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A particle that is at the origin of coordinates at exactly \(t=0\) vibrates about the origin along the \(y\) -axis with a frequency of 20 \(\mathrm{Hz}\) and an amplitude of \(3.0 \mathrm{~cm}\). Write out its equation of motion in centimeters.

Short Answer

Expert verified
The equation of motion is \( y(t) = 3.0 \sin(40\pi t) \).

Step by step solution

01

Understand the Problem

The problem describes a particle vibrating along the y-axis with a given frequency and amplitude. The goal is to write the equation of motion based on this information.
02

Identify Key Parameters

The amplitude of the motion is given as 3.0 cm, and the frequency is 20 Hz. According to the problem, the motion occurs around the origin.
03

Determine the Type of Motion

The vibration suggests simple harmonic motion (SHM), which follows a sinusoidal pattern described by the equation: \[ y(t) = A \sin(\omega t + \phi) \]where \(A\) is the amplitude, \(\omega\) is the angular frequency, and \(\phi\) is the phase constant.
04

Calculate Angular Frequency

The angular frequency \(\omega\) is related to the frequency \(f\) by the equation:\[ \omega = 2\pi f \].Substitute \(f = 20\, \text{Hz}\) to find:\[ \omega = 2\pi \times 20 = 40\pi\, \text{rad/s} \].
05

Consider Initial Conditions

The particle is at the origin \(y = 0\) at \(t = 0\), suggesting a phase constant \(\phi\). Since sin(0) = 0, a suitable starting phase for a sine function is \(\phi = 0\) or \(\phi = \pi\), but \(\phi = 0\) makes it straightforward here.
06

Formulate the Equation of Motion

With the amplitude \(A = 3.0\, \text{cm}\), angular frequency \(\omega = 40\pi\, \text{rad/s}\), and phase \(\phi = 0\), we construct the equation:\[ y(t) = 3.0 \sin(40\pi t) \].
07

Verify Units and Context

Ensure that the units of amplitude (cm) and frequency (Hz) are correct and that the sine function describes SHM along the y-axis as requested.

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Key Concepts

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

Equation of Motion
In the context of simple harmonic motion, the equation of motion is essentially a mathematical formula that describes how a particle moves over time. This equation can be expressed in the form: \[ y(t) = A \sin(\omega t + \phi) \] - **\( y(t) \)** represents the position of the particle at time \( t \), along the y-axis in this case.- **\( A \)** is the amplitude of the motion, which is the maximum displacement from the origin. Here, it's given as 3.0 cm.- **\( \omega \)** is the angular frequency, which we'll discuss in more detail next.- **\( \phi \)** is the phase constant, indicating the initial state of the motion at \( t = 0 \). In this problem, since we're dealing with simple harmonic motion along the y-axis, understanding the parameters allows us to construct this equation which is essential to predict the particle's future positions.
Angular Frequency
Angular frequency \( \omega \) is a crucial component in the equation of motion for simple harmonic motion. It represents how many radians per second the particle oscillates through during its motion. Angular frequency and frequency \( f \) of oscillation are linked by the equation: \[ \omega = 2\pi f \] - For this exercise, the frequency \( f \) is given as 20 Hz.- Using the relationship, we find \( \omega = 2\pi \times 20 = 40\pi \) rad/s. Understanding angular frequency helps describe the tempo of the oscillation. It tells us essentially how rapid the cycles are occurring over time. This rapidity matters when determining how quickly the particle returns to the same position, which is fundamental to plotting its course over time.
Phase Constant
In simple harmonic motion, the phase constant \( \phi \) plays a role in determining the initial position of the particle. It's the offset at the starting point of time \( t = 0 \). It aligns the motion with its specific starting condition.For a motion starting at the origin, known as its equilibrium position, the phase constant is often zero. This is because we want sin(\( 0 \)) to also be zero, aligning perfectly with the origin when time \( t = 0 \). In scenarios where the particle needs to start at peak amplitude or at some other point on its path, \( \phi \) would be adjusted. Nonetheless, choosing \( \phi = 0 \) often simplifies problems such as the one given, where the particle starts exactly at the origin at \( t = 0 \). This means there's no phase shift needed in the sine function.
Amplitude and Frequency Relationship
In simple harmonic motion, amplitude and frequency are both fundamental characteristics, yet they describe different aspects of the motion. - **Amplitude (A):** Is the maximum displacement from the equilibrium position. It defines the extent of motion on either side of the origin. In this problem, the amplitude is 3.0 cm. - **Frequency (f):** Measures how often the cycles of the motion repeat per second. It's given by 20 Hz here, defining how quickly the motion repeats. Even though amplitude and frequency are measured independently, they're related in how they define the motion: amplitude provides the size, while frequency provides the pace. Both need to be known to fully understand the nature of a vibrational motion, allowing complete prediction of the particle's oscillation behavior over time.

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Most popular questions from this chapter

When a 400-g mass is hung at the end of a vertical spring, the spring stretches \(35 \mathrm{~cm}\). Determine the elastic constant of the spring. How much farther will it stretch if an additional 400 -g mass is hung from it? Use \(F_{\text {ext }}=k y\), where that force is the weight of the hanging mass: $$ F_{e \times 1}=m g=(0.400 \mathrm{~kg})\left(9.81 \mathrm{~m} / \mathrm{s}^{2}\right)=3.92 \mathrm{~N} $$ Therefore. $$ k=\frac{F_{\mathrm{eat}}}{y}=\frac{3.92 \mathrm{~N}}{0.35 \mathrm{~m}}=11.2 \mathrm{~N} / \mathrm{m} \quad \text { or } \quad 11 \mathrm{~N} / \mathrm{m} $$ Once the elastic constant is known, we can determine how the spring will behave. With an additional 400-g load, the total force stretching the spring is \(7.84 \mathrm{~N}\). Then $$ y=\frac{F}{k}=\frac{7.84 \mathrm{~N}}{11.2 \mathrm{~N} / \mathrm{m}}=0.70 \mathrm{~m}=2 \times 35 \mathrm{~cm} $$

A particle vibrates according to the equation \(x=20 \cos 16 t\), where \(x\) is in centimeters. Find its amplitude, frequency, and position at exactly \(t=0 \mathrm{~s}\).

A stretched wire vibrates in SHM such that 1000 cycles takes \(2.00\) s. Determine its oscillatory frequency and period.

A "seconds pendulum" beats seconds; that is, it takes \(1 \mathrm{~s}\) for half a cycle. (a) What is the length of a simple "seconds pendulum" at a place where \(g=9.80 \mathrm{~m} / \mathrm{s}^{2} ?(b)\) What is the length there of \(a\) pendulum for which \(T=1.00 \mathrm{~s}\) ?

A pendulum is timed as it swings back and forth. The clock is started when the bob is at the left end of its swing. When the bob returns to the left end for the 90 th return, the clock reads \(60.0 \mathrm{~s}\). What is the period of vibration? The frequency?

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