Problem 60 - Precalculus: A Concise Course

Question

For the simple harmonic motion described by the trigonometric function, find (a) the maximum displacement, (b) the frequency, (c) the value of \(d\) when \(t=5,\) and (d) the least positive value of \(t\) for which \(d=0 .\) Use a graphing utility to verify your results. $$ d=\frac{1}{64} \sin 792 \pi t $$

Step-by-Step Solution

Answer

a) The maximum displacement is \(1/64\), b) The frequency is 198 Hz, c) The value of \(d\) when \(t=5\) is \(1/64\), d) The least positive value of \(t\) for which \(d=0\) is 0.00253 seconds.

Step 1: Identify the parameters in the function

For a simple harmonic motion described by the equation \(d=A \sin(\omega t)\), \n- \(A=1/64\) is the maximum displacement (amplitude), \n- \(\omega = 792\pi\) is the angular frequency in rad/s.

Step 2: Find the maximum displacement

The maximum displacement \(A\) is given directly by the equation. It will be equal to \(1/64\).

Step 3: Find the frequency

The frequency \(f\) is related to the angular frequency \(\omega\) by the relation \(\omega = 2\pi f\). Rearranging for \(f\), you get \(f = \omega/(2\pi)\), substituting \(\omega = 792\pi\), you get \(f = 198\) Hz.

Step 4: Find the value of \(d\) when \(t=5\)

Substitute \(t=5\) into the equation \(d=1/64 \sin(792\pi t)\), you get \(d = 1/64 \sin(792 \pi * 5) = 1/64\). Note that the sine function is periodic with period \(2\pi\), and \( \sin(792\pi * 5)\) will be same as \( \sin(0)\), which is 0.

Step 5: Find the least positive value of \(t\) for which \(d=0\)

For sine function, \( \sin (\theta) = 0\) when \(\theta=n\pi\), where \(n\) is an integer. Therefore, \(t = n/(2f)\) when \(d=0\). Since we're looking for the least positive \(t\), \(n=1\). Substituting \(f=198\), find \(t=1/(2*198) = 0.00253\) seconds.

Key Concepts

Maximum Displacement

In simple harmonic motion, the maximum displacement is often called the amplitude. It represents the furthest point the oscillating object moves from its equilibrium position. In our trigonometric function, the amplitude is straightforward to find. It is given as the coefficient of the sine function. Therefore, for the equation \(d = \frac{1}{64} \sin(792 \pi t)\), the maximum displacement \(A\) is \(\frac{1}{64}\).
This simply means that the oscillating object can move as far as \(\frac{1}{64}\) units from where it normally rests or starts, at the maximum height of its movement.

Frequency

Frequency is a crucial aspect of simple harmonic motion. It tells us how many complete cycles occur per second. In our problem, the relationship between frequency \(f\) and angular frequency \(\omega\) is vital. Angular frequency is given by \(\omega = 792\pi\), and frequency can be found using the formula \(f = \frac{\omega}{2\pi}\).
By substituting the angular frequency, we calculate the frequency as \(f = \frac{792\pi}{2\pi} = 198\) Hz.
This indicates the object completes 198 oscillations every second, which reflects a very fast-moving harmonic system.

Trigonometric Function

A trigonometric function like sine or cosine describes the repetitive nature of simple harmonic motion. It captures the wave-like behavior of the oscillation.

• In the equation \(d = \frac{1}{64} \sin(792\pi t)\), the sine function \(\sin(792\pi t)\) models this periodic motion.
• Sine functions range between \(-1\) and 1. This property helps the amplitude (\(\frac{1}{64}\)) to scale the range of motion.
• When you plug in different time values \(t\), you get the displacement \(d\) at any point in the object's cycle.
• The periodic nature implies that the motion repeats every cycle, with periods calculated by \(T = \frac{1}{f}\), where \(f\) is the frequency.

Thus, trigonometric functions are essential for describing how the displacement changes over time.

Angular Frequency

Angular frequency is another key element in understanding simple harmonic motion. It connects the frequency of the oscillation to the trigonometric function. The angular frequency \(\omega\) is measured in radians per second and governs how fast an object moves through its cycles.
For our equation, \(\omega = 792\pi\). This value tells us the rate at which the angle \(\theta\) in the sine function \(\sin(\theta)\) changes as time progresses, impacting how rapidly the object oscillates.

• It relates directly to the period \(T\) of the motion by \(T = \frac{2\pi}{\omega}\).
• The larger the \(\omega\), the smaller the period, indicating a faster oscillation.

Understanding angular frequency helps us grasp the speed and timing of the entire oscillation process.