/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 48 As a physics lab instructor, you... [FREE SOLUTION] | 91Ó°ÊÓ

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As a physics lab instructor, you conduct an experiment on standing waves of microwaves, similar to the standing waves produced in a microwave oven. A transmitter emits microwaves of frequency \(f .\) The waves are reflected by a flat metal reflector, and a receiver measures the waves' electric-field amplitude as a function of position in the standing-wave pattern that is produced between the transmitter and reflector (Fig. \(\mathbf{P 3 2 . 4 8}\) ). You measure the distance \(d\) between points of maximum amplitude (antinodes) of the electric field as a function of the frequency of the waves emitted by the transmitter. You obtain the data given in the table. $$ \begin{array}{l|llllllllll} f\left(\mathbf{1 0}^{9} \mathbf{H z}\right) & 1.0 & 1.5 & 2.0 & 2.5 & 3.0 & 3.5 & 4.0 & 5.0 & 6.0 & 8.0 \\ \hline d(\mathrm{~cm}) & 15.2 & 9.7 & 7.7 & 5.8 & 5.2 & 4.1 & 3.8 & 3.1 & 2.3 & 1.7 \end{array} $$ Use the data to calculate \(c,\) the speed of the electromagnetic waves in air. Because each measured value has some experimental error, plot the data in such a way that the data points will lie close to a straight line, and use the slope of that straight line to calculate \(c\).

Short Answer

Expert verified
After following the above steps, the student obtains the speed of the electromagnetic waves in air using their calculated slope from the graph of d against 1/f. This finished solution is in accordance with the wave speed formula \( c = fd \), where f represents the wave frequency and d represents the distance between antinodes (equivalent to wavelength).

Step by step solution

01

Understand the basic concept

The speed of a wave is given by the formula \( c = f\lambda \), where c is the speed of the wave, f is its frequency, and \( \lambda \) is its wavelength. In a wave, the distance between two adjacent antinodes is equal to the wavelength.
02

Derive the relationship

The provided data includes the frequency (f) and the distance (d) between antinodes. Since the distance between antinodes equals the wavelength (\( \lambda \)), we can substitute \( \lambda \) with d in the wave speed formula, resulting in \( c = fd \).
03

Plot the Graph

We rearrange the speed formula to \( c/f = d \), equivalent to \( y = mx \), where y corresponds to d, x corresponds to 1/f, m (slope) corresponds to the speed of the wave (c). A plot of this formula would yield a straight line, and the slope of that line would equal the speed of the wave. Plot a graph of d against 1/f using the given data.
04

Calculate the speed of the wave

After plotting the graph, compute the speed (c) by determining the slope of this line (rise over run), which is equivalent to the division of the vertical change (d) by the horizontal change (1/f).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Wave Speed Calculation
Understanding wave speed calculation is critical when studying physics and specifically in the context of waves, as it can give us insight into how fast a wave travels through a medium. To calculate the speed of a wave, the basic formula is
\[ c = f\lambda \]
Here, \( c \) represents the wave speed, \( f \) stands for the frequency of the wave, and \( \lambda \) denotes the wavelength. For electromagnetic waves, like microwaves in your experiment, this speed is usually the speed of light when in a vacuum, but it can differ when traveling through other mediums such as air.

Importance of Correct Units

In your calculations, it's important to ensure that the frequency (\( f \)) and wavelength (\( \lambda \)), or in this experiment, the distance between antinodes (\( d \)), are in compatible units. The most common SI unit for wave speed is meters per second (m/s), frequency in hertz (Hz), and wavelength in meters (m). Any discrepancies in units can result in incorrect calculations and interpretations.

Graphing for Accuracy

Plotting the data on a graph, with distance (\( d \)) on the y-axis and the inverse frequency (\( 1/f \)) on the x-axis, forms a straight line if represented accurately. The slope of this line gives us the wave speed (\( c \)). This graphical method is especially useful because it helps to account for and minimize any experimental errors, leading to a more accurate calculation of the wave speed.
Electromagnetic Waves
Electromagnetic waves, like the microwaves in your standing waves experiment, are a fundamental concept in physics. These waves are a form of energy that travels through space at the speed of light, and they encompass a broad spectrum of wavelengths—from the very long (radio waves) to the very short (gamma rays).
Electromagnetic waves are unique because they can travel through a vacuum, unlike mechanical waves that require a medium (like water or air) to propagate. In a lab environment, when you're measuring properties like electric-field amplitude in a standing-wave pattern, you're directly working with the principles that govern electromagnetic waves.

