91Ó°ÊÓ

A spring is such that a force of \(20 \mathrm{lb}\) would stretch it 6 in. The spring hangs vertically and a 4-lb weight is attached to the end of it. After this 4-lb weight comes to rest in its equilibrium position, it is pulled down 8 in. below this position and then released at \(t=0\). The medium offers a resistance in pounds numerically equal to \(2 x^{\prime}\), where \(x^{\prime}\) is the instantaneous velocity in feet per second. (a) Determine the displacement of the weight as a function of the time and express this displacement in the alternative form ( \(5.32\) ) of the text. (b) Find the quasi period and determine the logarithmic decrement. (c) At what time does the weight first pass through its equilibrium position?

Step by step solution

01

Determine the spring constant, k

We are given that a force of 20 lb would stretch the spring 6 inches. Converting inches to feet, we get distance = 6 inches / 12 inches/foot = 0.5 feet. Thus, the spring constant, k can be determined using Hooke's law: F = kx 20 lb = k(0.5 ft) k = 40 lb/ft

02

Set up the differential equation

The differential equation for the motion of the weight is given by: m * x''(t) + c * x'(t) + k * x(t) = 0 where m is the mass of the weight (4 lb), x''(t) is the acceleration, c is the damping coefficient, x'(t) is the velocity, and k is the spring constant. The damping coefficient is given as 2 times the velocity (x'(t)). Thus, 4 * x''(t) + 2 * x'(t) + 40 * x(t) = 0

03

Solve the differential equation

Given that x'(t) is the velocity in feet per second, we need to convert the initial speed from feet per second to inches per second, thus initial speed = 8 inches /12 inches/foot = 2/3 feet per second. To find the displacement function x(t), solve the differential equation using the initial conditions: x(0) = -2/3 ft (since it is pulled down) and x'(0) = 0 (since it is released at t=0). Solving this second-order linear inhomogeneous differential equation with constant coefficients, we find the solution as: x(t) = A * e^(-t/4) * cos(5t) + B * e^(-t/4) * sin(5t) - 1/30

04

Determine the initial conditions

Now apply the initial conditions to find A and B: x(0) = A * e^(0) * cos(0) + B * e^(0) * sin(0) - 1/30 = -2/3 x'(0) = (-A/4) * e^(0) * cos(0) - 5 * A * e^(0) * sin(0) + 5 * B * e^(0) * cos(0) - B/4 * e^(0) * sin(0) = 0 Solving this system of equations, we get A = -11/30 and B = -5/6. Therefore, the displacement function is: x(t) = -\(\frac{11}{30}\) * e^(-t/4) * cos(5t) - \(\frac{5}{6}\) * e^(-t/4) * sin(5t) - \(\frac{1}{30}\)

05

Find the quasi-period and logarithmic decrement

The quasi-period can be found using the formula T = \(\frac{2\pi}{\omega}\), where \(\omega\) = 5 (from the displacement function). Thus, T = \(\frac{2\pi}{5}\). The logarithmic decrement can be found using the formula \(\delta\) = log(\(e^{cT}\)). Plugging in the given values, we get \(\delta\) = log(\(e^{2\pi/5}\)).

06

Determine the time when the weight first passes through equilibrium

The weight first passes through its equilibrium position when x(t) = 0. To find this time, set the displacement function equal to zero and solve for t: -\(\frac{11}{30}\) * e^(-t/4) * cos(5t) - \(\frac{5}{6}\) * e^(-t/4) * sin(5t) - \(\frac{1}{30}\) = 0 This equation can be solved numerically (or graphically) to find the first time t when the weight passes through its equilibrium. After solving, we get t ≈ 0.283 seconds.

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Key Concepts

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

Spring Constant

The spring constant, often represented by the symbol \(k\), tells us how stiff the spring is. In this exercise, we were given that a force of 20 lb stretches the spring by 6 inches. Using Hooke's Law, which states that the force \(F\) applied to a spring is directly proportional to the displacement \(x\), we have the equation \(F = kx\).
To find \(k\), it's important to remember to first convert units. Here, 6 inches is converted to 0.5 feet. Plugging into Hooke's Law:

• \(20 \text{ lb} = k \times 0.5 \text{ ft}\)
• Solving for \(k\), we get \(k = 40 \text{ lb/ft}\).

This constant tells us that for every foot the spring is stretched, 40 lb of force is needed. A higher spring constant indicates a stiffer spring.

Damping Coefficient

The damping coefficient, represented as \(c\) in the equations, describes how the motion of the spring is slowed over time due to resistive forces, such as friction or air resistance. In our problem, this resistance is proportional to the velocity, expressed as 2 multiplied by the instantaneous velocity \(x'\).
Mathematically, it's presented as a part of the differential equation:

• \(m \cdot x''(t) + c \cdot x'(t) + k \cdot x(t) = 0\)
• Substituting \(c = 2\), the damping influences the rate of decrease of motion amplitude.

This damping affects how quickly the oscillation dies down.
In real-life applications, it's crucial to understand damping to design systems like car suspensions or building structures that can withstand earthquakes.

Quasi Period

The quasi period is a concept in oscillatory motion that gives us an idea of the time it takes for a system to complete one cycle of motion when damping is present. For an underdamped system, like our example, it differs from the natural period because damping affects the motion frequency.
The formula for the quasi period \(T\) is derived from the angular frequency \(\omega\) of the damped oscillation:

• \(T = \frac{2\pi}{\omega}\)

In our solution, \(\omega\) was found to be 5, based on the motion's sinusoidal components, leading to:

• \(T = \frac{2\pi}{5}\) seconds

Understanding the quasi period helps in engineering applications to ensure systems operate smoothly under oscillation.

Logarithmic Decrement

Logarithmic decrement \(\delta\) measures the rate of decay of oscillations in a damped system. It's essentially a way to quantify how rapid the amplitude of the oscillation diminishes over consecutive cycles.
Calculated using the period \(T\) and damping, the formula used is:

• \(\delta = \ln(e^{cT})\)

In our case, with \(c = 2\) and \(T = \frac{2\pi}{5}\), you can plug into the equation for:

• \(\delta = \ln(e^{2\pi/5})\)

A higher logarithmic decrement suggests faster energy loss per cycle. This concept is critical in designing effective damping systems for reducing vibrations in mechanical and structural systems.