Sketch the path of a particle that moves in the \(x y\) plane according to \(x=x_{\mathrm{m}} \cos (\omega t-\pi / 2)\) and \(y=2 x_{\mathrm{m}} \cos \omega t\)

Short Answer

Expert verified
The path of the particle in the xy-plane will be an ellipse with semi-major axis \(2x_{\mathrm{m}}\) (along the y-coordinate) and semi-minor axis \(x_{\mathrm{m}}\) (along the x-coordinate), and the direction of the movement will be counter-clockwise.

Step by step solution

01

Analyze the given equations

The given equations are \(x=x_{\mathrm{m}} \cos (\omega t-\pi / 2)\) and \(y=2 x_{\mathrm{m}} \cos \omega t\). Notice that the x equation has a phase shift of \(\pi / 2\) radians as compared to the y equation, meaning the particle's motion in x direction is delayed by \(\pi / 2\) radians compared to its motion in y direction.
02

Plot function y(x)

The goal is to sketch the path in the xy-plane, so instead of having the equations in terms of time \(t\), we need a function y(x). We can achieve this by substituting \(\omega t\) from the \(x\) equation into the \(y\) equation. Because \(\cos (\omega t-\pi / 2) = \sin \omega t\), we can rewrite the \(x\) equation as \(x=x_{\mathrm{m}} \sin \omega t\). Solving this equation for \(\omega t\) gives us \(\omega t = \arcsin (x / x_{\mathrm{m}})\). Substituting this into the \(y\) equation gives us \(y = 2 x_{\mathrm{m}} \cos (\arcsin (x / x_{\mathrm{m}}))\). This is the function we will use to sketch the path on the xy-plane.
03

Sketch the function y(x)

Upon plotting the function \(y = 2 x_{\mathrm{m}} \cos (\arcsin (x / x_{\mathrm{m}}))\), the path will look like an ellipse, with the particle moving in the counter-clockwise direction. The amplitude of the \(x\) and \(y\)-coordinates reflects the semi-axes of the ellipse. Thus the semi-major axis is \(2x_{\mathrm{m}}\) (along the y-coordinate) and the semi-minor axis is \(x_{\mathrm{m}}\) (along the x-coordinate). These observations together complete the sketch of the particle's path in the xy-plane.

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Key Concepts

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

Harmonic Motion
Harmonic motion refers to a type of periodic oscillatory motion where an object moves back and forth through an equilibrium position. This motion is characterized by its sinusoidal wave pattern, often described using trigonometric functions like sine and cosine.

In the context of our exercise, the particle's path in the xy-plane is defined by a set of equations that are characteristic of harmonic motion. For the x-coordinate, the equation is given by
\(x=x_{\mathrm{m}} \cos (\omega t-\pi / 2)\).
Here, \(x_{\mathrm{m}}\) represents the maximum displacement in the x-direction, \(\omega\) is the angular frequency, and \(t\) is the time variable. The term \(\pi / 2\) introduces a phase shift, which will be discussed in the next section. The particle is periodically oscillating in the x-direction according to this cosine function.

For the y-coordinate, it follows a similar pattern but without the phase shift:
\(y=2 x_{\mathrm{m}} \cos \omega t\).
This indicates that the maximum displacement in the y-direction is twice that of the x-direction, and the motion is purely cosine, suggesting that when the x-coordinate is at its maximum displacement, the y-coordinate is at zero, which is consistent with harmonic motion behavior.
Phase Shift
Phase shift in harmonic motion refers to when the oscillation of a wave is offset by a certain amount from another wave or a reference point. Phase shifts are typically measured in radians or degrees and indicate a starting point that is different from the standard position.

In our exercise, the equation for the x-coordinate has a phase shift. This is expressed as
\(\cos (\omega t-\pi / 2)\),
which is equivalent to
\(\sin \omega t\).
The phase shift of \(\pi / 2\) or 90 degrees delays the x motion relative to the y motion. This means that when the particle's y-coordinate starts moving from its maximum value, the x-coordinate motion is just beginning from its equilibrium (zero displacement).

