Problem 65
Question
An object weighing \(W\) pounds is suspended from a ceiling by a steel spring. The weight is pulled downward (positive direction) from its equilibrium position and released (see figure). The resulting motion of the weight is described by the function \(y=\frac{1}{2} e^{-t / 4} \cos 4 t,\) where \(y\) is the distance in feet and \(t\) is the time in seconds \((t>0)\). (a) Use a graphing utility to graph the function. (b) Describe the behavior of the displacement function for increasing values of time \(t\).
Step-by-Step Solution
Verified Answer
The function \(y=\frac{1}{2} e^{-t / 4} \cos(4t)\) will show damped oscillations when graphed, with the oscillations decreasing over time due to the damping factor \(\frac{1}{2} e^{-t / 4}\).
1Step 1: Understanding the Function
The function \(y=\frac{1}{2} e^{-t / 4} \cos(4t)\) describes the motion of the weight. This is a damped harmonic motion function, which models situations where an oscillating motion is slowed down by a frictional (or damping) force. In this equation, the cosine term represents the oscillating nature of the motion, and the exponential term accounts for the damping effect.
2Step 2: Graphing the Function
We want to plot the function \(y=\frac{1}{2} e^{-t / 4} \cos(4t)\) for \(t > 0\). To do so, we would use a graphing utility. Values from \(t=0\) to any positive time \(t\) can be plotted. This will give us a curve which shows how the displacement \(y\) changes with time \(t\).
3Step 3: Interpreting the Graph
Once the graph is plotted, we look for trends in the function's behavior as \(t\) increases. Pay attention to whether the oscillations are increasing, decreasing or remaining constant. In this case, due to the exponential decay term \(\frac{1}{2} e^{-t / 4}\), as \(t\) gets larger, this term gets smaller (decays), leading to smaller oscillations over time.
Key Concepts
Oscillating MotionDisplacement FunctionExponential Decay
Oscillating Motion
Oscillating motion refers to the repetitive back and forth movement of an object around a central point, known as equilibrium. In the context of the given function, this motion is expressed by the term \( \cos(4t) \), where the cosine function indicates the oscillatory pattern.
The '4' inside the cosine function is the angular frequency. It signifies how quickly the oscillation cycles through its motion. A higher value would mean the object oscillates more frequently in the same amount of time.
Oscillation has a typical pattern: it starts strong with wide swings, and in the absence of damping would continue indefinitely. However, in reality, factors like air resistance cause oscillations to eventually subside.
The '4' inside the cosine function is the angular frequency. It signifies how quickly the oscillation cycles through its motion. A higher value would mean the object oscillates more frequently in the same amount of time.
Oscillation has a typical pattern: it starts strong with wide swings, and in the absence of damping would continue indefinitely. However, in reality, factors like air resistance cause oscillations to eventually subside.
Displacement Function
The displacement function tells us how far the weight is from its equilibrium position at any time \( t \). In our case, it is modeled by the equation \( y=\frac{1}{2} e^{-t / 4} \cos(4t) \).
The function gives a value of \( y \) for each \( t \), representing the current position of the weight. When plotted, this function will show the path of the oscillating motion over time.
Key points about displacement functions include:
The function gives a value of \( y \) for each \( t \), representing the current position of the weight. When plotted, this function will show the path of the oscillating motion over time.
Key points about displacement functions include:
- Balance point: The weight oscillates around a central median point, which is zero here.
- Cosine represents the upward and downward swings around the equilibrium.
Exponential Decay
Exponential decay is a crucial concept that describes how quantities decrease rapidly initially and gradually over time. In the context of our motion, it is represented by the term \( \frac{1}{2} e^{-t / 4} \).
Let's break it down:- The term \( e^{-t / 4} \) represents the rate at which the oscillations die down. The negative exponent signifies a decrease.
- At \( t=0 \), the term is at its maximum, \( \frac{1}{2} \), and as \( t \) increases, \( e^{-t / 4} \) approaches zero.
Let's break it down:- The term \( e^{-t / 4} \) represents the rate at which the oscillations die down. The negative exponent signifies a decrease.
- At \( t=0 \), the term is at its maximum, \( \frac{1}{2} \), and as \( t \) increases, \( e^{-t / 4} \) approaches zero.
- This exponential factor multiplies the oscillating motion part, dampening its amplitude.
- Over time, as \( t \) increases, the oscillation becomes less noticeable, and the object slows to a stop.
Other exercises in this chapter
Problem 65
A ball that is bobbing up and down on the end of a spring has a maximum displacement of 3 inches. Its motion (in ideal conditions) is modeled by \(y=\frac{1}{4}
View solution Problem 65
Use trigonometric identities to transform one side of the equation into the other \((0
View solution Problem 65
Find the reference angle \(\theta^{\prime} .\) Sketch \(\theta\) in standard position and label \(\boldsymbol{\theta}^{\prime}\). $$\theta=-292^{\circ}$$
View solution Problem 65
Find the exact value of the expression. Use a graphing utility to verify your result. (Hint: Make a sketch of a right triangle.) \(\sin \left(\arctan \frac{4}{3
View solution