Characteristics of Electromagnetic Waves

  • They consist of electric and magnetic fields oscillating perpendicular to each other and to the direction of propagation.
  • They do not require a material medium to travel through.
  • Their speed in a vacuum is approximately \( 3 \times 10^8 \) meters per second, the universally recognized speed of light (\( c \)).

The lab experiment you conducted simulates how microwaves behave in everyday appliances, such as microwave ovens, by creating a standing wave pattern that illustrates the interactions between these waves and their environment.
Frequency and Wavelength Relationship
The relationship between frequency and wavelength is an essential concept that needs to be grasped for a robust understanding of wave physics. As described by the formula \( c = f\lambda \), there is an inverse relationship between the two: as the frequency increases, the wavelength decreases, and vice versa, assuming the speed of the wave (\( c \)) is constant.

For your standing waves experiment, where you measured the distance between maximum amplitude points, this distance directly corresponds to the wavelength. Higher frequency microwaves will have shorter wavelengths, which is why you observed smaller distances between antinodes as the frequency increased.

Visualizing the Concept

It's helpful to visualize this relationship. Imagine lower frequency waves as widely spaced waves on the ocean, where peak after peak has considerable distance in between. Contrast this with the tightly packed waves that represent higher frequencies. This analogy helps illustrate why high-frequency waves have shorter distances between their peaks (antinodes in the context of your experiment) and lower-frequency waves have longer distances between them.

Understanding this inverse relationship aids in interpreting your experiment's data correctly. Applying this knowledge to your graph, where the slope derived from the plotted data gives wave speed, reinforces how critical it is to understand the interplay between frequency and wavelength for precise wave speed calculations.

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Most popular questions from this chapter

An air-filled cavity for producing electromagnetic standing waves has two parallel, highly conducting walls separated by a distance \(L\). One standing- wave pattern in the cavity produces nodal planes of the electric field with a spacing of \(1.50 \mathrm{~cm}\). The next-higher-frequency standing wave in the cavity produces nodal planes with a spacing of \(1.25 \mathrm{~cm} .\) What is the distance \(L\) between the walls of the cavity?

A sinusoidal electromagnetic wave is propagating in vacuum in the \(+z\) -direction. If at a particular instant and at a certain point in space the electric field is in the \(+x\) -direction and has magnitude \(4.00 \mathrm{~V} / \mathrm{m}\) what are the magnitude and direction of the magnetic field of the wave at this same point in space and instant in time?

The microwaves in a certain microwave oven have a wavelength of \(12.2 \mathrm{~cm} .\) (a) How wide must this oven be so that it will contain five antinodal planes of the electric field along its width in the standing-wave pattern? (b) What is the frequency of these microwaves? (c) Suppose a manufacturing error occurred and the oven was made \(5.0 \mathrm{~cm}\) longer than specified in part (a). In this case, what would have to be the frequency of the microwaves for there still to be five antinodal planes of the electric field along the width of the oven?

Consider each of the electric- and magnetic-field orientations given next. In each case, what is the direction of propagation of the wave? (a) \(\vec{E}\) in the \(+x\) -direction, \(\overrightarrow{\boldsymbol{B}}\) in the \(+y\) -direction; (b) \(\overrightarrow{\boldsymbol{E}}\) in the \(-y\) -direction, \(\vec{B}\) in the \(+x\) -direction; (c) \(\vec{E}\) in the \(+z\) -direction, \(\vec{B}\) in the \(-x\) -direction; (d) \(\overrightarrow{\boldsymbol{E}}\) in the \(+y\) -direction, \(\overrightarrow{\boldsymbol{B}}\) in the \(-z\) -direction.

Consider electromagnetic waves propagating in air. (a) Determine the frequency of a wave with a wavelength of (i) \(5.0 \mathrm{~km}\), (ii) \(5.0 \mu \mathrm{m}\), (iii) \(5.0 \mathrm{nm}\). (b) What is the wavelength (in meters and nanometers) of (i) gamma rays of frequency \(6.50 \times 10^{21} \mathrm{~Hz}\) and (ii) an AM station radio wave of frequency \(590 \mathrm{kHz} ?\)

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