Problem 3 - Basic Engineering Mathematics

Question

Determine a sinusoidal expression for \(y_{1}-y_{2}\) when \(y_{1}=4 \sin \omega t\) and \(y_{2}=3 \sin (\omega t-\pi / 3)\)

Step-by-Step Solution

Answer

The sinusoidal difference is \(y_1 - y_2 = \sqrt{37} \sin(\omega t + \delta)\) where \(\delta = \tan^{-1}(\frac{3\sqrt{3}}{5})\).

Step 1: Understand the Problem

We need to find a single sinusoidal expression that represents the difference between the two given sinusoidal functions: \(y_1 - y_2\). The given functions are \(y_1 = 4 \sin \omega t\) and \(y_2 = 3 \sin(\omega t - \frac{\pi}{3})\). We will express this difference as a combined sinusoidal function.

Step 2: Use the Sinusoidal Difference Formula

The difference of two sinusoidal functions \(A \sin(\omega t) - B \sin(\omega t - \phi)\) can be expressed using the identity: \[ A \sin x - B \sin(x - \phi) = C \sin(x + \delta) \]where \[ C = \sqrt{A^2 + B^2 + 2AB \cos \phi} \] \[ \tan \delta = \frac{B \sin \phi}{A - B \cos \phi} \]For our problem, \(A = 4\), \(B = 3\), and \(\phi = \frac{\pi}{3}\).

Step 3: Calculate the Amplitude C

To find \(C\), use the formula: \[ C = \sqrt{A^2 + B^2 + 2AB \cos \phi} \].Substitute \(A = 4\), \(B = 3\), and \(\cos \phi = \cos(\frac{\pi}{3}) = \frac{1}{2}\):\[ C = \sqrt{4^2 + 3^2 + 2(4)(3)(\frac{1}{2})} \]\[ C = \sqrt{16 + 9 + 12} = \sqrt{37} \]

Step 4: Calculate the Phase Shift \(\delta\)

Use the formula for \(\tan \delta\): \[ \tan \delta = \frac{B \sin \phi}{A - B \cos \phi} \]Substituting given values, where \(\sin \phi = \sin(\frac{\pi}{3}) = \frac{\sqrt{3}}{2}\):\[ \tan \delta = \frac{3(\frac{\sqrt{3}}{2})}{4 - 3(\frac{1}{2})} \]\[ \tan \delta = \frac{3\sqrt{3}/2}{4 - 1.5} = \frac{3\sqrt{3}/2}{2.5} \]\[ \tan \delta = \frac{3\sqrt{3}}{5} \]Calculate \(\delta\) using \(\tan^{-1}\).

Step 5: Write the Combined Expression

Combine everything to find the expression for \(y_1 - y_2\). The expression is: \[ y_1 - y_2 = \sqrt{37} \sin(\omega t + \delta) \]Use the previously calculated values \(C = \sqrt{37}\) and \(\delta = \tan^{-1}(\frac{3\sqrt{3}}{5})\). This is the resultant sinusoidal expression.

Key Concepts

Sinusoidal Functions

Sinusoidal functions are fundamental in mathematics and science, particularly in fields like physics and engineering. These functions, typically in the forms of sine and cosine, describe oscillations and wave patterns. They are defined mathematically as functions that can represent regular periodic phenomena. A basic sinusoidal function is written as:

• \( y = A \, \sin(\omega t + \phi) \)

Where:

• \( A \) is the amplitude, determining the peak value.
• \( \omega \) is the angular frequency, representing the rate of oscillation.
• \( \phi \) is the phase shift, indicating the horizontal translation of the wave.

These functions are incredibly versatile, used to model everything from the pulsing of a star to alternating current in electrical grids.
Understanding how to manipulate and combine these functions is crucial for precise modeling and analysis of natural and technological processes.

Amplitude Calculation

Amplitude in sinusoidal functions is a critical value as it reflects the maximum extent of oscillation. The amplitude directly influences the height of the wave's peaks and troughs. To understand this better, consider this function:

• \( y = A \, \sin(\omega t) \)

In this formula, \( A \) is the amplitude. It is visually the distance from the midpoint to the peak of the graph. In the context of combining sinusoidal functions like in our exercise, amplitude calculations require combining multiple amplitudes using trigonometric identities. For two different functions like \( y_1 \) and \( y_2 \), calculate using:

• \( C = \sqrt{A^2 + B^2 + 2AB \cos \phi} \)

This formula stems from considering the vector sum of two sinusoidal functions.
Understanding and using this is important when seeking to consolidate functions and predict resultant wave behaviors.

Phase Shift Calculation

Phase shift is another vital element in the analysis of sinusoidal functions. It represents the horizontal displacement of the wave. If phase changes, the entire wave moves along the x-axis. The phase shift can be challenging to comprehend, but let's break it down using this function:

• \( y = A \, \sin(\omega t + \phi) \)

Here, \( \phi \) is the phase shift. It determines how far to the right or left the standard sine function shifts. Positive values push the graph left, while negative values push it right.

When combining functions like in our exercise, finding the new phase shift, \(\delta\), involves:

• \( \tan \delta = \frac{B \sin \phi}{A - B \cos \phi} \)

Once \(\tan \delta\) is found, use the inverse tangent function, \( \tan^{-1} \), to identify the actual phase angle \(\delta\).
This is crucial for accurately describing the resultant wave position and understanding its temporal behavior.

Trigonometric Identities

Trigonometric identities are algebraic expressions involving trigonometric functions that are true for every single value of the occurring variables. They allow significant simplification of the complex operations seen in sinusoidal function manipulation. Some of the core identities are:

• The Pythagorean Identity: \( \sin^2 \theta + \cos^2 \theta = 1 \)
• The Sum-to-Product Identities: \( \sin \alpha + \sin \beta = 2 \sin \frac{\alpha + \beta}{2} \cos \frac{\alpha - \beta}{2} \)
• The Double Angle Formulas: \( \sin 2\theta = 2 \sin \theta \cos \theta \)

In the context of sinusoidal function differences as seen in our exercise, identities help reframe expressions like . For instance, the difference formula \( A \sin x - B \sin(x - \phi) \) helps in simplifying and expressing in common sinusoidal forms as \( C \sin(x+\delta) \) through understood identities.
Mastery of these identities can deepen understanding of trigonometric functions and facilitate complex calculations and transformations.