(E(t)=5 \sin 100 t \mathrm{~V}, \mathrm{a}\) resistor of \(10 \Omega\), an inductor of \(0.05 \mathrm{H}\), and a capacitor of \(2 \times 10^{-4}\) farads. If the initial current and the initial charge on the capacitor are both zero, find the charge on the capacitor at any time \(t>0\).

A circuit has in series a constant electromotive force of \(100 \mathrm{~V}\), a resistor of \(10 \Omega\), and a capacitor of \(2 \times 10^{-4}\) farads. The switch is closed at time \(t=0\), and the charge on the capacitor at this instant is zero. Find the charge and the current at time \(t>0\)

A 6-lb weight is hung on the lower end of a coil spring suspended from the ceiling. The weight comes to rest in its equilibrium position, thereby stretching the spring 4 in. Then beginning at \(t=0\) an external force given by \(F(t)=27 \sin 4 t-3 \cos 4 t\) is applied to the system. If the medium offers a resistance in pounds numerically equal to three times the instantaneous velocity, measured in feet per second, find the displacement as a function of the time.

A 10 -lb weight is hung on the lower end of a coil spring suspended from the ceiling, the spring constant of the spring being \(20 \mathrm{lb} / \mathrm{ft}\). The weight comes to rest in its equilibrium position, and beginning at \(t=0\) an external force given by \(F(t)=10 \cos 8 t\) is applied to the system. The medium offers a resistance in pounds numerically equal to \(5 x^{\prime}\), where \(x^{\prime}\) is the instantaneous velocity in feet per second. Find the displacement of the weight as a function of the time.

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