Problem 5 - Physics

Question

Equation \(16.3, y=A \sin (2 \pi f t-2 \pi x / \lambda),\) gives the mathematical representation of a wave oscillating in the \(y\) direction and traveling in the positive \(x\) direction. Let \(y\) in this equation equal the electric field of an electromagnetic wave traveling in a vacuum. The maximum electric field is \(A=156 \mathrm{~N} / \mathrm{C},\) and the frequency is \(f=1.50 \times 10^{8} \mathrm{~Hz} .\) Plot a graph of the electric field strength versus position, using for \(x\) the following values: 0 , \(0.50,1.00,1.50,\) and \(2.00 \mathrm{~m} .\) Plot this graph for \((\) a) a time \(t=0 \mathrm{~s}\) and \((\mathrm{b})\) a time \(t\) that is one-fourth of the wave's period.

Step-by-Step Solution

Answer

Electric fields vary sinusoidally: from \(0\) to \(-156\) at \(t=0\) and from \(156\) to \(-156\) at \(t=T/4\).

Step 1: Understand the Wave Equation

The equation given is \( y = A \sin(2\pi ft - 2\pi x / \lambda) \), where \( A \) is the amplitude of the wave, \( f \) is the frequency, \( t \) is the time, \( x \) is the position, and \( \lambda \) is the wavelength. For an electromagnetic wave in a vacuum, the wavelength is related to the speed of light and frequency by \( \lambda = \frac{c}{f} \).

Step 2: Calculate the Wavelength

Given that the speed of light \( c = 3 \times 10^8 \) m/s and the frequency \( f = 1.50 \times 10^8 \) Hz, the wavelength \( \lambda \) can be calculated using \( \lambda = \frac{c}{f} = \frac{3 \times 10^8}{1.50 \times 10^8} = 2 \) m.

Step 3: Calculate Period for Time Calculation

The period \( T \) of the wave is given by \( T = \frac{1}{f} = \frac{1}{1.50 \times 10^8} \approx 6.67 \times 10^{-9} \) s.

Step 4: Electric Field at Time t=0 s

For \( t = 0 \) s, the equation simplifies to \( y = A \sin(-2\pi x / \lambda) \). Substitute \( A = 156 \) N/C and \( \lambda = 2 \) m. Evaluate this for \( x = 0, 0.50, 1.00, 1.50, 2.00 \) m to obtain respective \( y \) values.

Step 5: Calculate Positions and Corresponding Electric Fields at Time t = 0 s

Calculate \( y \) at each \( x \) value with \( y = 156 \sin(-\pi x) \):- \( x = 0 \): \( y = 156 \sin(0) = 0 \)- \( x = 0.5 \): \( y = 156 \sin(-\pi/4) = -110.42 \)- \( x = 1.0 \): \( y = 156 \sin(-\pi/2) = -156 \)- \( x = 1.5 \): \( y = 156 \sin(-3\pi /4) = -110.42 \)- \( x = 2.0 \): \( y = 156 \sin(-\pi) = 0 \)

Step 6: Electric Field at Time t = \( T/4 \)

At \( t = T/4 = 1.67 \times 10^{-9} \) s, the equation becomes \( y = 156 \sin(2\pi f (T/4) - 2\pi x / \lambda) \). Evaluate again for \( x = 0, 0.50, 1.00, 1.50, 2.00 \) m.

Step 7: Calculate Positions and Corresponding Electric Fields at Time t = \( T/4 \)

Using \( y = 156 \sin(\pi/2 - \pi x) \):- \( x = 0 \): \( y = 156 \sin(\pi/2) = 156 \)- \( x = 0.5 \): \( y = 156 \sin(\pi/4) = 110.42 \)- \( x = 1.0 \): \( y = 156 \sin(0) = 0 \)- \( x = 1.5 \): \( y = 156 \sin(-\pi/4) = -110.42 \)- \( x = 2.0 \): \( y = 156 \sin(-\pi/2) = -156 \)

Step 8: Graph the Results

Plot two graphs:1. Electric field vs. position at \( t = 0 \) s with points: (0,0), (0.5,-110.42), (1.0,-156), (1.5,-110.42), (2.0,0).2. Electric field vs. position at \( t = T/4 \) with points: (0,156), (0.5,110.42), (1.0,0), (1.5,-110.42), (2.0,-156).

Key Concepts

Wave Equation

The wave equation describes how waves propagate in various directions. In this exercise, the equation used is\[ y = A \sin (2 \pi f t - 2 \pi x / \lambda) \]where:

• \( y \) is the wave’s oscillation along the y-direction,
• \( A \) is the maximum amplitude,
• \( f \) is the frequency,
• \( t \) is time,
• \( x \) is the position along the x-direction,
• \( \lambda \) is the wavelength.

For electromagnetic waves in a vacuum, the wavelength is related to the speed of light, \( c \), and the frequency by the relationship \( \lambda = \frac{c}{f} \). This wave equation represents how an electromagnetic wave's electric field oscillates in space and time. It's essential for predicting how waves move and interact with the environment.

Wavelength

Wavelength, denoted by \( \lambda \), is a measure of the distance between two identical points in consecutive cycles of a wave, such as crest to crest or trough to trough. In the context of electromagnetic waves traveling in a vacuum, the wavelength can be calculated using:\[ \lambda = \frac{c}{f} \]where:

• \( \lambda \) is the wavelength,
• \( c = 3 \times 10^8 \text{ m/s} \) is the speed of light,
• \( f \) is the frequency.

In this particular exercise, the frequency provided is \( 1.50 \times 10^8 \text{ Hz} \), leading to a wavelength of \( 2 \text{ m} \). Understanding wavelength is crucial since it affects the wave's energy and ability to interact with matter.

Frequency

Frequency, symbolized by \( f \), indicates how many wave cycles pass through a point in one second. It's measured in Hertz (Hz). For electromagnetic waves, such as those in this exercise, frequency determines both the wavelength and the type of electromagnetic wave (radio, microwave, light, etc.). In the wave equation:\[ y = A \sin (2 \pi f t - 2 \pi x / \lambda) \]frequency \( f \) is essential for understanding the oscillation speed of the wave. Using the wave's frequency, you can determine the period \( T \) of the wave, which is the time it takes to complete one cycle:\[ T = \frac{1}{f} \]In this exercise, the frequency is \( 1.50 \times 10^8 \text{ Hz} \), resulting in a period of approximately \( 6.67 \times 10^{-9} \text{ s} \).Understanding frequency allows insight into how quickly a wave oscillates and impacts signal processing and communication technologies.

Electric Field

The electric field is a physical field that surrounds electrically charged particles and exert force on other charged particles in the field. In electromagnetic waves, such as those described here, the electric field is directionally oscillating. The maximum electric field strength, or amplitude \( A \), given here is \( 156 \text{ N/C} \). This maximum value describes how strong the electric component of the wave is at its peak. It is essential to calculate how the wave behaves over time and space. In the wave equation \( y = A \sin (2 \pi f t - 2 \pi x / \lambda) \), the electric field variable \( y \) changes with both position and time, showing how the wave's electric component moves and varies. Graphing the electric field strength against position for different times shows the propagation characteristics of the electromagnetic wave, providing crucial information on its interaction with other objects within the wave's transmission path.