Problem 6 - College Trigonometry

Question

Suppose an object weighing 10 pounds is suspended from the ceiling by a spring which stretches 2 feet to its equilibrium position when the object is attached. (a) Find the spring constant \(k\) in \(\frac{\mathrm{lbs} .}{\mathrm{ft} .}\) and the mass of the object in slugs. (b) Find the equation of motion of the object if it is released from 1 foot below the equilibrium position from rest. When is the first time the object passes through the equilibrium position? In which direction is it heading? (c) Find the equation of motion of the object if it is released from 6 inches above the equilibrium position with a downward velocity of 2 feet per second. Find when the object passes through the equilibrium position heading downwards for the third time.

Step-by-Step Solution

Answer

(a) The spring constant is 5 lbs/ft and mass is 0.31 slugs. (b) First equilibrium crossing: 0.391 seconds, upward. (c) Third downward crossing: 1.57 seconds.

Step 1: Determine the Spring Constant

The spring constant \(k\) can be found using Hooke's Law, which is \(F = kx\), where \(F\) is the force applied and \(x\) is the displacement. Given, the weight of the object \(F = 10\) lbs and it stretches \(x = 2\) feet, so \(10 = k \times 2\). Solving for \(k\), we find:\[ k = \frac{10}{2} = 5 \text{ lbs/ft} \]

Step 2: Calculate the Mass of the Object

The weight of the object is 10 pounds. To find its mass in slugs, use the formula \( m = \frac{W}{g} \), where \(W\) is weight in pounds and \(g = 32.2\, \text{ft/s}^2\) is the acceleration due to gravity. Then, \(m = \frac{10}{32.2} \approx 0.31\) slugs.

Step 3: Establishing the Differential Equation of Motion (Case a)

When released from 1 foot below the equilibrium, the equation of motion is given by the D'Alembert's principle for a simple harmonic oscillator: \(m\frac{d^2y}{dt^2} + ky = 0\). Substituting \(m = 0.31\) slugs and \(k = 5\) lbs/ft:\[ 0.31 \frac{d^2y}{dt^2} + 5y = 0 \] Simplifying, we get:\[ \frac{d^2y}{dt^2} + \frac{5}{0.31}y = 0 \approx \frac{d^2y}{dt^2} + 16.13y = 0 \]

Step 4: Solve the Differential Equation (Case a)

The solution to the equation \( \frac{d^2y}{dt^2} + 16.13y = 0 \) is \( y(t) = A\cos(\sqrt{16.13}t) + B\sin(\sqrt{16.13}t) \). Given that the object is released 1 foot below the equilibrium, so initially, \( y(0) = -1 \), and it starts from rest, \( \dot{y}(0) = 0 \). These initial conditions provide:\( A = -1 \) and \( B = 0 \), so the equation is:\[ y(t) = -\cos(\sqrt{16.13}t) \]

Step 5: Find when the Object Passes Through Equilibrium (Case a)

The object passes through the equilibrium position when \( y(t) = 0 \). Solving \(-\cos(\sqrt{16.13}t) = 0\) gives:\( \cos(\sqrt{16.13}t) = 0 \Rightarrow \sqrt{16.13}t = \frac{\pi}{2} \,\text{or}\, \frac{3\pi}{2}, \ldots \) For the first crossing, when heading upwards (since the derivative changes sign from negative to positive):\[ \sqrt{16.13}t = \frac{\pi}{2} \Rightarrow t \approx \frac{\pi}{2\sqrt{16.13}} \approx 0.391 \text{ seconds} \]

Step 6: Establish Equation of Motion and Conditions (Case b)

For release from 6 inches above, initial position is \( y(0) = 0.5 \) feet above the equilibrium. Initial velocity \( \dot{y}(0) = -2 \text{ ft/s} \). The equation remains \( y(t) = A\cos(\sqrt{16.13}t) + B\sin(\sqrt{16.13}t) \). Substituting initial conditions gives:\( A = 0.5 \) and solving \( \dot{y}(0) = -2 \) gives \( B = \frac{-2}{\sqrt{16.13}} \approx -0.50 \).Thus, the equation is:\[ y(t) = 0.5\cos(\sqrt{16.13}t) - 0.50\sin(\sqrt{16.13}t) \]

Step 7: Determine Time for Third Equilibrium Passage (Case b)

