AP Physics 2 - Other Wave Concepts

Which of the following parameters will increase when the frequency of a sound wave is decreased?

I. Period
II. Wavelength
III. Amplitude

I and II

I, II, and III

I and III

II and III

II only

Explanation

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:

$f=\frac{1}{T}$

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:

$v=f \lambda$

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.

Which of the following waves can propagate in a vacuum?

I. Sound waves

II. X-ray waves

III. Radio waves

II and III

I only

II only

III only

I, II, and III

Explanation

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.

Determine the frequency of electromagnetic radiation of wavelength $1000\textup{ nm}$ .

$3*10^{14}\textup{ Hz}$

$9*10^{14}\textup{ Hz}$

$1*10^{14}\textup{ Hz}$

$800\textup{ Hz}$

$6*10^{14}\textup{ Hz}$

Explanation

Using the following equation:

$c=\lambda* u$

Converting to and plugging in values:

$3.0*10^8\frac{m}{s}=1000*10^{-9}m* u$

$u=3*10^{14}Hz$

Which of the following electromagnetic waves has the longest wavelength?

Radio waves

Microwaves

Visible light

X-rays

Gamma rays

Explanation

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

Radio
Micro
Infrared
Red
Orange
Yellow
Green
Blue
Indigo
Violet
Ultraviolet
X-ray
Gamma

A team of engineers decides to drill a hole all the way through the diameter of the Earth. A wave (with wavelength ) is sent through the hole to a receiver on the other end who, upon receiving the wave, immediately sends the wave back through the hole to the original sender. The original sender says it took the wave exactly half a second to return to him after he sent it. The hole is approximately long. What is the frequency of the wave?

$1.49*10^{14}s^{-1}$

$8.57*10^{14}s^{-1}$

$18.2s^{-1}$

$6.73*10^{-15}s^{-1}$

$5.2*10^7s^{-1}$

Explanation

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.

$s=\frac{d}{t}$

$s=\frac{(2*13*10^6)}{.5}$

$s=5.2*10^7\frac{m}{s}$

Now we can use $s=\lambda* u$ to find the frequency.

$\lambda$ -wavelength

-frequency

$5.2*10^7=(350*10^{-9})* u$

$u=1.49*10^{14}s^{-1}$

*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 $\approx5.77$ .

What is defined by the term wavelength?

From the crest of one wave to the subsequent crest of the next wave.

How far a wave travels in one second.

How fast the waves travel in one second.

The number of waves going past in one second.

The number of seconds it takes for one full wave to pass.

Explanation

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.

For open pipes, the formula for wave patterns at any given time can be given by a Fourier Sine Series which is given as the infinite sum:

$y(x)=\sum_{n=1}^{\infty} sin(\frac{n\pi x}{l})$ when is an integer, and is the length of the pipe. Each individual value of is called a harmonic.

What is the wavelength of the fundamental harmonic?

$\frac{l}{2}$

$\frac{l}{2\pi}$

Explanation

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

$W=\frac{2\pi}{b}=\frac{2\pi}{{\frac{\pi }{l}}}=2l$ , where is the frequency of the fundamental harmonic.

As a wave propagates from one medium to another, the speed decreases by a factor of two (halves). Which of the following is true about the wave in the second medium?

Wavelength halves

Frequency halves

Frequency doubles

Wavelength doubles

No change

Explanation

$v=\lambda f$ , where is the velocity of the wave, $\lambda$ 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.

As you gradually turn down the light on a dimmer switch, you notice that it shows a red glow the instant before it turns off. Why does this happen?

Compared to other colors, red light has less energy.

Compared to other colors, red light travels through air faster.

Compared to other colors, red light travels through air slower.

Compared to other colors, red light has more energy.

Explanation

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

$E=\frac{hc}{\lambda }$

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.

Consider an electromagnetic wave travelling through a medium where the speed of light is . If the strength of the electric field is $21000\frac{N}{C}$ , and the speed of light is $3 \cdot 10^8\frac{m}{s}$ , what is the magnetic field strength?

$7.45 \cdot 10^{-5} T$

$1.55 \cdot 10^{-5} T$

$9.41 \cdot 10^{-5} T$

$6.45 \cdot 10^{-4} T$

$9.31 \cdot 10^{-4} T$

Explanation

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 ,

$\frac{E}{B}=v$

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.