Problem 54 - Precalculus: A Concise Course

Question

Find a model for simple harmonic motion satisfying the specified conditions. \(\begin{array}{ll}\text { Displacement } (t=0) & \text { Amplitude }\end{array}\) Period $$ \begin{array}{lll} 0 & 3 \text { meters } & 6 \text { seconds } \end{array} $$

Step-by-Step Solution

Answer

The required model for the simple harmonic motion is \(y = 3 \cos(\frac{\pi}{3}t + \frac{\pi}{2})\).

Step 1: Identify given values

The problem provides the following values: Initial displacement = 0 m (since displacement is given at t=0), Amplitude A = 3 m, and Period T = 6 seconds.

Step 2: Calculate the Angular Frequency

The angular frequency is related to period by the formula \( w = \frac{2\pi}{T} \). Substituting the values we get \( w = \frac{2\pi}{6} \) = \( \frac{\pi}{3} \) rad/sec.

Step 3: Determine the Phase Constant

The initial displacement is 0. Using the fact that \(\cos{0} = 1\), and knowing that displacement y at \(t=0\) must equal 0, the only value possible for the phase constant to satisfy the equation \(y=A \cos(wt+\phi)\) at \(t=0\) is \(\phi = \pi/2\) or \(\phi=3\pi/2\). However, as SHM starts at the equilibrium (y=0), motion continues in positive y direction first, \(\phi = \pi/2\). For \(y=0\) and \(wt=0\), \(\cos(\phi) = 0\), which is correct for \(\phi=\pi/2\).

Step 4: Substitute the values

Substitute the known values into the equation for A, w, and φ. The equation thus becomes \(y = 3 \cos(\frac{\pi}{3}t + \frac{\pi}{2})\).

Key Concepts

Angular Frequency

Angular frequency is a pivotal term in understanding simple harmonic motion (SHM). It is defined as the rate at which an object travels through its cycle of motion, and it's a measure of how often the motion repeats itself per unit of time. Expressed in radians per second (rad/s), it is given by the formula \( \omega = \frac{2\pi}{T} \) where \( T \) is the period of oscillation—the time it takes for one complete cycle of motion.

For our exercise, with a given period of 6 seconds, the calculation of angular frequency is simple: \( \omega = \frac{2\pi}{6} = \frac{\pi}{3} \) rad/s. This result tells us how quickly the object is oscillating in its harmonic motion. It's essential to realize that the larger the angular frequency, the faster the oscillation, and conversely, a smaller angular frequency indicates slower movement.

Phase Constant

The phase constant, often symbolized by \( \phi \), is a value that represents the initial angle of the sinusoidal function that models an SHM system. It effectively shifts the wave either left or right along the time axis, setting the starting position of the oscillating object.

In SHM, the phase constant determines at which point in its cycle the motion begins. If the initial displacement is zero, as it is at the start of our exercise where the displacement \( t=0 \) is 0 meters, the phase constant will adjust the cosine function so that its value starts at zero at \( t=0 \). The calculated phase constant for our example is \( \phi = \frac{\pi}{2} \) or \( \phi = 3\pi/2 \), but \( \frac{\pi}{2} \) is the appropriate value as the motion proceeds in the positive direction first. It's like setting the starting line for the oscillating object's motion path.

Amplitude

Amplitude in the context of simple harmonic motion is the maximum displacement from the equilibrium position. It is the 'height' of the wave and signifies the extent of the oscillation. Higher amplitudes mean the object is moving further from its rest position during its motion.

In our exercise, the amplitude is provided as 3 meters which means at the peak of its oscillation, either positive or negative, the object will be 3 meters away from the equilibrium. It's important to note that amplitude is always a positive value and does not provide information on direction, only the magnitude of the maximum displacement.

Period of Oscillation

The period of oscillation, designated by \( T \), is the duration of time it takes for a complete cycle of harmonic motion. This period remains constant for a given system unless there's a change in the system's parameters like the mass or the spring stiffness in case of a mass-spring oscillator.

From our example, with a specified period of 6 seconds, one can infer the slow or fast nature of the oscillation. A six-second period means that it takes the object 6 seconds to return to its starting point after one complete motion cycle. Understanding the period is crucial because it relates to other aspects of SHM such as the frequency \( f = \frac{1}{T} \) and angular frequency, and helps indicate how dynamic or gentle the motion is.