The object first crosses equilibrium when \( y(t) = 0 \). Use \(0.5\cos(\sqrt{16.13}t) - 0.50\sin(\sqrt{16.13}t) = 0\), which implies:\( \tan(\sqrt{16.13}t) = 1 \). This results in \( \sqrt{16.13}t = \frac{\pi}{4} + n\pi \) for integer \(n\). To find the third downward crossing where \(n=5\), solve:\[ \sqrt{16.13}t = \frac{\pi}{4} + 5\pi \]\[ t = \frac{\pi}{4\sqrt{16.13}} + \frac{5\pi}{\sqrt{16.13}} \approx 1.57 \text{ seconds} \]

Key Concepts

Hooke's Law

When dealing with spring motion problems, understanding Hooke's Law is crucial. Hooke's Law states that the force exerted by a spring is directly proportional to the displacement from its equilibrium position. This relationship can be expressed as:\[ F = kx \]where:

• \( F \) is the force applied (usually in pounds or newtons),
• \( k \) is the spring constant (units depend on the measurement system, such as lbs/ft),
• \( x \) is the displacement of the spring from its equilibrium position.

In the given problem, a spring stretches 2 feet under a force of 10 pounds, allowing us to find the spring constant \( k \). This relationship, \( 10 = k \times 2 \), solves to \( k = 5 \) lbs/ft. Understanding this concept provides the foundation for solving how objects move when attached to springs, crucial for many physics applications.

Differential Equations in Physics

Differential equations are a key tool in describing physical phenomena, especially in motion problems. They allow us to express the relationship between various motion factors and time, such as position, velocity, and acceleration. In the context of spring motion involving a simple harmonic oscillator, we often encounter a second-order linear differential equation of the form:\[ m\frac{d^2y}{dt^2} + ky = 0 \]Here:

• \( m \) is the mass of the object (in slugs, a term for mass in the imperial system),
• \( \frac{d^2y}{dt^2} \) is the acceleration,
• \( k \) is the spring constant,
• \( y \) is the displacement from equilibrium.

This equation describes the motion of an object on a spring. Solving such an equation gives us insights into how the object will behave over time, such as oscillation frequency and amplitude. These equations are foundational for predicting the behavior of more complex physics systems.

Simple Harmonic Motion

Simple harmonic motion (SHM) relates to periodic oscillations where the restoring force is proportional to the displacement. In the context of spring problems, SHM provides a framework to understand how an object moves back and forth around its equilibrium position. The general solution for SHM can be written as:\[ y(t) = A\cos(\omega t) + B\sin(\omega t) \]where:

• \( y(t) \) is the displacement at time \( t \),
• \( A \) and \( B \) are constants determined by initial conditions,
• \( \omega \) is the angular frequency, related to the spring constant and mass as \( \omega = \sqrt{\frac{k}{m}} \).

The oscillatory nature of SHM means that an object will continuously pass through its equilibrium point in a predictable pattern. Understanding these principles helps in determining when and in which direction the object moves through the equilibrium, essential for solving problems like the ones presented.

D'Alembert's Principle

D'Alembert's principle is fundamental in transforming dynamic motion problems into static ones, making them easier to analyze with classical mechanics. It states that the sum of the differences between the forces acting on a system and the inertial forces is zero. Mathematically, for a spring system, it simplifies as:\[ m\frac{d^2y}{dt^2} + ky = 0 \]This principle effectively means that the dynamics of a system can be studied by considering the equilibrium of forces, incorporating inertial forces into the system's analysis. This reduction allows problems that involve motion to be tackled using techniques from static systems.For spring motion, D'Alembert's principle helps in deriving the familiar second-order differential equation, critical in determining the behavior of the object attached to the spring. It bridges the gap between classical mechanics and differential equations, offering a deep understanding of motion dynamics.

Initial Conditions in Physics

Initial conditions specify the state of a system at the onset of observation, forming the boundary conditions necessary to solve differential equations. For any motion problem, knowing the initial position and velocity is essential. These conditions allow us to determine the constants in the SHM equation, such as \( A \) and \( B \).For instance, if an object is released from rest at a particular position, the initial conditions might be:

• \( y(0) \) (initial position)
• \( \dot{y}(0) \) (initial velocity)

In the problem explored, different scenarios with varying initial conditions lead to distinct equations of motion. Adjusting for these initial circumstances thereafter lets us predict precisely how and when certain motion features occur, like passing through the equilibrium position. Understanding these starting points is vital for analyzing the subsequent behavior of the system.