Sketch the function \(x(t)=110 \cos \left[2 \pi(60) t-\frac{\pi}{8}\right]\) versus \(t .\) Repeat for \(x(t)=5 \cos \left[2 \pi\left(16 \times 10^{6}\right) t+\frac{\pi}{4}\right]\) and \(x(t)=2 \cos \left[\frac{2 \pi}{10^{-3}}\left(t-\frac{10^{-3}}{8}\right)\right] .\) For each function, determine \(A, \omega, T, f, \varphi\), and \(\tau .\) Label your sketches carefully.

The amplitude of a sinusoidal function represents the peak value of the wave. It tells you how far the wave oscillates above and below its central value. In the function given by the equation \[x(t) = 110 \cos\left[2 \pi(60) t - \frac{\pi}{8}\right]\], the amplitude is 110. This means the maximum and minimum values of the function are 110 and -110, respectively. In the second function \[x(t) = 5 \cos\left[2 \pi (16 \times 10^{6}) t + \frac{\pi}{4}\right]\], the amplitude is 5, and in the third function \[x(t) = 2 \cos\left[\frac{2\pi}{10^{-3}}\left(t-\frac{10^{-3}}{8}\right)\right]\], the amplitude is 2. Amplitude is crucial because it indicates the strength or intensity of the signal.

Angular frequency, denoted by \(\omega\), is a measure of how quickly the sinusoidal wave oscillates. It's given by the parameter multiplied by \(2\pi\) in the function’s argument. In the first function \[x(t) = 110 \cos\left[2 \pi (60) t - \frac{\pi}{8}\right]\], \(\omega\) equals \(2\pi(60)\). In the second function, \(\omega\) is \(2\pi(16 \times 10^{6})\), and for the third \[x(t) = 2 \cos\left[\frac{2\pi}{10^{-3}}\left(t - \frac{10^{-3}}{8}\right)\right]\], \(\omega\) is \(\frac{2\pi}{10^{-3}}\). The higher the angular frequency, the more cycles the wave completes in a given time period. Angular frequency helps us understand how rapidly the signal is oscillating.

The period of a sinusoidal function is the time it takes for one complete cycle of the wave to occur. It is denoted by \(T\) and is calculated as the inverse of the frequency, \(T = \frac{1}{f}\). In the first example function, the frequency \(f\) is 60 Hz, leading to a period of \(T = \frac{1}{60} \text{ s}\). For the second function with \(f = 16 \times 10^{6} \text{ Hz}\), the period is \(T = \frac{1}{16 \times 10^{6}} \text{ s}\). In the third case, \(f\) is 1000 Hz, giving a period of \(T = 10^{-3} \text{ s}\). Understanding the period is crucial because it determines the duration of one full oscillation.

The phase shift of a sinusoidal function indicates how much the wave is shifted horizontally from its standard position. It is denoted by \(\varphi\) and represented in the argument of the cosine function. For example, in the function \[x(t) = 110 \cos\left[2\pi(60)t - \frac{\pi}{8}\right]\], the phase shift \(\varphi = -\frac{\pi}{8}\). In the second function \[x(t) = 5 \cos\left[2 \pi(16 \times 10^{6})t + \frac{\pi}{4}\right]\], \(\varphi = \frac{\pi}{4}\). For the third function, it is calculated as \(\varphi = \frac{\pi}{4}\). The phase shift helps us understand the starting point of the wave on the time axis.

Time shift, denoted by \(\tau\), indicates the delay or advancement of the wave in time. It is calculated as \(\tau = -\frac{\varphi}{\omega}\). For the first function \[x(t) = 110 \cos\left[2 \pi(60)t - \frac{\pi}{8}\right]\], it becomes \(\tau = \frac{1}{480} \text{ s}\). In the second function \[x(t) = 5 \cos\left[2 \pi(16 \times 10^{6})t + \frac{\pi}{4}\right]\], the time shift is \(\tau = -\frac{1}{128 \times 10^{6}} \text{ s}\). In the third function, it is directly given as \(\tau = \frac{10^{-3}}{8} \text{ s}\). Understanding time shift is essential for knowing how the wave has been displaced along the time axis.