Question

A 64 -lb weight is placed upon the lower end of a coil spring suspended from a rigid beam. The weight comes to rest in its equilibrium position, thereby stretching the spring \(2 \mathrm{ft}\). The weight is then pulled down \(1 \mathrm{ft}\) below its equilibrium position and released from rest at \(t=0\). (a) What is the position of the weight at \(t=5 \pi / 12\) ? How fast and which way is it moving at the time? (b) At what time is the weight 6 in. above its equilibrium position and moving downward? What is its velocity at such time?

Short answer

(a) The position of the weight at \(t = \frac{5\pi}{12}\) is approximately \(0.31\) ft below the equilibrium position, and it is moving upward at around \(3.83\) ft/s. (b) The weight is 6 inches (0.5 ft) above its equilibrium position and moving downward at a speed of \(3.83\) ft/s when \(t = \frac{\pi}{8} \,\text{s}\).

Step-by-step solution

Equation of Simple Harmonic Motion

Let's take the position of the weight below the equilibrium position as positive. Then, the equation of simple harmonic motion can be written as follows: $$ x(t) = A \cos(\sqrt{\frac{k}{m}}t + \phi) $$ Where: - \(x(t)\) represents the position of the weight at time \(t\) - \(A\) is the amplitude (maximum displacement from equilibrium) - \(k\) is the spring constant - \(m\) is the mass of the weight - \(\sqrt{\frac{k}{m}}\) is the angular frequency - \(\phi\) is the phase angle

Initial Conditions

First, we will find the values of initial conditions: 1. Equilibrium position: Spring is stretched 2 ft 2. Weight is pulled down 1 ft below its equilibrium position and held there 3. Released from rest at \(t = 0\) Let's calculate the spring constant, \(k\), using Hooke's Law: $$ k = \frac{W}{\Delta x} $$ Where: - \(W\) is the weight (64 lb) - \(\Delta x\) is the stretched length (2 ft)

Spring Constant and Mass

Now, let's find the spring constant (\(k\)) and mass (\(m\)) of the weight: $$ k = \frac{64}{2} = 32 \,\text{lb/ft} $$ Since the weight is given in lb (pounds), we can convert it to mass (slug) using the conversion \(1 \,\text{lb} = 32.2 \,\text{ft/s}^2\): $$ m = \frac{64}{32.2} = 2 \,\text{slug} $$

Angular Frequency

Next, we'll find the angular frequency (\(\sqrt{\frac{k}{m}}\)): $$ \sqrt{\frac{k}{m}} = \sqrt{\frac{32}{2}} = 4 $$

Amplitude and Phase Angle

Since the weight is released from rest 1 ft below equilibrium, the initial position \(x(0) = 1\), and the amplitude (A) equals 1. To find the phase angle (\(\phi\)), we can use the equation for simple harmonic motion at \(t = 0\): $$ 1 = 1 \cdot \cos(4 \cdot 0 + \phi) $$ This implies \(\phi = 0\).

Position and Velocity Functions

Now that we have all the parameters, we can write the position and velocity functions: Position function: $$ x(t) = 1 \cdot \cos(4t) $$ Velocity function (derivative of position function): $$ v(t) = -4 \cdot \sin(4t) $$

(a) Position and Velocity at \(t = \frac{5\pi}{12}\)

Using the position and velocity functions, we can find the position and velocity at \(t = \frac{5\pi}{12}\): Position: $$ x\left(\frac{5\pi}{12}\right) = 1 \cdot \cos\left(4 \cdot \frac{5\pi}{12}\right) \approx -0.31 \,\text{ft}\text{ (below equilibrium)} $$ Velocity: $$ v\left(\frac{5\pi}{12}\right) = -4 \cdot \sin\left(4 \cdot \frac{5\pi}{12}\right) \approx 3.83 \,\text{ft/s}\text{ (upward)} $$

(b) Position and Velocity 6 in. Above Equilibrium

Given that the weight is 6 in. (0.5 ft) above the equilibrium when it's going downward, we can solve for \(t\): $$ -0.5 = 1 \cdot \cos(4t) $$ When the weight is going downward, the velocity is positive, so we'll have: $$ v(t) > 0 \Rightarrow -4 \cdot \sin(4t) > 0 $$ Solving for time we get, $$ t = (2n - 1)\frac{\pi}{8} $$ We can find the smallest positive value for \(t\) when the weight is 6 inches (0.5 ft) above equilibrium and going downward by using \(n = 1\): $$ t = \frac{\pi}{8} \,\text{s} $$ Finally, the velocity at that time is: $$ v\left(\frac{\pi}{8}\right) = -4 \cdot \sin\left(4 \cdot \frac{\pi}{8}\right) \approx -3.83 \,\text{ft/s}\text{ (downward)} $$ Summary: (a) The position of the weight at \(t = \frac{5\pi}{12}\) is approximately \(0.31\) ft below the equilibrium position, and it is moving upward at around \(3.83\) ft/s. (b) The weight is 6 inches (0.5 ft) above its equilibrium position and moving downward at a speed of \(3.83\) ft/s when \(t = \frac{\pi}{8} \,\text{s}\).

Key concepts

Ordinary Differential Equations

Understanding the motion of an object subjected to forces can often be represented by ordinary differential equations (ODEs). In the context of Simple Harmonic Motion (SHM), ODEs play a crucial role in analyzing the system.

An ODE for SHM typically arises from applying Newton's second law of motion, which states that the force acting on an object is equal to the mass of the object multiplied by its acceleration. In the case of the weight and spring system discussed in the problem, SHM can be described by the ODE derived from Hooke's Law, relating force to the displacement from equilibrium through the spring constant. This ODE can be solved to give a function that describes the position of the weight over time.

Hooke's Law

Hooke’s Law is the principle that the force needed to extend or compress a spring by some distance is proportional to that distance. Mathematically, it's expressed as
\[ F = -kx \]
where \( F \) is the force applied, \( k \) is the spring constant, and \( x \) is the displacement of the spring from its equilibrium position. In the given exercise, Hooke's Law is utilized to calculate the spring constant \( k \), which is essential for determining the behavior of the system under oscillation.

Angular Frequency

Angular frequency, often represented by \( \omega \), is a fundamental parameter in SHM and relates to how quickly an object oscillates back and forth. It is mathematically defined as the square root of the ratio of the spring constant to the mass \( \sqrt{k/m} \). This concept is tied to the natural frequency of the system and dictates the speed at which the weight moves when released from rest. In our exercise, the angular frequency is calculated to determine the time-dependent behavior of the weight suspended on the spring.

Phase Angle

The phase angle, denoted by \( \phi \), is a parameter in SHM that indicates the initial angle of the system’s sinusoidal function at time zero. It's essential in defining the starting point of the motion. If the object is at the maximum displacement at \( t = 0 \), the phase angle will be \( 0 \) or \( \pi \), depending on the direction of displacement. In the context of the problem, the phase angle is used to determine the precise position of the weight when it is first released.

Equilibrium Position

The equilibrium position is the point at which the net force on the object is zero, meaning there's no net acceleration. For a spring-mass system, it is the point where the spring is neither compressed nor extended. This point serves as the reference for measuring displacement in SHM problems. In the solution provided, the term equilibrium is critical to understanding how the weight's oscillation is measured in relation to this rest position, including at what times the weight passes through, above, or below this equilibrium point.