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Chapter 32: Problem 48

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鈥攆rom 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 electromagnetic wave with frequency \(5.70 \times 10^{14} \mathrm{~Hz}\) propagates with a speed of \(2.17 \times 10^{8} \mathrm{~m} / \mathrm{s}\) in a certain piece of glass. Find (a) the wavelength of the wave in the glass; (b) the wavelength of a wave of the same frequency propagating in air; (c) the index of refraction \(n\) of the glass for an electromagnetic wave with this frequency; (d) the dielectric constant for glass at this frequency, assuming that the relative permeability is unity.

An electromagnetic wave has an electric field given by \(\overrightarrow{\boldsymbol{E}}(y, t)=\left(3.10 \times 10^{5} \mathrm{~V} / \mathrm{m}\right) \hat{k} \cos \left[k y-\left(12.65 \times 10^{12} \mathrm{rad} / \mathrm{s}\right) t\right] .(\mathrm{a}) \mathrm{In}\) which direction is the wave traveling? (b) What is the wavelength of the wave? (c) Write the vector equation for \(\boldsymbol{B}(y, t)\)

Radio station WCCO in Minneapolis broadcasts at a frequency of \(830 \mathrm{kHz}\). At a point some distance from the transmitter, the magnetic- field amplitude of the electromagnetic wave from \(\mathrm{WCCO}\) is \(4.82 \times 10^{-11} \mathrm{~T}\). Calculate (a) the wavelength; (b) the wave number; (c) the angular frequency; (d) the electric-field amplitude.

An electromagnetic wave with frequency \(65.0 \mathrm{~Hz}\) travels in an insulating magnetic material that has dielectric constant 3.64 and relative permeability 5.18 at this frequency. The electric field has amplitude \(7.20 \times 10^{-3} \mathrm{~V} / \mathrm{m} .\) (a) What is the speed of propagation of the wave? (b) What is the wavelength of the wave? (c) What is the amplitude of the magnetic field?

Electromagnetic waves propagate much differently in conductors than they do in dielectrics or in vacuum. If the resistivity of the conductor is sufficiently low (that is, if it is a sufficiently good conductor), the oscillating electric field of the wave gives rise to an oscillating conduction current that is much larger than the displacement current. In this case, the wave equation for an electric field \(\vec{E}(x, t)=E_{y}(x, t) \hat{\jmath}\) propagating in the \(+x\) -direction within a conductor is $$ \frac{\partial^{2} E_{y}(x, t)}{\partial x^{2}}=\frac{\mu}{\rho} \frac{\partial E_{y}(x, t)}{\partial t} $$ where \(\mu\) is the permeability of the conductor and \(\rho\) is its resistivity. (a) \(\mathrm{A}\) solution to this wave equation is \(E_{y}(x, t)=E_{\max } e^{-k_{\mathrm{C}} x} \cos \left(k_{\mathrm{C}} x-\omega t\right)\) where \(k_{\mathrm{C}}=\sqrt{\omega \mu / 2 \rho}\). Verify this by substituting \(E_{y}(x, t)\) into the above wave equation. (b) The exponential term shows that the electric field decreases in amplitude as it propagates. Explain why this happens. (Hint: The field does work to move charges within the conductor. The current of these moving charges causes \(i^{2} R\) heating within the conductor, raising its temperature. Where does the energy to do this come from?) (c) Show that the electric-field amplitude decreases by a factor of \(1 / e\) in a distance \(1 / k_{\mathrm{C}}=\sqrt{2 \rho / \omega \mu},\) and calculate this distance for a radio wave with frequency \(f=1.0 \mathrm{MHz}\) in copper (resistivity \(1.72 \times 10^{-8} \Omega \cdot \mathrm{m} ;\) permeability \(\mu=\mu_{0}\) ). Since this distance is so short, electromagnetic waves of this frequency can hardly propagate at all into copper. Instead, they are reflected at the surface of the metal. This is why radio waves cannot penetrate through copper or other metals, and why radio reception is poor inside a metal structure.
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