Question

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 \(t>0\).

Short answer

For a critically damped motion, the damping constant \(a\) is calculated as \(a = \sqrt{4mk} = 32 \, \text{lb sec/ft}\) based on the given conditions. Furthermore, under critical damping, the displacement of the weight from its equilibrium position decreases monotonically for all \( t > 0 \) as the system returns to equilibrium without oscillating.

Step-by-step solution

Determine the spring constant \( k \)

The spring constant \( k \) is given by Hooke's law: \( F = k x \), where \( F \) is the force and \( x \) is the displacement. Given that the 64-lb weight stretches the spring \(\frac{1}{3}\) foot in its equilibrium position, we obtain: \( k = \frac{64}{1/3} = 192 \, \text{lb/ft} \).

Calculate the Mass \( m \)

The weight given is 64 lbs, which is the force exerted due to gravity. So the mass \( m \) of the object can be obtained using the equation \( W = m g \), where \( g \) is the acceleration due to gravity. Taking \( g \) as \( 32 \, \text{ft/sec}^2 \), we get \( m = \frac{W}{g} = \frac{64}{32} = 2 \, \text{slugs} \).

Calculate the value of \( a \) for critical damping

The condition for critical damping is that the damping constant \( a \) is equal to \( \sqrt{4 m k} \). Plugging in the computed values for \( m \) and \( k \), we find \( a = \sqrt{4 \times 2 \times 192} = 32 \, \text{lb sec/ft} \).

Describe the behavior of displacement in critical damping

For critically damped motion, the system returns to equilibrium as quickly as possible without oscillating back and forth about this position. This implies that the displacement from equilibrium is always decreasing monotonically once the system is released. That is, for any time \( t > 0 \), the magnitude of displacement is less than (or equal to) the magnitude of displacement at an earlier time. Hence, the displacement of the weight from its equilibrium position decreases monotonically for all \( t > 0 \).

Key concepts

Critical Damping

Critical damping is a specific condition in the study of oscillatory systems, like springs and weights, where the system returns to equilibrium as quickly as possible without oscillating. Imagine a car's shock absorber: it ideally brings the car back to its original position swiftly after hitting a bump, without bouncing. This is analogous to critical damping.
In mathematical terms, a critically damped system has a damping constant equal to \(\sqrt{4mk}\), where \(m\) is the mass and \(k\) is the spring constant.

• It is "critically" timed, which means any less damping would result in oscillations.
• Any more damping would make the return slower than necessary.

For our spring problem, critical damping ensures the weight decreases its displacement from equilibrium monotonically, meaning it reduces steadily without increasing again.

Hooke's Law

Hooke's Law describes the relationship between the force exerted on a spring and the displacement the spring experiences. It's a foundational principle in physics, particularly in mechanics and materials science. The formula is given by \( F = kx \), where \( F \) is the force applied, \( k \) is the spring constant, and \( x \) is the displacement from the rest position.

• This law helps in determining how stiff or compliant a spring is.
• The more force you apply, the more a spring stretches, provided it doesn't reach its elastic limit.

In our problem, when a 64-lb weight stretches the spring by \( \frac{1}{3} \) foot, we use Hooke's Law to find the spring constant \( k \). By rearranging the formula, we find \( k \) to be 192 lb/ft, showing how much force is required to achieve a certain stretch in the spring.

Equilibrium Position

An equilibrium position is where all forces acting on a system are balanced, leading to a state of rest. In mechanical systems, particularly those involving springs, it is the position where the spring force is equal to the gravitational force pulling on a weight.

• At equilibrium, there's no net movement because forces cancel each other out.
• If disturbed, the system may oscillate or return to this position depending on damping factors.

For the given exercise, the equilibrium position is where the weight naturally rests after stretching the spring by \( \frac{1}{3} \) foot. This is the reference point for analyzing how the spring behaves when the weight is displaced further or when subjected to damping forces.

Damping Constant

The damping constant \( a \) in oscillatory systems quantifies the resistance force opposing motion due to factors like friction or air resistance. It is a crucial parameter in determining how fast energy is lost from the system, affecting the rate at which oscillations diminish over time.

• A higher damping constant means the system quickly loses energy, reducing oscillations.
• A lower damping constant could result in prolonged, unchecked oscillations.

In our spring example, the damping constant must be chosen to achieve critical damping, where \( a = \sqrt{4mk} \). This ensures the weight returns to the equilibrium without oscillation, depicted by the constant being 32 lb sec/ft. The magnitude of \( a \) determines how the displacement decreases over time, creating a smooth transition back to equilibrium.