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Magazine Chapter 4: Problem 32
In Exercises 25-40, graph the given sinusoidal functions over one period. $$ y=\sin (0.5 x) $$
Short Answer
Expert verified
Graph the sine wave over one period from 0 to \(4\pi\) with amplitude 1 and peaks at \(x=\pi\) and \(x=3\pi\).
Step by step solution
01
Identify the Function's Type
The function given is a sinusoidal function, specifically a sine function, represented as \( y = \sin(0.5x) \). Since it's a sine function, we expect a periodic wave pattern.
02
Determine the Amplitude and Frequency
In the function \( y = \sin(0.5x) \), the coefficient in front of \( x \) inside the sine function is 0.5, which affects the frequency of the sine wave but not the amplitude. The amplitude (the maximum deviation from the horizontal axis) of a basic sine function \( y = \sin x \) is 1, and it remains the same for \( y = \sin(0.5x) \). The frequency here is affected by the coefficient 0.5, indicating that the wave stretches horizontally.
03
Calculate the Period
The period of a sinusoidal function \( y = \sin(kx) \) is determined by the formula \( \frac{2\pi}{k} \). For \( y = \sin(0.5x) \), \( k = 0.5 \). Substitute \( k \) into the formula: \( \frac{2\pi}{0.5} = 4\pi \). This means the wave completes one full cycle over an interval of \( 4\pi \).
04
Graph the Function Over One Period
To graph \( y = \sin(0.5x) \) over one period, start at \( x = 0 \). The key points of a sine wave occur at one-quarter, halfway, three-quarters, and full completion of the period:- At \( x = 0 \), \( y = \sin(0) = 0 \).- At \( x = \frac{1}{4}\times4\pi = \pi \), \( y = \sin(0.5\times\pi) = 1 \).- At \( x = \frac{1}{2}\times4\pi = 2\pi \), \( y = \sin(0.5\times2\pi) = 0 \).- At \( x = \frac{3}{4}\times4\pi = 3\pi \), \( y = \sin(0.5\times3\pi) = -1 \).- At \( x = 4\pi \), \( y = \sin(0.5\times4\pi) = 0 \).Plot these points on a coordinate plane, and draw the smooth curve of the sine wave connecting them.
05
Label Key Characteristics
On your graph, label the x-axis with points from \( 0 \) to \( 4\pi \). Mark the peaks and troughs of the wave as \( y = 1 \) and \( y = -1 \) respectively, which are the maximum and minimum values. Note that these values reflect the amplitude of the original sine function \( y = \sin x \), which has not changed for \( y = \sin(0.5x) \).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Amplitude
In a sinusoidal function like the sine function given by \( y = \sin(0.5x) \), the amplitude is a key feature that describes the height of the wave. The amplitude is the maximum distance from the central axis or the horizontal line where the function is centered. For a basic sine function \( y = \sin(x) \), the amplitude is always 1. This remains constant unless a coefficient is directly applied to the sine function itself (e.g., \( y = 2\sin(x) \) would have an amplitude of 2).
The equation \( y = \sin(0.5x) \) has no coefficient applied to the function itself, meaning the amplitude stays at 1. The sine wave will reach a maximum value of 1 and a minimum value of -1. This determines the peaks and troughs on the graph, which correspond to the points where the function reaches these values. The amplitude is crucial for understanding how "tall" or "short" the waves of a sinusoidal function are.
The equation \( y = \sin(0.5x) \) has no coefficient applied to the function itself, meaning the amplitude stays at 1. The sine wave will reach a maximum value of 1 and a minimum value of -1. This determines the peaks and troughs on the graph, which correspond to the points where the function reaches these values. The amplitude is crucial for understanding how "tall" or "short" the waves of a sinusoidal function are.
Frequency
While the amplitude tells us about the height of the sine wave, the frequency is concerned with how often the cycle of the wave repeats itself in a given interval.
In the equation \( y = \sin(0.5x) \), the coefficient in front of \( x \) is 0.5, and this directly affects the frequency. Generally, the frequency dictates how rapidly the sine wave oscillates. Here, because the coefficient is a fraction less than 1, the wave "stretches" out, meaning it oscillates more slowly than the base sine function \( y = \sin(x) \).
The frequency, in this case, is reciprocal to this coefficient when compared to the standard sine wave. A standard sine wave \( y = \sin(x) \) completes one full cycle from 0 to \( 2\pi \). Thus, when the coefficient is 0.5, the frequency decreases, causing the wave to complete its cycle over a longer interval.
In the equation \( y = \sin(0.5x) \), the coefficient in front of \( x \) is 0.5, and this directly affects the frequency. Generally, the frequency dictates how rapidly the sine wave oscillates. Here, because the coefficient is a fraction less than 1, the wave "stretches" out, meaning it oscillates more slowly than the base sine function \( y = \sin(x) \).
The frequency, in this case, is reciprocal to this coefficient when compared to the standard sine wave. A standard sine wave \( y = \sin(x) \) completes one full cycle from 0 to \( 2\pi \). Thus, when the coefficient is 0.5, the frequency decreases, causing the wave to complete its cycle over a longer interval.
Period
The period of a sinusoidal function signifies the length of one full cycle of the wave. To calculate the period, the formula \( \frac{2\pi}{k} \) is used, where \( k \) is the coefficient of \( x \) within the function. This formula helps determine over what interval the wave will complete one complete cycle.
For \( y = \sin(0.5x) \), we substitute \( k = 0.5 \) into the formula, resulting in \( \frac{2\pi}{0.5} = 4\pi \). This longer period indicates that the cycle of the wave takes twice as long compared to the standard sine function, which has a period of \( 2\pi \).
The extended length of the period translates visually into the graph showing a stretched wave, allowing it to cover more of the x-axis before it begins to repeat the same pattern. The period is key for understanding the horizontal layout of a wave on a graph.
For \( y = \sin(0.5x) \), we substitute \( k = 0.5 \) into the formula, resulting in \( \frac{2\pi}{0.5} = 4\pi \). This longer period indicates that the cycle of the wave takes twice as long compared to the standard sine function, which has a period of \( 2\pi \).
The extended length of the period translates visually into the graph showing a stretched wave, allowing it to cover more of the x-axis before it begins to repeat the same pattern. The period is key for understanding the horizontal layout of a wave on a graph.
