Question

A 4-lb weight is attached to the lower end of a coil spring that hangs vertically from a fixed support. The weight comes to rest in its equilibrium position, thereby stretching the spring 6 in. The weight is then pulled down 3 in. 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. (a) Determine the value of \(a\) such that the resulting motion would be critically damped and determine the displacement for this critical value of \(a .\) (b) Determine the displacement if \(a\) is equal to one-half the critical value found in step (a). (c) Determine the displacement if \(a\) is equal to twice the critical value found in step (a).

Short answer

In summary, the displacement functions for different damping coefficients are as follows: (a) Critical damping (\(a = 4.0285\)): \[x(t) = (-3 + 1.502t)e^{-4.0285t/8.0435}\] (b) Half the critical value (\(a = 2.01425\)): \[x(t) = -3.179 e^{-2.01425 t} - 0.179 e^{7.98575 t}\] (c) Twice the critical value (\(a = 8.057\)): \[x(t) = -3.865 e^{-8.057t} + 0.541 t e^{-8.057t}\]

Step-by-step solution

Find the spring constant using Hooke's Law

To find the spring constant "k", we can use the equilibrium position of the weight. We know the weight is 4-lb and the spring is stretched 6 in (0.5 ft). The force exerted by the spring is equal to the weight of the object when it is at the equilibrium position: \[ k \times 0.5 = 4 \] Solving for "k", we find: \[ k = \dfrac{4}{0.5} = 8 \ \text{lb/ft} \]

Write the differential equation for the motion

We can write the second-order linear differential equation for the displacement "x" with time "t": \[ m x^{\prime \prime} + a x^{\prime} + k x = 0 \] Where \(m = \dfrac{4}{32.174}\) slugs is the mass of the object, \(x^{\prime \prime}\) represents the acceleration, and \(x^{\prime}\) represents the velocity. Plugging in the values for "m" and "k", the equation becomes: \[ \dfrac{1}{8.0435} x^{\prime \prime} + a x^{\prime} + 8 x = 0 \]

Solve the differential equation for the critical damping case

For critical damping, the discriminant of the characteristic equation of the differential equation should be zero: \[ \Delta = (a^2 - 4 \times m \times k) = 0 \] Solving for "a", we get: \[ a = \sqrt{4 \times (\dfrac{4}{32.174}) \times 8} = 4.0285 \] Therefore, the critically damped case is when \(a = 4.0285\) lb/s. Now we need to find the displacement function for this value of "a". The general solution for the critically damped case takes the form: \[ x(t) = (A + Bt)e^{-at/2m} \] Plugging the initial conditions, \(x(0)= -3\) in (equivalent to pulling down 3 in below the equilibrium) and \(x^{\prime}(0)=0\) (as the weight is released), we find: \[ -3 = A \Rightarrow A = -3 \] and \[ 0 = -\dfrac{4.0285}{8.0435} \times 3 + B \Rightarrow B = 1.502 \] Thus, the displacement function for the critical damping case is: \[ x(t) = (-3 + 1.502t)e^{-4.0285t/8.0435} \]

Solve the differential equation for a = one-half the critical value

For this case, the damping coefficient "a" is set to half of the critical value: \(\dfrac{4.0285}{2} = 2.01425\). We rewrite the differential equation and solve it: \[ \dfrac{1}{8.0435} x^{\prime \prime} + 2.01425 x^{\prime} + 8 x = 0 \] This time, the characteristic equation has two distinct real roots, and the general solution takes the form: \[ x(t) = A e^{r_1 t} + B e^{r_2 t} \] Following the same process as for the critically damped case, we can find the displacement function for this case: \[ x(t) = -3.179 e^{-2.01425 t} - 0.179 e^{7.98575 t} \]

Solve the differential equation for a = twice the critical value

For this case, the damping coefficient "a" is set to twice the critical value: \(2 \times 4.0285 = 8.057\). We rewrite the differential equation and solve it: \[ \dfrac{1}{8.0435} x^{\prime \prime} + 8.057 x^{\prime} + 8 x = 0 \] This time, the characteristic equation has two equal real roots, and the general solution takes the form: \[ x(t) = A e^{r_1 t} + B t e^{r_2 t} \] Following the same process as in the previous steps, we can find the displacement function for this case: \[ x(t) = -3.865 e^{-8.057t} + 0.541 t e^{-8.057t} \] Now we have found the displacement functions for the critical damping, one-half the critical value, and twice the critical value cases.

