Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

The displacement \(d\) (in meters) of an object at time \(t\) (in seconds) is given. (a) Describe the motion of the object. (b) What is the maximum displacement from its rest position? (c) What is the time required for one oscillation? (d) What is the frequency? $$ d(t)=4 \sin (2 t) $$

Short Answer

Expert verified
The motion is periodic. The max displacement is 4 meters. The period is \(\pi\) seconds, and the frequency is \(\frac{1}{\pi}\) Hz.

Step by step solution

01

- Identify the equation for displacement

The displacement equation given is:\[d(t) = 4 \sin(2t)\]
02

- Describe the motion

The given equation represents a sinusoidal motion, specifically a sine wave. This indicates that the object oscillates back and forth around its rest position in a periodic manner.
03

- Find the maximum displacement

The maximum value of \(\sin(2t)\) is 1, so the maximum displacement \(d_{max}\) is:\[d_{max} = 4 \times 1 = 4 \text{meters}\]
04

- Calculate the time period (time for one oscillation)

The general formula for the displacement is \(d(t) = A \sin(\omega t)\). Here, \(\omega\) (angular frequency) is 2. The time period \(T\) is given by:\[T = \frac{2\pi}{\omega} = \frac{2\pi}{2} = \pi \text{seconds}\]
05

- Determine the frequency

The frequency \(f\) is the reciprocal of the time period \(T\):\[f = \frac{1}{T} = \frac{1}{\pi} \text{Hz}\]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Displacement Equation
The first core concept in harmonic motion is the displacement equation. The given equation for displacement is \(d(t) = 4 \sin(2t)\). This equation tells us how the position of the object changes over time.
The variable \(d(t)\) represents the displacement at time \(t\). The expression \(4 \sin(2t)\) means that the object’s displacement follows a sine wave.
The number 4 is the amplitude, and it shows the maximum displacement, meaning the object moves 4 meters away from its rest position at the highest point.
The 2 inside the sine function (\(2t\)) is related to the angular frequency, which we will discuss later. This influences how quickly the object oscillates back and forth. Understanding the displacement equation is vital to grasping the behavior of oscillatory motion.
Sinusoidal Motion
Sinusoidal motion describes a type of repetitive, wave-like movement, commonly represented by sine or cosine functions. In our example, the displacement equation \(d(t) = 4 \sin(2t)\) represents sinusoidal motion.
This means the object moves in a specific, smooth, and periodic pattern. Think of how a pendulum swings back and forth or how a mass on a spring moves up and down.
Sinusoidal motion is essential in various phenomena, including mechanical vibrations, sound waves, and even electrical signals.
The key takeaway is that the object's motion is predictable and repeats itself after a constant interval. Recognizing sinusoidal motion helps in understanding the object's behavior over time.
Angular Frequency
Angular frequency, represented by the Greek letter \(\omega\) (omega), plays a critical role in determining how rapidly the object oscillates. In the equation \(d(t) = 4 \sin(2t)\), the angular frequency \(\omega\) is 2.
Angular frequency links directly to the periodic nature of the motion. Its unit is radians per second (rad/s).
To understand how quickly the object completes a cycle, it's helpful to compare angular frequency to the more familiar concept of frequency, which we'll cover in a bit. The formula \(\omega = 2\pi f\) demonstrates that angular frequency and frequency are directly related.
Thus, a higher angular frequency means faster oscillations. In our example, \(\omega = 2\) implies moderate speed, ensuring we can see clear cycles of motion.
Time Period
The time period, denoted as \(T\), shows how long it takes for the object to complete one full cycle of motion. Calculating the time period involves the formula \(T = \frac{2\pi}{\omega}\).
Using our previously identified angular frequency of 2, we get \(T = \frac{2\pi}{2} = \pi\) seconds. This value means the object returns to its starting position every \(\pi\) seconds.
The time period is inversely related to frequency, as the faster an object oscillates, the shorter each individual cycle becomes. Knowing the time period allows us to predict when the object will reach specific points in its path.
Frequency
Frequency, represented as \(f\), indicates how many cycles an object completes per second. It's measured in Hertz (Hz). Calculating frequency involves taking the reciprocal of the time period: \(f = \frac{1}{T}\).
Using our time period of \(\pi\) seconds, we find \(f = \frac{1}{\pi}\) Hz, meaning the object completes about \(0.318\) cycles every second.
Higher frequencies imply more rapid oscillations. Frequency is crucial in everyday applications like tuning musical instruments and optimizing communication signals.
Understanding both the time period and frequency offers a complete picture of the object's motion, enabling precise predictions and applications in practical scenarios.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Colossus Added to Six Flags St. Louis in \(1986,\) the Colossus is a giant Ferris wheel. Its diameter is 165 feet; it rotates at a rate of about 1.6 revolutions per minute; and the bottom of the wheel is 15 feet above the ground. Find a function that relates a rider's height \(h\) above the ground at time \(t\). Assume the passenger begins the ride at the bottom of the wheel.

In Problems 23-26, graph each damped vibration curve for \(0 \leq t \leq 2 \pi\). $$ d(t)=e^{-t / \pi} \cos (2 t) $$

Clint is building a wooden swing set for his children. Each supporting end of the swing set is to be an A-frame constructed with two 10 -foot-long 4 by 4 's joined at a \(45^{\circ}\) angle. To prevent the swing set from tipping over, Clint wants to secure the base of each A-frame to concrete footings. How far apart should the footings for each A-frame be?

In Problems \(47-52,\) the function \(d\) models the distance (in meters) of the bob of a pendulum of mass \(m\) (in kilograms) from its rest position at time \(t\) (in seconds) is given. The bob is released from the left of its rest position and represents a negative direction. (a) Describe the motion of the object. Be sure to give the mass and damping factor. (b) What is the initial displacement of the bob? That is, what is the displacement at \(t=0 ?\) (c) Graph the motion using a graphing utility. (d) What is the displacement of the bob at the start of the second oscillation? (e) What happens to the displacement of the bob as time increases without bound? $$ d(t)=-20 e^{-0.7 t / 40} \cos \left(\sqrt{\left(\frac{2 \pi}{5}\right)^{2}-\frac{0.49}{1600}} t\right) $$

Solve each triangle. $$ a=6, \quad b=4, \quad C=60^{\circ} $$

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free