Problem 23 - Precalculus Mathematics for Calculus

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 \(19-22,\) and of the form \(y=k e^{-c t}\) sin \(\omega t\) in Exercises \(23-26\) (b) Graph the function. $$k=7, \quad c=10, \quad p=\pi / 6$$

Step-by-Step Solution

Answer

The damped motion is modeled by the function \(y = 7 e^{-10t} \cos(12t)\).

Step 1: Determine the Angular Frequency

Given the period \(p = \frac{\pi}{6}\), we need to calculate the angular frequency \(\omega\). Recall that frequency \(f\) is related to period by \(f = \frac{1}{p}\). Thus:\[ f = \frac{1}{\frac{\pi}{6}} = \frac{6}{\pi} \]Angular frequency \(\omega\) is equivalent to \(2\pi f\), so:\[ \omega = 2\pi \times \frac{6}{\pi} = 12 \]

Step 2: Choose the Function Form

Since this exercise requests a function of the form \(y=k e^{-c t} \cos \omega t\), we will use this to model the damped harmonic motion.

Step 3: Substitute Given Values into the Function

With \(k = 7\), \(c = 10\), and \(\omega = 12\) (calculated in Step 1), we plug these values into the function:\[ y = 7 e^{-10t} \cos(12t) \]

Step 4: Graph the Function

Using a graphing tool, plot the function \(y = 7 e^{-10t} \cos(12t)\). The graph will show oscillations dampening over time, with exponentially decaying envelope lines due to the \(e^{-10t}\) term. The oscillations are fast with a frequency related to \(\omega = 12\).

Key Concepts

Amplitude

Amplitude in the context of damped harmonic motion refers to the maximum displacement from the equilibrium position of the oscillating system. It indicates the peak value that the oscillating motion can reach. In undamped harmonic motion, the amplitude remains constant. However, in damped harmonic motion, the amplitude decreases over time due to the damping effect.

For example, if you consider a swinging pendulum, when you initially pull it back and let go, the first swing is typically the highest. This height corresponds to the initial amplitude. As time progresses, the swings become shorter and shorter, which means the amplitude is reducing over time.

Now, in the function used for modeling damped harmonic motion, like the one in the given problem, amplitude is represented by the constant \(k\). In our exercise, \(k = 7\), meaning the maximum initial displacement of the motion is 7 units. However, given the presence of damping (which we'll cover next), this amplitude diminishes exponentially over time according to the term \(e^{-10t}\). This results in less pronounced oscillations as time goes forward.

Damping Constant

The damping constant is a crucial parameter that characterizes the rate at which the oscillations of a system decrease over time. This constant, typically represented as \(c\), affects how quickly the energy is lost in the system due to forces like friction or air resistance.

If we consider an everyday example, it acts similarly to how brakes function in a vehicle—they slow down the motion over time. In our damped harmonic motion equation, the damping constant \(c = 10\) influences the exponential part of the model: \(e^{-10t}\).

• Large damping constants imply a rapid decrease in amplitude, leading to a quick halt in the oscillations.
• Smaller damping constants mean the system takes longer to stop, allowing it to oscillate for a more extended period but with gradually decreasing amplitude.

The exponential decay term \(e^{-10t}\) ensures that, over time, the amplitude of the system's oscillation decreases. This exponential decay causes the oscillations' prominence to taper off, depicting the practical scenario where, for instance, a car comes to a gradual stop.

Angular Frequency

Angular frequency, denoted by \(\omega\), is a concept that explains how fast an object is oscillating. It relates to the frequency, \(f\), and gives us the rate at which the object circles around a reference point in radians per second.

The formula to find the angular frequency from the frequency is \(\omega = 2\pi f\). In our problem, we derived \(\omega\) from the period \(p = \frac{\pi}{6}\) by finding \(f\) with the formula \(f = \frac{1}{p} = \frac{6}{\pi}\). Then, \(\omega = 2\pi \times \frac{6}{\pi} = 12\).

• High angular frequency means rapid oscillations. The system cycles back and forth quickly.
• Low angular frequency indicates slow oscillations and a longer period to complete each cycle.

In our damped harmonic function, \(\omega = 12\) tells us that the oscillations occur rapidly even though they decrease in amplitude. It's akin to a system being swung quickly back and forth but with less intensity over time due to damping. This fast oscillation, however, won't last as the damping constant ensures the amplitude diminishes quickly, showcasing how different parameters influence a system's behavior.