Problem 41 - Calculus Concepts

Question

Motion Laws When a spring is stretched and then released, it oscillates according to two laws of physics: Hooke's law and Newton's second law. These two laws combine to form the following differential equation in the case of free, undamped oscillation: $$ m \frac{d^{2} x}{d t^{2}}+k x=0 $$ where \(m\) is the mass of an object attached to the spring, \(x\) is the distance the spring is stretched beyond its standard length with the object attached (its equilibrium point), \(t\) is time, and \(k\) is a constant associated with the strength of the spring. Consider a spring with \(k=15\) from which is hung a 30 -pound weight. The spring with the weight attached stretches to its equilibrium point. The spring is then pulled 2 feet farther than its equilibrium and released. a. Write a differential equation describing the acceleration of the spring with respect to time \(t\) measured in seconds. Use the fact that mass \(=\frac{\text { weight }}{\mathrm{g}},\) where \(g\) is the gravi- tational constant, 32 feet per second per second. b. Write a particular solution for this differential equation. Use the fact that when the spring is first released, its velocity is \(0 .\) c. Graph this solution over several periods and explain how to interpret the graph. d. How quickly is the mass moving when it passes its equilibrium point?

Step-by-Step Solution

Answer

The differential equation is \( \frac{d^2 x}{dt^2} + 16x = 0 \), with solution \( x(t) = 2 \cos(4t) \). Speed at equilibrium is 8 ft/s.

Step 1: Calculate the Mass

First, we need to convert the weight of the object into mass. The object weighs 30 pounds, so we use the formula for mass: \( m = \frac{\text{weight}}{g} \). Given \( g = 32 \text{ ft/s}^2 \): \[ m = \frac{30}{32} = \frac{15}{16} \text{ slugs (units of mass in Imperial system)} \]

Step 2: Establish the Differential Equation

With \( m = \frac{15}{16} \) and \( k = 15 \), plug these values into the differential equation: \[ \frac{15}{16} \frac{d^2 x}{d t^2} + 15x = 0 \] Simplifying the equation, we get: \[ \frac{d^2 x}{dt^2} + 16x = 0 \] This is the differential equation describing the system's acceleration.

Step 3: Solve the Differential Equation

The solution to the differential equation \( \frac{d^2 x}{dt^2} + 16x = 0 \) is of the form: \[ x(t) = A \cos(\omega t) + B \sin(\omega t) \] Here, \( \omega = \sqrt{16} = 4 \), so: \[ x(t) = A \cos(4t) + B \sin(4t) \]

Step 4: Apply Initial Conditions

Initially, the spring is 2 feet stretched beyond equilibrium (\( x(0) = 2 \)), and its velocity is 0 (\( \frac{dx}{dt} \big|_{t=0} = 0 \)). Using \( x(0) = 2 \), we find: \[ A \cos(0) + B \sin(0) = 2 \] Therefore, \( A = 2 \). Differentiating \( x(t) \) gives: \( \frac{dx}{dt} = -4A \sin(4t) + 4B \cos(4t) \). At \( t = 0 \), \( -4A \cdot 0 + 4B \cdot 1 = 0 \rightarrow B = 0 \). This makes \( x(t) = 2 \cos(4t) \).

Step 5: Graph the Solution

The graph of \( x(t) = 2 \cos(4t) \) is a cosine wave with amplitude 2, period \( \frac{2\pi}{4} = \frac{\pi}{2} \), and oscillates over time. The graph will demonstrate how the displacement varies periodically around the equilibrium (x=0) point.

Step 6: Calculate Speed at Equilibrium

To find the speed as the spring passes through equilibrium, compute \( v(t) = \frac{dx}{dt} = -8 \sin(4t) \). At equilibrium points (where \( x=0 \)), \( \,\text{cos}(4t) \) must be zero, meaning \( t = \frac{\pi}{8} + n\frac{\pi}{4} \) for integer \( n \), leading to maximum speed. The speed at these points is \( v(\frac{\pi}{8}) = \pm8 \) ft/s.

Key Concepts

Hooke's Law

Hooke's Law is a fundamental principle in physics that describes how springs behave when they are stretched or compressed. It states that the force exerted by a spring is directly proportional to the distance it is stretched from its equilibrium position. This can be mathematically expressed as: \( F = -kx \). Here, \( F \) is the force exerted by the spring, \( k \) is the spring constant which measures the stiffness of the spring, and \( x \) is the displacement from the equilibrium position.

Hooke's Law only applies when the deformations are elastic, meaning the material returns to its original shape. It provides a linear relationship between force and displacement. In our exercise, the spring constant \( k = 15 \) indicates how strongly the spring resists stretching. The negative sign in Hooke's Law represents that the force exerted by the spring is in the opposite direction to the displacement.

Key Points of Hooke's Law:

• The relationship is linear and only valid for small deformations.
• The spring constant \( k \) is unique to each spring and remains constant within the elastic limit.
• Beyond the elastic limit, permanent deformation occurs, and Hooke's Law no longer applies.

Newton's Second Law

Newton's Second Law is a cornerstone of classical mechanics, providing the fundamental equation of motion that relates the net force acting on an object to its mass and acceleration. The law is commonly stated as \( F = ma \), where \( F \) is the net force, \( m \) is the mass of the object, and \( a \) is its acceleration.

In our context, when a weight is attached to a spring, the motion of the spring-mass system can be modeled using Newton's Second Law. Combining it with Hooke's Law gives us the differential equation of motion: \( mrac{d^{2} x}{d t^{2}} + kx = 0 \). Here, \( rac{d^{2} x}{d t^{2}} \) represents the mass's acceleration.

Core Concepts of Newton's Second Law:

• The acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass.
• If the forces are balanced (net force is zero), the object maintains its state of motion.
• In our case, the oscillatory motion results from the interplay of the spring's restorative force and the mass's inertia.

Oscillatory Motion

Oscillatory motion refers to a repetitive back-and-forth movement around an equilibrium position. In this context, the spring-weight system exhibits simple harmonic motion, a type of oscillatory motion where the restoring force is directly proportional to the displacement and acts in the opposite direction.

The solution to the differential equation \( rac{d^{2} x}{dt^{2}} + 16x = 0 \), explained in the exercise, mathematically describes the oscillations as \( x(t) = 2 \cos(4t) \). This indicates that the system oscillates with an amplitude of 2 feet and angular frequency of 4 rad/s.

Features of Oscillatory Motion:

• Amplitude: The maximum extent of the motion, 2 feet in our example.
• Frequency: The number of oscillations per unit time, related to the angular frequency \( \omega = 4 \).
• Period: The time taken for one complete oscillation, given by \( \frac{2\pi}{\omega} = \frac{\pi}{2} \) seconds in this exercise.
• Energy is conserved as potential energy in the spring converts to kinetic energy and vice versa.

Oscillatory motion is critical in understanding various physical systems and engineering applications such as clocks, musical instruments, and earthquake analysis.