/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 3 The position of a particle is gi... [FREE SOLUTION] | 91影视

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Chapter 15: Problem 3

The position of a particle is given by the expression \(x=(4.00 \mathrm{m}) \cos (3.00 \pi t+\pi),\) where \(x\) is in meters and \(t\) is in seconds. Determine (a) the frequency and period of the motion, (b) the amplitude of the motion, (c) the phase constant, and (d) the position of the particle at \(t=0.250 \mathrm{s}\).

Short Answer

Expert verified
The motion's frequency is 1.5 Hz, the period is 0.67 s, the amplitude is 4.00 m, the phase constant is \( \pi \) rad, and the particle's position at t=0.250 s is 0 m.

Step by step solution

01

Unnamed Variables

Let's commit the given function to different variables. \( x(t) = A \cos(\omega t + \phi) \) where \( x(t) \) is the displacement, \( A \) is the amplitude, \( \omega \) is the angular frequency, \( t \) is the time, and \( \phi \) is the phase constant. In the given function, let's compare it to the mentioned convention and name the values. The given expression is \( x = (4.00 m) \cos(3.00 \pi t + \pi) \)
02

Determining Variables

In comparing with \( x(t) = A \cos(\omega t + \phi) \), \( A = 4.00m \) is the amplitude, \( \omega = 3.00 \pi rad/sec \) is the angular frequency, and \( \phi = \pi \) is the phase constant.
03

Frequency and Period

The angular frequency \( \omega \) relates to the frequency \( f \) through the equation \( \omega = 2\pi f \) or \( f = \omega / 2\pi \). Plugging \( \omega = 3\pi \) into this latter equation gives \( f = (3\pi rad/s) / 2\pi = 1.5 Hz \). The period \( T \) is the inverse of the frequency, hence \( T = 1/f = 1/1.5s = 0.667s \).
04

Amplitude

The amplitude \( A \) is a coefficient in the equation and signifies the maximum displacement from the equilibrium position, which is \( A = 4.00m \).
05

Phase Constant

The phase constant is already given in the equation as \( \phi = \pi \) rad.
06

Position at t=0.250s

We can determine the particle's position at a given time by substituting \( t = 0.250s \) into the original equation: \( x = 4.00m \cos(3\pi*0.250s + \pi) \). Simplifying the expression inside the cosine gives \( x = 4.00m \cos(\pi/2) \), which simplifies to \( x = 4.00m * 0 = 0m \). So, at \( t = 0.250s \), the particle is at the equilibrium position.

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Key Concepts

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

Frequency and Period
Understanding the frequency and period of motion is key to analyzing simple harmonic motion. The frequency, denoted as \( f \), tells us how many complete cycles of motion occur in one second. It is quantified in Hertz (Hz). The angular frequency \( \omega \) is related to \( f \) by the equation \( \omega = 2\pi f \). Given \( \omega = 3\pi \) rad/s from the problem, we determine the frequency:
  • \( f = \frac{\omega}{2\pi} = \frac{3\pi}{2\pi} = 1.5 \) Hz.
The period \( T \) represents the time for one complete cycle. It is the inverse of frequency, calculated as:
  • \( T = \frac{1}{f} = \frac{1}{1.5} = 0.667 \) seconds.
This tells us each cycle of the particle's motion completes in approximately 0.667 seconds.
Amplitude
Amplitude refers to the maximum extent of a vibration or displacement from the equilibrium position. In simple harmonic motion, it is represented by \( A \) and can be easily observed as the coefficient before the cosine function in the equation.
  • From the expression \( x = (4.00 \text{ m}) \cos(3\pi t + \pi) \), we readily see that \( A = 4.00 \text{ m} \).
This indicates the particle oscillates 4 meters from its central rest position. The amplitude remains constant for undamped simple harmonic motion, reflecting the energy of the system.
Phase Constant
The phase constant \( \phi \) is an angle that determines the initial position of the particle at \( t=0 \). It shifts the wave along the time axis without altering its shape.
  • In our expression, \( \phi = \pi \).
This implies that for \( t = 0 \), the function \( \cos(\phi) = \cos(\pi) = -1 \), indicating it starts at its maximum negative amplitude. The phase constant is essential for solving problems where the initial conditions need precise modeling.
Position of a Particle
The position function gives us the exact location of the particle at any point in time within its oscillatory cycle. To find the position at \( t = 0.250 \text{ s} \), we substitute it into the position equation:
  • \( x = 4.00 \text{ m} \cos(3\pi(0.250 \text{ s}) + \pi) \).
Simplifying, we have:
  • Inside the cosine function, it becomes \( \cos(\frac{3\pi}{4} + \pi) = \cos(\frac{\pi}{2}) \).
Since \( \cos(\frac{\pi}{2}) = 0 \), the equation simplifies to:
  • \( x = 4.00 \text{ m} \times 0 = 0 \text{ m} \).
Thus, at \( t = 0.250 \text{ s} \), the particle is at the equilibrium position, neither displaced positively nor negatively.

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Most popular questions from this chapter

A "seconds pendulum" is one that moves through its equilibrium position once each second. (The period of the pendulum is precisely 2 s.) The length of a seconds pendulum is \(0.9927 \mathrm{m}\) at Tokyo, Japan and \(0.9942 \mathrm{m}\) at Cambridge, England. What is the ratio of the free-fall accelerations at these two locations?

The initial position, velocity, and acceleration of an object moving in simple harmonic motion are \(x_{i}, v_{i},\) and \(a_{i} ;\) the angular frequency of oscillation is \(\omega .\) (a) Show that the position and velocity of the object for all time can be written as,$$\begin{array}{l}x(t)=x_{i} \cos \omega t+\left(\frac{v_{i}}{\omega}\right) \sin \omega t \\\v(t)=-x_{i} \omega \sin \omega t+v_{i}\cos \omega t\end{array}$$ (b) If the amplitude of the motion is \(A\), show that,$$v^{2}-a x=v_{i}^{2}-a_{i} x_{i}=\omega^{2} A^{2}$$.

A \(1.00-\mathrm{kg}\) glider attached to a spring with a force constant of \(25.0 \mathrm{N} / \mathrm{m}\) oscillates on a horizontal, frictionless air track. At \(t=0\) the glider is released from rest at \(x=-3.00 \mathrm{cm} .\) (That is, the spring is compressed by \(3.00 \mathrm{cm} .)\) Find (a) the period of its motion, (b) the maximum values of its speed and acceleration, and (c) the position, velocity, and acceleration as functions of time.

A particle executes simple harmonic motion with an amplitude of \(3.00 \mathrm{cm} .\) At what position does its speed equal half its maximum speed?

After a thrilling plunge, bungee-jumpers bounce freely on the bungee cord through many cycles (Fig. P15.22). After the first few cycles, the cord does not go slack. Your little brother can make a pest of himself by figuring out the mass of each person, using a proportion which you set up by solving this problem: An object of mass \(m\) is oscillating freely on a vertical spring with a period \(T\). An object of unknown mass \(m^{\prime}\) on the same spring oscillates with a period.\(T^{\prime} .\) Determine (a) the spring constant and (b) the unknown mass.
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