Understanding phase shifts is crucial because it affects the trajectory of the particle's movement, which in this case results in the formation of an elliptical path rather than a simple circular one. This is directly associated with the differing behaviors of the x and y coordinates due to the inclusion of the phase shift.
Elliptical Path
An elliptical path is a closed curve resembling a stretched-out circle, where two axes of symmetry are distinct, known as the semi-major axis and the semi-minor axis.

The particle described in this exercise follows an elliptical path due to its sine and cosine functions with differing amplitudes and the introduced phase shift. This disparity causes the particle's motion to stretch further in one direction, in this case, the y-axis has a larger maximum displacement than the x-axis.

Mathematically, the particle's elliptical path in our exercise is shown by plotting the function obtained from trigonometric substitution: \(y = 2 x_{\mathrm{m}} \cos (\arcsin (x / x_{\mathrm{m}}))\).
With the semi-major axis being \(2x_{\mathrm{m}}\) and the semi-minor axis being \(x_{\mathrm{m}}\), this equation graphs out the shape of an ellipse in the xy-plane. The amplitude of the functions corresponds to the axes of the ellipse, with the particle moving counterclockwise along this path.
Trigonometric Substitution
Trigonometric substitution is a technique used to simplify equations, where one trigonometric function is expressed in terms of another, often to aid in solving integrals or, as in our case, sketching paths.

For our exercise, we utilize trigonometric substitution to create a function for \(y\) in terms of \(x\). Initially, we have the equations for \(x\) and \(y\) as functions of time, \(t\). However, to sketch a path in the xy-plane, we require a relationship between \(x\) and \(y\) exclusively.

We substitute \(\omega t\) from the x equation into the y equation, realize that \(\cos (\omega t-\pi / 2)\) is equal to \(\sin \omega t\), and solve for \(\omega t\). This allows us to write the y component in terms of x, leading to the relation \(y = 2 x_{\mathrm{m}} \cos (\arcsin (x / x_{\mathrm{m}}))\).
Not only does this substitution provide us with a direct equation to graph, but it also reflects the underlying relationship between the horizontal and vertical components of our particle's elliptical path.

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Most popular questions from this chapter

An oscillator consists of a block of mass \(512 \mathrm{~g}\) connected to a spring. When set into oscillation with amplitude \(34.7 \mathrm{~cm}\), it is observed to repeat its motion every \(0.484 \mathrm{~s}\). Find \((a)\) the period, \((b)\) the frquency, \((c)\) the angular frequency, \((d)\) the force constant, \((e)\) the maximum speed, and \((f)\) the maximum force exerted on the block.

An oscillator consists of a block attached to a spring \((k=\) \(456 \mathrm{~N} / \mathrm{m}) .\) At some time \(t\), the position (measured from the equilibrium location), velocity, and acceleration of the block are \(x=0.112 \mathrm{~m}, v_{x}=-13.6 \mathrm{~m} / \mathrm{s}, a_{x}=-123 \mathrm{~m} / \mathrm{s}^{2} .\) Calcu- late \((a)\) the frequency, \((b)\) the mass of the block, and \((c)\) the amplitude of oscillation.

Find the legnth of a simple pendulum whose period is \(1.00 \mathrm{~s}\) at a location where \(g=9.82 \mathrm{~m} / \mathrm{s}^{2}\)

An automobile can be considered to be mounted on four springs as far as vertical oscillations are concerned. The springs of a certain car of mass \(1460 \mathrm{~kg}\) are adjusted so that the vibrations have a frequency of \(2.95 \mathrm{~Hz}\). (a) Find the force constant of each of the four springs (assumed identical). (b) What will be the vibration frequency if five persons, averaging \(73.2\) kg each, ride in the car?

(a) When the displacement is one-half the amplitude \(x_{m}\), what fraction of the total energy is kinetic and what fraction is potential in simple harmonic motion? (b) At what displacement is the energy half kinetic and half potential?

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