Question

A 10-kg object suspended from the end of a vertically hanging spring stretches the spring \(9.8 \mathrm{~cm}\). At time \(t=0\), the resulting spring-mass system is disturbed from its rest state by the given applied force, \(F(t)\). The force \(F(t)\) is expressed in newtons and is positive in the downward direction; time is measured in seconds. (a) Determine the spring constant, \(k\). (b) Formulate and solve the initial value problem for \(y(t)\), where \(y(t)\) is the displacement of the object from its equilibrium rest state, measured positive in the downward direction. (c) Plot the solution and determine the maximum excursion from equilibrium made by the object on the \(t\)-interval \(0 \leq t<\infty\) or state that there is no such maximum. $$ F(t)=\left\\{\begin{array}{cl} 20, & 0 \leq t \leq \frac{\pi}{2} \\ 0, & \frac{\pi}{2}

Short answer

(a) Find the spring constant, k. k = 1000 N/m. (b) Solve the initial value problem for y(t), the displacement from equilibrium. (1) For 0 ≤ t ≤ π/2, the equation of motion is: \(10 \frac{d^2y}{dt^2} + 1000y = 1000 \sin(t)\) (2) For π/2 < t < ∞, the equation of motion is: \(10 \frac{d^2y}{dt^2} + 1000y = 0\) Using the initial conditions y(0) = 0 and dy/dt(0) = 0, and solving for y(t), we get: (1) For 0 ≤ t ≤ π/2, y(t) = -0.01(10cos(t)-10-9sin(t)) (2) For π/2 < t < ∞, y(t) = -0.01(1000sin(t-π/2)e^(-100t)) (c) Plot the solution and find the maximum excursion. Upon plotting the displacement function y(t) from 0 to ∞, we observe that the maximum excursion from its equilibrium position occurs at t = π/2, and thus the maximum displacement is: y(π/2) ≈ -0.029 meters. The maximum excursion from the equilibrium is approximately 2.9 cm.

Step-by-step solution

Find the equilibrium extension of the spring

The spring stretches 9.8 cm = 0.098 m when a 10-kg mass is suspended from it.

Use Hooke's Law

Hooke's Law states \(F = -kx\). In equilibrium, the force due to gravity equals the spring force. Therefore, \(F_{gravity}=mg\), where \(m\) is the mass of the object and \(g\) is gravitational acceleration (approximately 9.81 m/s^2). So, \(mg=-kx\). We need to solve for \(k\). $$ k = \frac{mg}{x} = \frac{10 \times 9.81}{0.098} \approx 1000 \, N/m $$ **Step 2: Formulate and solve the initial value problem for y(t)**

Write down the equation of motion

The equation of motion is given by \(m\frac{d^2y}{dt^2} + ky = F(t)\), where \(m\) is the mass (10 kg), \(k\) is the spring constant (1000 N/m) and \(F(t)\) is the applied force.

Solve the equation of motion for two cases

We have to solve the given equation of motion separately for two cases: (1) \(0\leq t\leq \frac{\pi}{2}\), and (2) \(\frac{\pi}{2}<t<\infty\). Solve the equation with the given force function for each case and find an expression for y(t).

Apply the initial conditions

At \(t=0\), the object is in equilibrium, so \(y(0)=0\) and \(\frac{dy}{dt}(0)=0\). We will use these initial conditions to find the constants of integration. **Step 3: Plot the solution and determine the maximum excursion from equilibrium**

Plot the solution

Plot the displacement function y(t) found in Step 2 for the interval \([0,\infty)\).

Find the maximum excursion

Determine the maximum excursion from equilibrium by analyzing the plot and finding the maximum value of y(t) on the given interval. If there is no maximum excursion, state that there is no such maximum.

Key concepts

Understanding Differential Equations

Differential equations are fundamental in math and physics because they describe how things change over time, which is essential for understanding real-world phenomena. A classic example is the motion of a mass attached to a spring. The equation tells us how the position of the mass changes in response to various forces, and finding this equation is central to solving the problem at hand.

Here, the differential equation models the motion of a 10-kg object on a spring. With the given force, we can predict the object's position at any time. Understanding how to set up and solve these equations is critical in fields ranging from engineering to economics.

Spring Constant Calculation

The spring constant, symbolized as k, is a measure of a spring's stiffness. A simple experiment where a known weight is suspended from the spring can find this value. By measuring how much the spring stretches, we apply Hooke's Law to calculate k. The larger k is, the stiffer the spring. This constant is crucial in solving problems involving spring motion, as it affects how the spring and attached mass move.

The calculation used in the solution takes gravity (9.81 m/s²) and the weight's extension of the spring (0.098 m) to find a spring constant of roughly 1000 N/m, which tells us this particular spring is relatively stiff.

Grasping Hooke's Law

What is Hooke's Law?

Hooke's Law is the principle that the force needed to extend or compress a spring is proportional to the distance it is stretched or compressed. Mathematically, it is expressed as F = -kx, where F is the force applied to the spring, x is the displacement of the spring from its equilibrium position, and k is the spring constant.

In this context, the law helps us determine the spring's resistance to the weight of the object. Given the equilibrium state, when the gravitational force equals the restorative spring force, we can determine the constant for a specific spring.

Equation of Motion

Developing the Equation

The equation of motion for an object on a spring is derived from Newton's second law of motion, which states that the sum of the forces is equal to the mass of the object times its acceleration. Here, the equation is m \( \frac{d^2y}{dt^2} \) + ky = F(t).

The spring's force is characterized by Hooke's Law, described above, while the applied force F(t) can change with time. In our problem, we consider two scenarios for F(t), leading to different forms of the equation of motion. This forms the basis for predicting how the object will move over time.

Maximum Excursion

Finding the Peak

The maximum excursion is the greatest distance the object moves from the equilibrium position during its motion. Identifying this point in a spring-mass system tells us how far the object will travel under specific forces. After plotting the displacement y(t) against time, we can examine the graph to find this maximum value.

In our problem, the plotted solution should show us oscillation patterns if they exist. The highest peak reached in the plot represents the maximum excursion, denoting the greatest extent of stretch or compression the spring undergoes due to the applied force.