Problem 14 - Algebra 2

Question

PHYSICS For Exercises \(13-15\) , use the following information. A weight its attached to a spring and suspended from the ceiling. At equilibrium, the weight is located 4 feet above the floor. The weight is pulled down 1 foot and released. Write the equation for the height \(h\) of the weight above the floor as a function of time \(t\) seconds.

Step-by-Step Solution

Answer

The height function is: \(h(t) = 4 + \cos(\omega t + \pi)\).

Step 1: Understand the Problem

We have a spring with a weight attached to it, which is at equilibrium 4 feet above the ground. If the weight is disturbed by pulling it down 1 foot, it will oscillate up and down over time. We need to write an equation to describe the height as a function of time.

Step 2: Determine the Spring's Rest Position

The spring's rest, or equilibrium, position for the weight is given as 4 feet above the floor. This is the height when the spring is not disturbed (i.e., not stretched or compressed).

Step 3: Identify Amplitude and Initial Conditions

The amplitude is how far the weight is pulled from its rest position. Since it is pulled 1 foot down, the amplitude, A, is 1 foot. This means initially, at time zero, the height will be 4 - 1 = 3 feet.

Step 4: Identify the Nature of Motion

The weight exhibits simple harmonic motion since it's oscillating about an equilibrium point after being pulled. This oscillation can be described by a sine or cosine function.

Step 5: Write the General Equation for Oscillation

The general form of the equation for oscillation due to simple harmonic motion is given by: \(h(t) = C + A \cos(\omega t + \phi)\) where C is the equilibrium position (4 ft), A is the amplitude (1 ft), \(\omega\) is the angular frequency, and \(\phi\) is the phase shift.

Step 6: Apply Initial Conditions

At time t = 0, the weight is at 3 feet: \(h(0) = 3 = 4 + 1 \cos(\phi)\). Solving for \(\phi\), we find \(\phi = \pi\) because \(\cos(\pi) = -1\).

Step 7: Finalize the Equation

The equation becomes \(h(t) = 4 + 1 \cos(\omega t + \pi)\). Assuming the typical period (taking \(\omega\) based on the period if given), we complete the function but given no exact frequency or period, we leave \(\omega\) until specified. Generally, the form could be simplified as \(\omega = 2\pi/T\) if we know period T.

Key Concepts

Amplitude

In simple harmonic motion, amplitude refers to the maximum distance an object moves from its equilibrium position. It represents the extent of oscillation. For our exercise, the weight is pulled 1 foot down from its equilibrium position. Thus, the amplitude is 1 foot. Simple harmonic motion is characterized by a regular back and forth movement, and the amplitude shows us the furthest point reached from the central, or equilibrium, position.

It's crucial to know the amplitude because it tells how much energy is stored in the motion. Greater the amplitude, the more vigorous the oscillation. In our problem, a 1-foot amplitude suggests a controlled, predictable motion that can be perfectly captured with a mathematical equation.

Equilibrium Position

The equilibrium position in oscillatory motion, like that described in our problem, is the point where the forces acting on the weight are balanced. For the weight attached to the spring, this position is 4 feet above the floor. This is where the spring neither stretches nor compresses due to extra forces.

Understanding the equilibrium position is vital because it serves as the baseline from which all oscillations are measured. Every movement of the weight before or after released energy swings around this midpoint. It's the typical rest position of the system when undisturbed.

Oscillation Equation

The oscillation equation models how the object moves over time, reflecting simple harmonic motion. Such an equation is commonly expressed in the form: \[ h(t) = C + A \cos(\omega t + \phi) \] where:

• \(h(t)\) is the height of the weight above the ground at time \(t\).
• \(C\) is the equilibrium position, which is 4 feet in this problem.
• \(A\) is the amplitude, here 1 foot.
• \(\omega\) is the angular frequency, which ties to how quickly the oscillations occur.
• \(\phi\) is the phase shift, affecting the initial position of the oscillator.

This mathematical formula shows how the height changes as time progresses. It enables predictions about the object's behavior in future time steps.

Phase Shift

Phase shift in the wave equation impacts where the oscillation starts. For our task, the weight starts 1 foot below equilibrium, marking an initial condition needing a phase shift. With the weight being at 3 feet when released, initially, the cosine function must show -1, leading to \[ \phi = \pi \] because \(\cos(\pi) = -1\).

This phase angle adjusts the wave's starting point, ensuring the model matches the real-world scenario observed. Without proper phase shift, the equation would not accurately reflect when and where the weight was first displaced. Understanding phase shifts helps tweak the standard equation to fit distinct situations.