Problem 21 - Algebra and Trigonometry

Question

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, find the following: a. the maximum displacement b. the frequency c. the time required for one cycle. $$ d=5 \cos \frac{\pi}{2} t $$

Step-by-Step Solution

Answer

a. The maximum displacement is 5 inches. b. The frequency is 0.25 Hz. c. The time required for one cycle is 4 seconds.

Step 1: Find maximum displacement

The maximum displacement (amplitude) in an SHM is the maximum distance the object moves from its equilibrium position, which is the coefficient of the cosine term in the given equation. Therefore, the maximum displacement A = 5 inches.

Step 2: Calculate the frequency

The angular frequency \(\omega\) represents how fast the motion is, and it is equal to \(2\pi f\), where \(f\) is the frequency of the motion. From the given equation \(\omega = \pi/2\). Therefore, we solve for \(f\): \(f = \omega / (2\pi) = (\pi/2) / (2\pi) = 0.25 Hz\).

Step 3: Determine the time required for one cycle

The time required for one complete cycle (period T) is the reciprocal of the frequency. Therefore, we calculate \(T = 1/f = 1/0.25 = 4 seconds.\).

Key Concepts

Understanding Maximum Displacement in Simple Harmonic Motion

In simple harmonic motion (SHM), the maximum displacement represents the furthest point an object moves from its starting or equilibrium position. This is also commonly referred to as the amplitude of the motion. It's like swinging a pendulum; the maximum displacement is how far it goes to one side.

In the given equation, \(d = 5 \cos \frac{\pi}{2} t\), the maximum displacement is determined by the coefficient in front of the cosine function. Here, it is 5 inches. This tells us the object swings 5 inches away from its equilibrium point at its peak.

This amplitude is crucial as it gives insight into the energy in the system; a greater amplitude generally means more energy is involved.

Frequency Calculation in Harmonic Motion

The frequency of an object's motion in SHM tells us how often the motion repeats in a second. It's like counting how many swings a pendulum makes in one second.

To find the frequency, we need to know the angular frequency, \(\omega\), from the equation, which here is \(\frac{\pi}{2}\). The relationship between angular frequency and frequency \(f\) is given by \(\omega = 2\pi f\). Solving for \(f\), we find:

• \(f = \frac{\omega}{2\pi} = \frac{\pi/2}{2\pi} = 0.25 \text{ Hz}\)

This means the object completes 0.25 cycles per second. Notice that frequency is always a positive value, representing cycles per second, or Hertz (Hz).

Knowing the frequency helps understand how rapid or slow the oscillation is.

Determining the Harmonic Motion Period

The period of an SHM is the time it takes for the object to complete one full cycle of motion. If we think of a pendulum again, it's the time it takes to swing forward and back to its starting point.

The period is directly related to the frequency by the formula \(T = \frac{1}{f}\). Using the frequency found earlier, \(f = 0.25 \text{ Hz}\), the period \(T\) becomes:

• \(T = \frac{1}{0.25} = 4 \text{ seconds}\)

This result means the complete cycle of motion takes 4 seconds. Grasping the concept of the period is essential, as it's a fundamental aspect of understanding the timing and rhythm of oscillations in SHM.