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Chapter 3: Problem 39

Given: $$F_{n}(t)=\left\\{\begin{array}{ll}\sin (\omega t) & 0

Short Answer

Expert verified
The function \(F_{n}(t)\) is a piecewise-defined function with two parts: 1) For interval \(0 < t < \pi / \omega\), it is a sine function with frequency \(\omega\), i.e., \(F_{n}(t) = \sin(\omega t)\). 2) For the interval \(\pi / \omega < t < 2 \pi / \omega\), it is a constant value of 0, i.e., \(F_{n}(t) = 0\). This can be visualized as a sine wave with frequency \(\omega\) for the first half of its period, followed by a flat line of 0 for the other half of the period.

Step by step solution

01

Functions for different intervals

\(F_{n}(t)\) is a piecewise-defined function, which means that it is defined differently for different intervals: 1. When \(0 < t < \pi / \omega\), the function is given by \(F_{n}(t) = \sin(\omega t)\). 2. When \(\pi / \omega < t < 2 \pi / \omega\), the function is given by \(F_{n}(t) = 0\).
02

Understanding and solving the sine function part

For the first interval, \(0 < t < \pi / \omega\), we are working with the sine function: \(F_{n}(t) = \sin(\omega t)\). Since this is a basic sine function, we should be familiar with its properties: - Amplitude: The sine function has a maximum value of 1 and a minimum value of -1, thus an amplitude of 1. - Frequency: The parameter \(\omega\) represents the angular frequency of the sine function. Higher values of \(\omega\) result in more oscillations over a given time period. - Phase: There is no phase shift in this function, as it starts at 0. We can identify the value of the function at any point \(t\) in the interval \(0 < t < \pi / \omega\) using the formula \(\sin(\omega t)\). Example: For \(\omega = 1\), and \(t = \pi / 4\), the value of the function is given by \(\sin(1 * \pi/4) = \sin(\pi / 4) \approx 0.707\).
03

Understanding and solving the 0 function part

For the second interval, \(\pi / \omega < t < 2 \pi / \omega\), the function is given by \(F_{n}(t) = 0\). This means that for any point in this interval, the function will have a value of 0. Example: For any \(t\) in the interval \(\pi / \omega < t < 2 \pi /\omega\), the value of the function is always 0.
04

Analyzing the complete function

Now that we have analyzed the two parts of the function separately, let's put everything together to visualize the complete function: 1. For interval \(0 < t < \pi / \omega\), the function resembles a sine function with a frequency determined by \(\omega\). 2. For the interval \(\pi / \omega < t < 2 \pi / \omega\), the function is a constant value of 0. This function can be visualized as a sine wave that lasts for half of its period, followed by a flat line of 0 for the other half of the period. The frequency of oscillation is determined by the parameter \(\omega\), which affects the speed of oscillation and the positioning of the interval where the function oscillates.

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Key Concepts

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

Sine Function
The sine function is a fundamental concept in trigonometry, often graphed as a smooth, wave-like curve. This function relates the angle of a right triangle to the ratio of the length of the opposite side to the length of the hypotenuse in a unit circle.

In mathematical terms, the sine of an angle \theta is written as \(\sin(\theta)\). The graph of the sine function is periodic, meaning it repeats its values in regular intervals, known as the period. This property is why the sine function is commonly used to model periodic phenomena such as sound waves, alternating current, and, as in our exercise, oscillating motions.

Properties of the Sine Function

  • Amplitude: This indicates the peak height of the wave and is typically measured from the centerline to the peak or trough of the wave.
  • Period: The distance over which the wave's shape repeats. For the sine function, the period is \(2\pi\) radians or 360 degrees.
  • Frequency: How often the wave repeats itself over a unit of time. The higher the frequency, the more waves there are in a given time frame.

Within the exercise, the sine function is used within one segment of a piecewise-defined function to model a specific behavior during that interval.
Angular Frequency
Angular frequency, often denoted by the Greek letter \(\omega\), is a quantitative expression for how fast an angle changes with time in applications involving periodic processes such as oscillations and waves.

It's connected to the frequency \(u\) of the wave by the relationship \(\omega = 2\pi u\), with \(u\) being the number of cycles per second (measured in Hertz). In essence, it tells us how quickly the sine function's angle argument increases, and consequently how fast the wave oscillates.

For the given function \(F_{n}(t) = \sin(\omega t)\), \(\omega\) shapes the rate at which the function undergoes its oscillatory pattern. As \(\omega\) increases, the oscillations become more frequent, leading to a higher pitch sound in acoustics or higher frequency waves in physics.
Oscillation Period
The oscillation period is the time it takes to complete one full cycle of a wave. It is often simply referred to as the 'period' of the function. In a mathematical function such as a sine wave, the period is the length of the interval after which the function values start repeating. For the standard sine function \(\sin(\theta)\), this interval is \(2\pi\) radians.

However, when the sine function is modified by an angular frequency \(\omega\), as in the exercise \(\sin(\omega t)\), the period adjusts accordingly. The period \(T\) of the function is found by the formula \(T = 2\pi / \omega\).

Understanding the period is crucial in applications ranging from engineering to music, as it defines the duration of one complete waveform and is inversely proportional to the frequency. The concept of the period in the exercise is important to identify the intervals in which the piecewise-defined function behaves differently - as an oscillating sine wave for half its period and as a constant value in the other half.

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Most popular questions from this chapter

Two masses \(m_{1}=100 \mathrm{g}\) and \(m_{2}=200 \mathrm{g}\) slide freely in a horizontal frictionless track and are connected by a spring whose force constant is \(k=0.5 \mathrm{N} / \mathrm{m}\). Find the frequency of oscillatory motion for this system.

A simple pendulum consists of a mass \(m\) suspended from a fixed point by a weightless, extensionless rod of length \(l\). Obtain the equation of motion and, in the approximation that \(\sin \theta \cong \theta,\) show that the natural frequency is \(\omega_{0}=\sqrt{g / l}\), where \(g\) is the gravitational field strength. Discuss the motion in the event that the motion takes place in a viscous medium with retarding force \(2 m \sqrt{g l} \dot{\theta}.\)

A damped linear oscillator, originally at rest in its equilibrium position, is subjected to a forcing function given by $$ \frac{F(t)}{m}=\left\\{\begin{array}{ll} 0, & t<0 \\ a \times(t / \tau), & 0\tau \end{array}\right. $$ Find the response function. Allow \(\tau \rightarrow 0\) and show that the solution becomes that for a step function.

Obtain the Fourier expansion of the function $$ F(t)=\left\\{\begin{array}{ll} -1, & -\pi / \omega

Obtain the response of a linear oscillator to the forcing function $$ \frac{F(t)}{m}=\left\\{\begin{array}{ll} 0, & t<0 \\ a \sin \omega t, & 0\pi / \omega \end{array}\right. $$
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