Problem 84 - Algebra and Trigonometry

Question

An object moves in simple harmonic motion described by \(d=6 \cos \frac{3 \pi}{2} t,\) where \(t\) is measured in seconds and \(d\) in inches. Find: a. the maximum displacement b. the frequency c. the time required for one cycle. (Section \(5.8, \text { Example } 8)\)

Step-by-Step Solution

Answer

The maximum displacement is 6 inches, the frequency is \( \frac{3}{4} Hz\), and the time required for one cycle is \( \frac{4}{3}\) seconds.

Step 1: Determine the maximum displacement

The maximum displacement is given by the amplitude of the cosine function, which is 6 in our equation \(d=6 \cos \frac{3\pi}{2} t\). So, the maximum displacement is 6 inches.

Step 2: Determine the frequency

The frequency \(\omega\) is given by the coefficient of \(t\) in the equation of motion. In our equation, the frequency is \(\frac{3pi}{2}\). Frequency in terms of Hz \(f\) can be obtained by dividing \(\omega\) by \(2\pi\). Therefore, frequency \(f=\frac{3\pi/2}{2\pi} = \frac{3}{4}\) Hz.

Step 3: Determine the time required for one cycle

The time required for one cycle, often referred to as the period \(T\), can be obtained by taking the reciprocal of the frequency. Hence, \(T= \frac{1}{f} = \frac{4}{3}\) seconds.

Key Concepts

Maximum Displacement

Understanding maximum displacement is crucial when studying simple harmonic motion (SHM). It's the utmost distance an object moves from its equilibrium position. In SHM, this is also known as the amplitude. Mathematically, it's represented by the highest value of the sine or cosine function in the equation of motion.

For instance, in the equation \(d=6 \cos \frac{3\pi}{2} t\), the coefficient before the cosine function, which is 6, indicates the maximum displacement. An object in SHM doesn't go beyond this point. Thus, for an object oscillating in SHM described by this specific equation, the maximum displacement is 6 inches. Simply put, at peak movement, the object will be 6 inches away from its central or resting position.

Frequency

Frequency is a fundamental concept in SHM that determines how often an object completes an oscillation cycle per unit of time. It is expressed in Hertz (Hz), which equates to one cycle per second.

In the context of the provided exercise, the frequency is associated with the constant that multiplies the time variable \(t\) inside the cosine function. From the equation \(d=6 \cos \frac{3\pi}{2} t\), we identify this constant as \(\frac{3\pi}{2}\). To obtain the frequency in Hz, we divide this value by \(2\pi\) resulting in \(f=\frac{3\pi/2}{2\pi} = \frac{3}{4}\) Hz. This implies that the object completes \(\frac{3}{4}\) of a cycle each second. In practice, the higher the frequency, the quicker the object oscillates back and forth.

Period of a Cycle

The period of a cycle, often symbolized by \(T\), is a measure of the time it takes for an object to complete one full oscillation in SHM. It's the inverse of frequency, providing insight into the duration of each individual cycle within a given period.

In the given problem, we calculate the period by taking the reciprocal of the frequency already determined. Given our frequency \(f=\frac{3}{4}\) Hz, the period can be expressed as \(T= \frac{1}{f} = \frac{4}{3}\) seconds. This indicates that for one complete oscillation, the object takes \(\frac{4}{3}\) seconds. Knowing the period is essential for understanding the timing and speed of the oscillations, and it is directly related to the energy involved in the system: slower oscillations (a longer period) generally coincide with lower energy, while faster oscillations (a shorter period) indicate higher energy.