Problem 68 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 67-70: The formula specifies the position of a point \(P\) that is moving harmonically on a vertical axis, where \(t\) is in seconds and \(d\) is in centimeters. Determine the amplitude, period, and frequency, and describe the motion of the point during one complete oscillation (starting at \(t=0\) ). $$ d=\frac{1}{3} \cos \frac{\pi}{4} t $$

Step-by-Step Solution

Answer

Amplitude: \(\frac{1}{3}\) cm, Period: 8 seconds, Frequency: 0.125 Hz; Motion: From max to min back to max in 8 seconds.

Step 1: Identify the Amplitude

The amplitude of harmonic motion in the equation \(d = A \cos(\omega t)\) can be directly identified as the coefficient of the cosine function. In the given formula, \(d = \frac{1}{3} \cos \frac{\pi}{4} t\), the amplitude \(A\) is \(\frac{1}{3}\).

Step 2: Determine the Angular Frequency and Period

The angular frequency \(\omega\) in the equation \(d = A \cos(\omega t)\) is the coefficient of \(t\) inside the cosine. Here, \(\omega = \frac{\pi}{4}\). The period \(T\) of the motion can be found using the relation \(T = \frac{2\pi}{\omega}\). Thus, \[T = \frac{2\pi}{\frac{\pi}{4}} = 8 \text{ seconds}\].

Step 3: Calculate the Frequency

The frequency \(f\) is the reciprocal of the period \(T\). Therefore, \[f = \frac{1}{T} = \frac{1}{8} = 0.125 \text{ Hz}\].

Step 4: Describe the Motion During One Complete Oscillation

During one complete oscillation starting at \(t = 0\), the point \(P\) will start from its maximum position \(d = \frac{1}{3}\), move to its equilibrium position \(d = 0\), continue to its minimum position \(d = -\frac{1}{3}\), return to the equilibrium, and finally go back to the maximum position at \(t = 8\) seconds.

Key Concepts

Amplitude

In harmonic motion, the **amplitude** is a fundamental concept. It describes how far the oscillating object moves from its equilibrium or resting position. This value gives us the maximum distance the point travels in either direction.
In our formula for the point's vertical movement, given by \(d = \frac{1}{3} \cos \frac{\pi}{4} t\), the amplitude is represented by the coefficient of the cosine function. Hence, the amplitude here is \(\frac{1}{3}\) centimeters.
This means that the point moves \(\frac{1}{3}\) centimeters away from its central or equilibrium position during its motion. Understanding the amplitude allows us to know the total range of motion of the point as it oscillates back and forth.

Period

The **period** of a harmonic motion is the time it takes for the oscillating point to complete one full cycle of its motion. This means moving from the starting position, through a maximum, back through the equilibrium, to a minimum, and finally back to the starting point.
To find the period, we need to understand the angular frequency \(\omega\), which is the coefficient of \(t\) within the cosine function. In our exercise, \(\omega = \frac{\pi}{4}\). The period \(T\) is calculated using the formula:

• \[T = \frac{2\pi}{\omega}\]

Plugging the values into the equation, we have \[T = \frac{2\pi}{\frac{\pi}{4}} = 8\] seconds.
This period of 8 seconds indicates the time taken for one complete oscillation cycle.

Frequency

The **frequency** of a harmonic motion is closely connected to the period. It describes how often the oscillation cycle occurs within a unit of time, usually in seconds. It is measured in Hertz (Hz), where 1 Hz means one cycle per second.
The frequency is the reciprocal of the period (\(T\)). For our scenario, where the period is 8 seconds, the frequency \(f\) is determined as follows:

• \[f = \frac{1}{T} = \frac{1}{8} = 0.125\] Hz

This means the point completes 0.125 oscillating cycles every second. Understanding frequency is important for explaining how fast or slow an oscillation occurs relative to time.