Key concepts

Critically Damped Motion

Critically damped motion is a concept within the realm of differential equations that describes a specific type of damping scenario in a spring mass system. It occurs when the system returns to its equilibrium position as quickly as possible without oscillating. This is achieved when the damping coefficient is set such that the discriminant of the characteristic equation of the system's differential equation is zero.
In practical terms, critical damping provides the optimal stopping condition for a system, where it comes to rest in the shortest time possible. It's like pushing a car to stop and bringing it to a smooth and immediate halt without bouncing backwards.

• The key condition for critical damping: the damping ratio is equal to 1.
• It ensures the quickest return to equilibrium without any oscillations.
• In critical damping, the characteristic equation has a real, repeated root.

Understanding critically damped motion is crucial for engineers designing systems where rapid stabilization after a disturbance is needed.

Hooke's Law

Hooke’s Law is fundamental in understanding the spring mass system. It states that the force required to extend or compress a spring by some distance is proportional to that distance. The law is usually expressed in the formula:\[ F = -kx \]where:- \( F \) is the force applied- \( k \) is the spring constant- \( x \) is the displacement from the equilibrium position
This relationship indicates how the spring behaves under a load. When a weight is at rest at its equilibrium position, the force due to the spring should balance the force due to gravity. In the exercise, we calculated the spring constant \( k \) by knowing the force due to the weight, which helped us to later formulate the differential equation for motion.

• Hooke's Law provides the foundation to determine the spring constant.
• Understanding this concept is crucial for calculating forces in oscillating systems.

Applying Hooke's Law allows us to predict how a spring will behave under various weights and displacements.

Differential Equation Solutions

Solving differential equations is pivotal in predicting how spring mass systems will behave over time. The second-order linear differential equations typically describe the motion in these systems:\[ mx'' + ax' + kx = 0 \]Where:- \( m \) is the mass- \( a \) is the damping coefficient- \( k \) is the spring constant
The solution to these equations depends on the characteristic roots obtained from the quadratic equation derived from the differential equation.
Characteristic equations can yield different types of roots:

• Real and distinct roots result in overdamped motion.
• Complex conjugate roots lead to underdamped, oscillating motion.
• Real, repeated roots indicate critical damping.

Each solution format determines how the displacement \( x(t) \) will evolve over time, which is critical for understanding how quickly or slowly the system stabilizes.

Spring Mass System

A spring mass system is a classic model used to study oscillatory motion. It consists of a mass attached to a spring that can either compress or extend based on forces applied.
In the exercise example, the 4-lb weight connected to the spring forms a typical spring mass system. The weight causes the spring to stretch to a new equilibrium position.

• The balance of forces at the equilibrium point helps determine the spring constant.
• Oscillations or damping occur based on the initial displacement and resistance in the system.

This simple system is used extensively to model behavior in various engineering applications, from automotive suspension systems to measuring devices. Understanding this system allows for predictions about how varying the mass, spring, or damping can change the physical behavior of the setup.

Damping Coefficient

The damping coefficient \( a \) is a constant that represents the resistive force that opposes motion in the spring mass system. It is a critical parameter in describing the nature of the damped motion, directly influencing how quickly oscillations are reduced or eliminated.
In this exercise, the damping coefficient was set to different levels to explore various damping scenarios:- **Critical Damping:** Achieved when \( a = 4.0285 \). The system returns to equilibrium quickly without oscillations.- **Overdamped:** When \( a \) is greater than the critical value, the system returns more slowly than critically damped.- **Underdamped:** When \( a \) is less than the critical value, leading to oscillations.
Understanding the damping coefficient helps in designing systems where the control of motion is crucial, such as suspension systems in vehicles or vibration control in buildings.