Problem 17
Question
An initial amplitude \(k,\) damping constant \(c,\) and frequency \(f\) or period \(p\) are given. (Recall that frequency and period are related by the equation \(f=1 / p . )\) (a) Find a function that models the damped harmonic motion. Use a function of the form \(y=k e^{-c t} \cos \omega t\) in Exercises \(17-20,\) and of the form \(y=k e^{-c t}\) sin \(\omega t\) in Exercises \(21-24\). (b) Graph the function. $$k=2, \quad c=1.5, \quad f=3$$
Step-by-Step Solution
Verified Answer
The function is \(y = 2 e^{-1.5t} \cos(6\pi t)\). Graph it as a decaying cosine wave.
1Step 1: Calculate Angular Frequency
First, we need to find the angular frequency \(\omega\). The frequency is given as \(f = 3\), and we know that \(\omega = 2\pi f\). Thus, \(\omega = 2\pi \times 3 = 6\pi\).
2Step 2: Write the Damped Harmonic Motion Function
Use the cosine form of the equation \(y = k e^{-ct} \cos(\omega t)\) because it belongs to Exercises 17-20 as stated in the problem. Substitute the given values: \(k = 2\), \(c = 1.5\), and \(\omega = 6\pi\). Thus, we have the function \(y = 2 e^{-1.5t} \cos(6\pi t)\).
3Step 3: Graph the Function
To graph the function \(y = 2 e^{-1.5t} \cos(6\pi t)\), notice that it is a product of an exponentially decaying envelope and a cosine wave. The amplitude of the cosine wave decreases over time due to the factor \(e^{-1.5t}\). Start by plotting the exponential decay \(2 e^{-1.5t}\) as an envelope curve. Then, superimpose the oscillating function \(\cos(6\pi t)\). The result will show decreasing oscillations within the exponentially decaying envelope.
Key Concepts
Angular FrequencyExponential DecayCosine Function
Angular Frequency
When dealing with oscillations and waves, understanding angular frequency is essential. Angular frequency, denoted by \(\omega\), measures how fast an object oscillates in a sinusoidal wave. It’s typically measured in radians per second. The relationship between angular frequency and regular frequency \(f\) is given by the formula:
- \(\omega = 2\pi f\).
- \(\omega = 2\pi \times 3 = 6\pi\).
Exponential Decay
Exponential decay is a process where a quantity decreases at a rate proportional to its current value. In mathematics, it is often expressed with an equation of the form \(e^{-ct}\) where \(c\) is a positive constant and \(t\) represents time. It describes how quantities such as radioactive substances diminish over time.
For damped harmonic motion, exponential decay indicates how the amplitude of oscillations reduces as time progresses. In the given exercise, this decay is represented in the function \(y = 2 e^{-1.5t} \cos(6\pi t)\), where the term \(e^{-1.5t}\) governs the decay rate. Specifically, the coefficient \(1.5\) determines how quickly the motion's amplitude decreases over time:
For damped harmonic motion, exponential decay indicates how the amplitude of oscillations reduces as time progresses. In the given exercise, this decay is represented in the function \(y = 2 e^{-1.5t} \cos(6\pi t)\), where the term \(e^{-1.5t}\) governs the decay rate. Specifically, the coefficient \(1.5\) determines how quickly the motion's amplitude decreases over time:
- Lower values of \(c\) lead to slower decay.
- Higher values of \(c\) produce rapid diminishment of amplitude.
Cosine Function
The cosine function is a fundamental concept in trigonometry and plays a vital role in understanding oscillatory motions such as waves and vibrations. It is defined as \(\cos(\theta)\), where \(\theta\) is the angle in radians.
In the context of damped harmonic motion, the cosine function models the oscillatory part of motion. The function \(y = 2 e^{-1.5t} \cos(6\pi t)\) incorporates the cosine wave \(\cos(6\pi t)\), which describes the periodic oscillations with angular frequency \(6\pi\). This means the wave completes:\vspace{-\baselineskip}
In the context of damped harmonic motion, the cosine function models the oscillatory part of motion. The function \(y = 2 e^{-1.5t} \cos(6\pi t)\) incorporates the cosine wave \(\cos(6\pi t)\), which describes the periodic oscillations with angular frequency \(6\pi\). This means the wave completes:\vspace{-\baselineskip}
- Every cycle in \(\frac{2\pi}{6\pi} = \frac{1}{3}\) seconds.
- \(\cos(\omega t)\) cycles between -1 and 1, affecting the amplitude of the wave.
Other exercises in this chapter
Problem 16
7–52 Find the period and graph the function. $$y=-3 \sec x$$
View solution Problem 17
Find the exact value of the trigonometric function at the given real number. $$ \text { (a) }\cos \left(-\frac{\pi}{4}\right) \quad \text { (b) } \csc \left(-\f
View solution Problem 17
Find the amplitude and period of the function, and sketch its graph. $$ y=-3 \sin 3 x $$
View solution Problem 17
\(13-18=\) The point \(P\) is on the unit circle. Find \(P(x, y)\) from the given information. The \(x\) -coordinate of \(P\) is \(-\sqrt{2} / 3\) and \(P\) lie
View solution