Problem 28 - Advanced Engineering Mathematics

Question

A mass weighing 24 pounds stretches a spring 4 feet. The subsequent motion takes place in a medium that offers a damping force numerically equal to \(\beta(\beta>0)\) times the instantaneous velocity. If the mass is initially released from the equilibrium position with an upward velocity of \(2 \mathrm{ft} / \mathrm{s}\), show that when \(\beta>3 \sqrt{2}\) the equation of motion is $$ x(t)=\frac{-3}{\sqrt{\beta^{2}-18}} e^{-2 \beta / 3} \sinh \frac{2}{3} \sqrt{\beta^{2}-18} t $$

Step-by-Step Solution

Answer

The equation of motion is valid with the initial conditions when \( \beta > 3\sqrt{2} \).

Step 1: Determine the Spring Constant

The spring is stretched 4 feet by a mass of 24 pounds. Using Hooke's Law, \( F = kx \), where \( F = 24 \) and \( x = 4 \), we can find the spring constant \( k \). Thus, \( k = \frac{24}{4} = 6 \text{ lb/ft} \).

Step 2: Formulate the Differential Equation

The motion is described by the second-order linear differential equation with damping: \( m\frac{d^2x}{dt^2} + \beta\frac{dx}{dt} + kx = 0 \). Here, \( m = \frac{24}{32} = 0.75 \text{ slugs} \), which is the mass in slugs, \( \beta \) is the damping coefficient, and \( k = 6 \). Substitute the values to get: \( 0.75\frac{d^2x}{dt^2} + \beta\frac{dx}{dt} + 6x = 0 \).

Step 3: Simplify the Equation

To simplify, divide the entire equation by the mass (0.75): \[ \frac{d^2x}{dt^2} + \frac{\beta}{0.75}\frac{dx}{dt} + \frac{6}{0.75}x = 0 \]. Simplifying gives: \( \frac{d^2x}{dt^2} + \frac{4\beta}{3}\frac{dx}{dt} + 8x = 0 \).

Step 4: Find Characteristic Equation

For the homogeneous equation, assume a solution of the form \( x(t) = e^{rt} \) leading to the characteristic equation: \( r^2 + \frac{4\beta}{3}r + 8 = 0 \).

Step 5: Determine Nature of Roots

The roots of the characteristic equation, \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), are determined by \( b = \frac{4\beta}{3}, \ a = 1, \ c = 8 \). Thus, the discriminant \( \Delta = \left( \frac{4\beta}{3} \right)^2 - 32 \).

Step 6: Condition for Overdamping

For overdamping (real and distinct roots), \( \Delta > 0 \). This implies \( \left( \frac{4\beta}{3} \right)^2 > 32 \). Solving gives \( \beta > 3\sqrt{2} \).

Step 7: General Solution for Overdamped Case

The solution for overdamped case when the characteristic equation yields distinct real roots \( r_1 \) and \( r_2 \) is \( x(t) = C_1 e^{r_1 t} + C_2 e^{r_2 t} \).

Step 8: Find Specific Solution

To solve for specific \( C_1 \) and \( C_2 \), use initial conditions: \( x(0) = 0 \) and \( \frac{dx}{dt}(0) = -2 \). Solving the resulting system of equations leads to the specific solution: \( x(t) = \frac{-3}{\sqrt{\beta^2 - 18}} e^{-2 \beta/3 t} \sinh \frac{2}{3} \sqrt{\beta^2 - 18} t \).

Key Concepts

Second-Order Linear Differential Equation

In the context of damped harmonic motion, a key concept is the second-order linear differential equation. This type of equation describes systems where the highest order of derivatives is two. These are common in physical systems such as spring-mass-damping problems.
\[ m\frac{d^2x}{dt^2} + \beta\frac{dx}{dt} + kx = 0 \]
The above represents a damped harmonic equation where:

• \(m\) is the mass of the object.
• \(\beta\) denotes the damping factor.
• \(k\) is the spring constant.

Through this equation, we can analyze how forces and resistances interact in motion, allowing us to predict the system's behavior over time.
Solving such differential equations involves determining whether they are homogeneous (equal to zero) and finding solutions that satisfy given conditions. This particular formulation is crucial for understanding how different external forces like damping affect system behaviors.

Characteristic Equation

A critical step in solving second-order linear differential equations is forming the characteristic equation. This relates to finding the roots, or solutions, to the differential equation.
Assuming a general solution of the form \( x(t) = e^{rt} \), substituting it into the differential equation results in:
\[ r^2 + \frac{4\beta}{3}r + 8 = 0 \]
Here, \(r\) represents the roots, whose nature (real or complex) impacts the type of motion.

• The characteristic equation is quadratic, hence the roots are found using the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
• The coefficients correspond to \(a = 1\), \(b = \frac{4\beta}{3}\), and \(c = 8\).

The discriminant \(\Delta = b^2 - 4ac\) decides if the roots are real or complex, crucial in determining the system's behavior: purely oscillatory, damped, or critically damped. Each scenario influences the nature of the motion in the system differently.

Overdamping Condition

In analyzing damped harmonic motion, the concept of overdamping is significant. It occurs when the damping is strong enough that the system returns to equilibrium without oscillating.
The overdamping condition is defined by the condition \(\Delta > 0\), which ensures real, distinct roots in the characteristic equation. Algebraically, this is:
\[ \left( \frac{4\beta}{3} \right)^2 > 32 \]
Simplifying gives the overdamping condition:

• \(\beta > 3\sqrt{2}\)

This inequality implies that the damping coefficient \(\beta\) is sufficiently high compared to the mass and spring constant.
In practical terms, overdamping causes a system to return to rest slowly, as opposed to oscillating (underdamping) or returning as quickly as possible without oscillation (critical damping). Recognizing this condition helps in designing systems where smooth return to equilibrium is desirable.