Other Wave Concepts - AP Physics 2

In an electromagnetic wave going through a vacuum, the ratio of and is the speed of light, ; however, this relation doesn't change when light goes through a medium. The ratio still is the speed of light, but instead of , it's the speed of light through the medium. Therefore, for velocity ,

In this problem, we're told that the speed of light through the medium is , so if we rearrange the equation to solve for and use in place of v, we'll get our answer.

Therefore, the strength of the magnetic field is .

Wavelength is indirectly related to the amount of energy in the wave (the frequency). The most energetic wves are gamma rays, while the least energetic are radio waves. This means that radio waves have the longest wavelength.

A helpful mnemonic for remembering the order of wavelengths in the electromagnetic spectrum from longest to shortest is "Raging Martians invaded Roy G Biv using x-ray guns."

In order, the letters stand for

1. Radio
2. Micro
3. Infrared
4. Red
5. Orange
6. Yellow
7. Green
8. Blue
9. Indigo
10. Violet
11. Ultraviolet
12. X-ray
13. Gamma

When light is refracted, the average speed is lower if the index of refraction is higher (and vice versa). In order to refract, the medium light travels through must have a changing index of refraction. Because the amount of energy the light carries must be constant during refraction, the frequency (which is proportional to energy) cannot change. Therefore, since and the frequency must stay the same when the speed changes, wavelength must also change.

When trying to find the minimum distance between two objects that makes them still distinguishable, we use the Rayleigh criterion equation.

is the minimum angle at which two objects can be resolved, is the wavelength of the wave being used, and is the diameter of the circular aperture (in this case, the antenna would be the aperture). An important part of equations like these is the small-angle approximation, which states, as the angle approaches zero,

This means that, if the angle is small enough, . In our problem, we're trying to find , so we can substitute for

Next, since we're not given the wavelength, but we are given the frequency, we can find the wavelength using the following equation:

Now, solve for .

Using

Where is the speed of light, is frequency and is wavelength

Converting to and plugging in values

Solving for

Using the following equation:

Converting to and plugging in values:

For this question, we're given the distance between wave peaks (wavelength) as well as how long it takes for each wave to hit the shore (period). We're then asked to use this information to solve for the speed of the waves.

First, we'll need to use an equation for the speed of a wave.

We know from the question stem that we have the wavelength, which is the distance between the peaks of each wave. But we do not have the frequency. However, we can solve for frequency by using the period of the wave as follows:

Which we can alternatively write as:

The period of the wave, , represents the amount of time it takes for a single wave to pass a given point, and thus has units of time. The inverse of this quantity is the frequency, which describes how many waves pass by a given point in a certain amount of time. Usually, the unit of time in these cases is seconds.

Thus, for every one second that passes, one-fifth of a wave will pass by any given point.

Now, we can plug this value into the equation for the speed of the wave to solve for our answer:

To answer this question, let's first recall that on the visible spectrum, red light has the longest wavelength. Due to this, red light also has the lowest energy. We can show this with the equation

As the dimmer is gradually turned down, less and less energy is being provided to the light bulb. Consequently, since red light has the least amount of energy of all the colors on the visible spectrum, we briefly see red light the instant before the light turns off completely. Note that the speed of a wave is determined by the medium. In this case, regardless of the wavelength, the medium is air, which determines the speed of any electromagnetic wave.

For this question, we need to consider wave characteristics. Specifically, we need to determine how a decrease in the frequency of a wave will alter its wavelength, period, and amplitude.

First, let's consider the period of the wave. The period is defined as the amount of time needed for one complete cycle of the wave to occur. Conversely, frequency is defined as the number of wave cycles completed within a given time frame. As such, period and frequency are inversely related to one another, as the following expression shows:

Therefore, the period of the wave will certainly increase as the frequency of the wave decreases.

Now, let's take a look at wavelength. We need to recall the speed of a wave is defined in terms of its wavelength and frequency according to the following equation:

As we can see from the above equation, frequency is inversely related to wavelength, just as it is with period. Therefore, as the frequency of a wave decreases, the wavelength will indeed rise.

