Problem 4 - Precalculus Mathematics for Calculus

Question

The given function models the displacement of an object moving in simple harmonic motion. (a) Find the amplitude, period, and frequency of the motion. (b) Sketch a graph of the displacement of the object over one complete period. $$y=3 \cos \frac{1}{2} t$$

Step-by-Step Solution

Answer

Amplitude: 3, Period: \(4\pi\), Frequency: \(\frac{1}{4\pi}\).

Step 1: Identify the Standard Form of the Function

The given function is \( y = 3 \cos \frac{1}{2} t \), which is in the form \( y = A \cos(Bt + C) + D \). Here, \( A = 3 \), \( B = \frac{1}{2} \), \( C = 0 \), and \( D = 0 \).

Step 2: Determine the Amplitude

The amplitude \( A \) of a cosine function is the absolute value of the coefficient of the cosine term. Therefore, the amplitude is \(|A| = |3| = 3\).

Step 3: Calculate the Period of the Motion

The period \( T \) of the cosine function is calculated using the formula \( T = \frac{2\pi}{B} \). Here, \( B = \frac{1}{2} \), so the period is \( T = \frac{2\pi}{\frac{1}{2}} = 4\pi \).

Step 4: Calculate the Frequency of the Motion

The frequency \( f \) is the reciprocal of the period. By using \( f = \frac{1}{T} \), we find that \( f = \frac{1}{4\pi} \).

Step 5: Sketch the Graph Over One Period

To sketch the graph, plot the function \( y = 3 \cos \frac{1}{2}t \) over the interval from \( t = 0 \) to \( t = 4\pi \). The graph starts at \( y = 3 \) when \( t = 0 \), reaches \( y = 0 \) at \( t = 2\pi \), and completes the cycle back to \( y = 3 \) at \( t = 4\pi \). This shows one complete cosine wave.

Key Concepts

Amplitude

In simple harmonic motion, the amplitude represents the maximum displacement from the rest position or equilibrium. It's essentially how far the object moves from its center position during its motion.
In the given function for displacement, which is modeled as \( y = 3 \cos \frac{1}{2} t \), the amplitude is given by the coefficient of the cosine function. Here, the coefficient is 3, which means the amplitude of the motion is \( 3 \).
This value indicates that the object swings 3 units above and 3 units below its equilibrium position during each cycle of the motion. The amplitude remains constant regardless of the specific point in the cycle. It defines the extremes of motion, ensuring that, at its peak, the displacement is at its greatest.

Period

The period of a simple harmonic motion is the time it takes to complete one full cycle of motion. Think of it as the time interval after which the motion starts repeating itself.
In mathematical terms, for a cosine function of the form \( y = A \cos(Bt + C) + D \), the period \( T \) is calculated using \( T = \frac{2\pi}{B} \). This formula stems from the properties of the cosine function, which completes a cycle every \( 2\pi \) radians.
In our function \( y = 3 \cos \frac{1}{2} t \), \( B = \frac{1}{2} \). Applying the formula, the period \( T \) becomes \( \frac{2\pi}{\frac{1}{2}} = 4\pi \).
This means the object repeats its motion every \( 4\pi \) units of time. Understanding the period helps in predicting the motion pattern and determining key time points like when the object returns to specific positions.

Frequency

Frequency is a measure of how often the motion repeats within a unit of time. It indicates how many complete cycles occur per time unit.
The frequency \( f \), in terms of the period \( T \), is calculated as \( f = \frac{1}{T} \). This reciprocal relationship shows that a longer period results in a lower frequency, and vice versa.
For the function \( y = 3 \cos \frac{1}{2} t \), the period \( T \) was found to be \( 4\pi \). Thus, the frequency \( f \) is \( \frac{1}{4\pi} \).
This value signifies that the motion completes \( \frac{1}{4\pi} \) cycles per unit time. Frequency is crucial for understanding how fast an object in harmonic motion cycles through its path. It helps in applications where synchronization or timing is key, such as in waves and vibrations throughout physics and engineering.