Problem 37 - Precalculus Essentials

Question

In Exercises \(37-40,\) an object moves in simple harmonic motion described by the given equation, where \(t\) is measured in seconds and d in inches In each exercise, graph one period of the equation. Then find the following: a. the maximum displacement b. the frequency c. the time required for one cycle d. the phase shift of the motion. Describe how (a) through (d) are illustrated by your graph. $$d=4 \cos \left(\pi t-\frac{\pi}{2}\right)$$

Step-by-Step Solution

Answer

a. The maximum displacement is 4 inches. b. The frequency is 0.5 Hz. c. The time for one cycle is 2 seconds. d. The phase shift is -1/2 of a cycle. These values are illustrated in the graph of the equation.

Step 1: Identify Key Values from the Given Equation

Comparing the given equation \(d = 4 \cos(\pi t - \frac{\pi}{2})\) with the general form \(d = A \cos(wt - φ)\), we can identify that: A \(= 4\) inches (maximum displacement or amplitude), \(w = π\) (angular frequency), and \(φ = -\frac{\pi}{2}\) (phase shift).

Step 2: Calculate the Frequency

The frequency \(f\) is given by \(f = \frac{w}{2π}\). Substituting \(w = π\), we get \(f = \frac{\pi}{2π} = \frac{1}{2}\) Hz. So the frequency of the motion is 0.5Hz.

Step 3: Find the Time for One Cycle

The time required for one complete cycle, also known as the period, is given by \(T = \frac{1}{f}\). Substituting the obtained value of \(f = 0.5\) Hz, we get \(T = \frac{1}{0.5} = 2\) seconds. Therefore, one complete cycle takes 2 seconds.

Step 4: Determine the Phase Shift

The value of φ gives us the phase shift. From step 1, we identified that \(φ = -\frac{\pi}{2}\). In normal terms, the phase shift is \(φ = - \frac{1}{2}\) of a cycle. Meaning, the motion has been shifted to left by half of the cycle.

Step 5: Graph the Motion

Plot the equation \(d = 4 \cos(\pi t - \frac{\pi}{2})\) on a graph for one complete cycle. Plot time \(t\) on the x-axis and displacement \(d\) on the y-axis. You can start at \(t = 0\) for convenience, and end at \(t = 2\) seconds, which is the time for one complete cycle obtained in step 3. The graph displays the features obtained: the maximum displacement at \(A = 4\) inches, the frequency as 0.5 oscillations per second, the time period of 2 seconds per cycle, and the phase shift of \(φ = -\frac{\pi}{2}\) showing the shift to the left by half a cycle.

Key Concepts

Maximum Displacement

The term maximum displacement, often denoted by the symbol 'A', is a core concept in simple harmonic motion (SHM). It represents the utmost distance that an object moves from its equilibrium position. In other words, it is the peak value of the object's oscillation.

When working with equations of SHM, such as the given equation from the exercise, we can identify maximum displacement directly from the coefficient of the cosine term. The equation provided was \(d = 4 \cos(\pi t - \frac{\pi}{2})\), from which we can determine that the maximum displacement is 4 inches. This value tells us how far the object is going to move from its central or rest position in either direction before reversing its course.
In the exercise's solution, we observed that the amplitude was the first value to be determined. When you graph this motion, the maximum displacement is visually indicated as the highest and lowest points that the graph reaches on the y-axis, which correlates with the farthest positions from the equilibrium in the physical motion of the object.

Frequency of Motion

Frequency of motion is a vital concept in the study of oscillations and waves. It's defined as the number of complete cycles or oscillations that occur per unit of time, usually expressed in Hertz (Hz).

In the context of SHM, the frequency tells us how often the object repeats its path back and forth in a given time frame. The equation from our exercise was used to calculate the frequency by finding the coefficient 'w' (angular frequency), and dividing it by \(2\pi\). Specifically, the frequency 'f' was calculated as \(f = \frac{w}{2\pi}\), leading to a value of 0.5 Hz.
When graphed, frequency can be seen as how many complete waves or cycles fit into a one second duration on the time axis. This helps us understand the rapidity of oscillations—higher frequency means more cycles per second, implying a faster motion.

Time Period

Time period, denoted as 'T', is one of the fundamental parameters describing simple harmonic motion. It represents the time taken to complete one full cycle of the motion. The time period is the reciprocal of frequency, given by the relation \(T = \frac{1}{f}\).

In the exercise, we determined the frequency to be 0.5 Hz, which yielded a time period of \(T = 2\) seconds. This value indicates that it takes 2 seconds for the object to return to its initial position after completing its path.
The time period is particularly useful when graphing simple harmonic motion as it allows us to demarcate one full oscillation cycle. On our graph, the length of one complete wave, from start to end, measured along the time axis, would span 2 seconds reflecting the calculated time period.

Phase Shift

Phase shift, in SHM, represents the horizontal shift of the wave or oscillation from its standard position. When comparing a motion to a regular cosine or sine wave, phase shift tells us how much the motion is shifted left or right on the graph, which affects the starting point of the motion in time.

From the equation \(d = 4 \cos(\pi t - \frac{\pi}{2})\), we decipher a phase shift of \(-\frac{\pi}{2}\), meaning our wave starts a quarter cycle earlier than a standard cosine wave without phase shift. This is because the cosine function has a phase shift of 0 at \(t = 0\).
In the graph, the phase shift is seen as the distance between the origin and where the first peak or valley of the motion occurs. A negative value, as seen in the exercise, indicates a shift to the left. To visualize this, you can imagine our regular cosine curve being pulled backward along the time axis by the phase shift amount.