Finally, let's look at amplitude. The amplitude of a wave is the magnitude of the difference between the extremes of the wave and its equilibrium position. In transverse waves, such as a rope, the amplitude is the maximum displacement of a particle of that rope from its equilibrium position in the direction perpendicular to the propagation of the wave. In longitudinal waves, such as sound, the amplitude is the maximum displacement of the medium from its equilibrium position in the direction parallel to the propagation of the wave.

Because there is no relationship between amplitude and frequency, a decrease in a wave's frequency will have no effect on that wave's amplitude. Thus, for this question, only wavelength and period are increased due to a decreased frequency.

We're given the wavelengths of two subsequent harmonics of a string fixed at both ends, and we're asked to solve for the length of the string.

First, we must used the equation for a standing wave that has a node at both ends.

Before we can solve for the length of the string, we need to first solve for the value of . We can do this by considering the wavelength of the two harmonics as follows.

Now, since the length of the string for either harmonic is the same length, we can set the two expressions above equal to each other.

Now, we can solve for the variable .

Now that we have the value for , we can plug this value into the original expression to calculate the length of the string.

We are given the distance the wave travels (twice the Earth's diameter), and the time it took for it to travel that distance. Using this we can determine the wave's speed.

Now we can use to find the frequency.

-wavelength

-frequency

*Side note-Let's assume the index of refraction of the hole is constant. In order for the wave to travel at that speed, the index of refraction must be .

In a vacuum, waves that travel at the speed of light propagate. Therefore, we need to determine which of the three types of waves have that speed. All parts of the electromagnetic spectrum travel at the speed of light (so X-ray and radio are true). Sound waves travel at the speed of sound, which is less than the speed of light. Therefore, that one is incorrect.

, where is the velocity of the wave, is the wavelength of the wave, and is the frequency. When the speed is cut in half, the frequency does not change! However, that means the wavelength is directly proportional to the speed and will also be cut in half.

Wavelength and frequency are inversely related. We define the wavelength of a wave as the distance between the two subsequent waves measured at the same point on the wave. Therefore, we can measure from crest to crest or trough to trough.

The fundamental harmonic is when , and therefore the wavelength can be given by:

, where is the frequency of the fundamental harmonic.

In this question, we're told that a certain sound is increases by decibals. We're asked to find how the intensity of the sound changes as a result.

To answer this, we need to understand that there is a logarithmic relationship between decibals and intensity. This can be shown by the following expression that relates the two.

We can rearrange this equation to isolate the intensity ratio of after and before.

This shows that the intensity increases by a factor of .

As a side note, for every decibals added, the intensity of the sound increases by a factor of .

This question is asking us to identify a statement that is not consistent with the wave theory of light.

In its beginning, light was traditionally viewed as being only a wave. In time, however, certain observations came to light (no pun intended) that did not agree with the wave theory. As a result, scientists needed to turn to a new theory that could account for the new observations.

This led to what is called the particle nature of light. When shining a beam of light on a metal plate, it was found that only certain frequencies of light could eject electrons from the metal. However, the wave theory of light alone could not explain this phenomenon, which came to be called the photoelectric effect. Instead, if light were to be treated like a particle, in which there are many discrete photons that behave like particles, the photoelectric effect can be adequately explained.

It's important to note, however, that this did not usurp the wave model. In fact, the wave model is able to explain several things that the particle model cannot. For instance, the diffraction, refraction, and interference of light are all consistent with the wave model, but not the particle model. Consequently, both models are valid and are invoked depending on the circumstance. This is what is known as the wave-particle duality of light.

In an electromagnetic wave, the electric and magnetic portions are proprtional to each other. The ratio of electric field to magnetic field is , the speed of light.

Using this information, we can determine the strength of one part given the other.

We have and , so now we multiply the two.

Therefore, the strength of the electric field is .

In an electromagnetic wave going through a vacuum, the ratio of and is the speed of light, ; however, this relation doesn't change when light goes through a medium. The ratio still is the speed of light, but instead of , it's the speed of light through the medium. Therefore, for velocity ,

In this problem, we're told that the speed of light through the medium is , so if we rearrange the equation to solve for and use in place of v, we'll get our answer.

Therefore, the strength of the magnetic field is .