Problem 49 - Precalculus: Functions and Graphs

Question

Lissajous figures are used in the study of electrical circuits to determine the phase difference \(\phi\) between a known voltage \(V_{1}(t)=A \sin (\omega t)\) and an unknown voltage \(V_{2}(\vec{t})=B \sin (\omega t+\phi)\) having the same frequency. The voltages are graphed parametrically as \(x=V_{1}(t)\) and \(y=V_{2}(t)\) If \(\phi\) is acute, then $$\phi=\sin ^{-1} \frac{y_{\mathrm{int}}}{y_{\max }}$$ where \(y_{\text {int }}\) is the nonnegative \(y\) -intercept and \(y_{\max }\) is the maximum \(y\) -value on the curve. (a) Graph the parametric curve \(x=V_{1}(t)\) and \(y=V_{2}(t)\) for the specified range of \(t\) (b) Use the graph to approximate \(\phi\) in degrees. $$\begin{aligned}V_{1}(t)=80 \sin (60 \pi t), & V_{2}(t)=70 \cos (60 \pi t-\pi / 3) & 0 \leq t \leq 0.035\end{aligned}$$

Step-by-Step Solution

Answer

\(\phi = 30\) degrees.

Step 1: Understand the Functions and their Relation

We are given two voltage functions: \(V_{1}(t) = 80\sin(60\pi t)\) for \(x\) and \(V_{2}(t) = 70\cos(60\pi t - \pi/3)\) for \(y\). These represent the x and y coordinates of a parametric plot and share the same frequency \(\omega = 60\pi\). We aim to understand the graph created and use it to find the phase difference \(\phi\).

Step 2: Transform \(V_2(t)\) into Sine

The function \(V_2(t)\) must be expressed in terms of sine to match the form given in the phase equation. Start by using the identity \(\cos(\theta) = \sin(\theta + \pi/2)\): \[V_2(t) = 70 \sin(60\pi t - \pi/3 + \pi/2) = 70 \sin(60\pi t + \pi/6)\]. This is now in the form \(B\sin(\omega t + \phi)\).

Step 3: Graph the Parametric Equations

Plot the parametric equations \(x = 80\sin(60\pi t)\) and \(y = 70\sin(60\pi t + \pi/6)\) for \(0 \leq t \leq 0.035\). This creates a Lissajous figure on the graph, visualizing how the two functions relate over time.

Step 4: Identify \(y_{\text{int}}\) and \(y_{\max}\) from the Graph

From the graph, determine the nonnegative \(y\)-intercept \(y_{\text{int}}\) and the maximum \(y\)-value \(y_{\max}\) of the Lissajous figure. Given the sine wave's properties, \(y_{\text{max}} = 70\), and from observation, \(y_{\text{int}}\) is the initial value, i.e., \(y_{\text{int}} = 70\times\sin(\pi/6) = 35\).

Step 5: Calculate the Phase Difference \(\phi\)

Apply the phase difference formula \(\phi = \sin^{-1}\left(\frac{y_{\text{int}}}{y_{\max}}\right)\). Substitute the values: \[\phi = \sin^{-1}\left(\frac{35}{70}\right) = \sin^{-1}\left(0.5\right)\]. This gives us \(\phi = \pi/6\) or 30 degrees.

Key Concepts

Parametric Equations

Parametric equations are a way to define a set of equations in terms of one or more variables, called parameters. In the study of Lissajous Figures, this concept is particularly useful because it allows us to graph complex, multi-dimensional curves by defining the x and y coordinates separately in terms of a common variable, typically time (t). By using parametric equations, we can better understand the relationship between two functions in terms of their oscillations and patterns over time.
This is particularly evident in our voltage functions:

• For the x-axis: \(x = V_1(t) = 80\sin(60\pi t)\)
• For the y-axis: \(y = V_2(t) = 70\cos(60\pi t - \pi/3)\)

These parametric equations allow us to graph and visualize the interaction between the two voltage functions, helping us explore how phase differences arise.

Phase Difference

The phase difference, often denoted by \(\phi\), is a crucial concept when dealing with wave functions like the sine and cosine functions in our voltage example. It represents the difference in phase between two periodic functions that are operating at the same frequency. In simpler terms, it tells us how much one wave is "ahead" or "behind" another.
This is critical in fields such as electronics and physics, where understanding the timing between signal peaks can impact how systems function. In the Lissajous Figure context, the phase difference is visualized as the distortion or shape of the figure itself. By evaluating the phase difference, we can determine the precise timing discrepancies between two synchronized systems based on the figure's symmetry.
In the given solution, this difference was calculated using the inverse sine function: \[\phi = \sin^{-1}\left(\frac{y_{\text{int}}}{y_{\max}}\right) = \sin^{-1}(0.5) = 30^\circ \text{ (or } \frac{\pi}{6} \text{ radians)}\] This value gives a clear quantification of their phase relationship.

Voltage Functions

Voltage functions define how voltage levels change over time and are often described by sine or cosine equations. In the context of Lissajous figures, they allow us to analyze AC (alternating current) circuits by plotting these functions parametrically. The voltage functions help simulate real-world alternating signals in electricity.
In our case, we have:

• \(V_{1}(t) = 80\sin(60\pi t)\): This explains how the first circuit's voltage fluctuates over time.
• \(V_{2}(t) = 70\cos(60\pi t - \pi/3)\): This represents another circuit but with a phase adjustment.

Voltage functions can be adjusted through amplitude (the value in front of the cosine or sine, representing how high the wave peaks) and the phase shift (which adjusts the starting angle), influencing how circuits are designed and understood.

Sine and Cosine Transformation

Sine and cosine transformations are mathematical manipulations that change the characteristics of wave functions without altering their fundamental shape. In this exercise, such transformations allow us to express one form in terms of another, making it easier to compare or compute phase differences.
For instance, converting from cosine to sine can highlight the timing and phase disparities between functions.

• The original cosine function \(V_2(t) = 70\cos(60\pi t - \pi/3)\) was transformed into a sine form:
• Using the identity \(\cos(\theta) = \sin(\theta + \pi/2)\), our function became:
• \(V_2(t) = 70\sin(60\pi t + \pi/6)\)

Such transformations are essential for solving problems involving phase comparisons or finding maxima and minima in oscillating phenomena. Recognizing the power of transforming sine and cosine forms helps in creating more intuitive insights into complex equations in physics and engineering.