### All College Physics Resources

## Example Questions

### Example Question #1 : Electromagnetics, Waves, And Optics

When a ray of light is reflected off of the surface of another medium, some of the energy from the light is transferred into this medium. As a result, which of the following is true about the reflected light ray?

**Possible Answers:**

It has a lower frequency and a higher wavelength

It has a higher frequency and a higher wavelength

It has a higher frequency and a lower wavelength

It has a lower frequency and a lower wavelength

**Correct answer:**

It has a lower frequency and a higher wavelength

For this question, we're told that light is reflected off of the surface of another medium. As a result, the light loses some of its energy, which is transferred into the medium where it reflected.

Because the light has lost some of its energy, we need to determine how the wavelength and frequency of the wave will be affected. To do this, we can recall the equation for the energy of a wave.

Where is Planck's constant and represents the frequency. We can see that when energy is decreased, the frequency also decreases. Thus, we can eliminate two of the answer choices.

To see how the wavelength changes, it's important to recall the relationship that wavelength and frequency have with each other.

Where is the speed of light and is the wavelength. Since the light is still traveling in the same medium, its speed will not change. Thus, as the frequency of the wave decreases, the wavelength has to increase. We can alternatively show how all of these variables are related as follows.

So, in summary, when the energy of light decreases, the frequency will decrease and the wavelength will increase.

### Example Question #1 : Electromagnetics, Waves, And Optics

Suppose that car A and car B are both traveling in the same direction. Car A is going and is sounding a horn with a frequency of . If car B is traveling at a speed of in front of car A, what frequency of sound does car B hear?

Note: The speed of sound in air is .

**Possible Answers:**

**Correct answer:**

The important concept being tested in this question is the Doppler effect. When a source of sound waves (or any other wave) is emitted from a source and is moving relative to some observer, the actual frequency of the wave will be different for the observer.

For starters, we'll need to use the Doppler equation.

The trickiest part about this question is to decide whether to add or subtract in both the numerator and denominator. To find the right sign orientation, it's helpful to do a quick thought experiment. Imagine that only one of them is moving and the other is stationary. Decide how that will affect the observed frequency; will it increase it or decrease it? Then repeat for the other one.

First, let's consider the detector (aka observer). We're told that car B, the detector, is traveling ahead of car A and also in the same direction. This means that car B is driving away from car A. So in this situation, is the frequency that car B hears expected to increase or decrease? The answer is that it will decrease. Since car B is traveling away from car A, each successive wave will take longer to reach car B. Hence, we will use subtraction in the numerator because that will make the observed frequency smaller.

Now let's apply this same logic to the denominator, which deals with the source of the sound waves. We know that car A is traveling in the same direction as car B and is behind. This means that car A is traveling toward car B. So from this perspective, each successive wave is expected to get closer together, thus making the time between each wave smaller and the frequency bigger. In the denominator, will adding or subtracting make the observed frequency bigger? The answer is subtraction. By making a smaller number in the denominator, the entire fraction becomes larger.

Keeping this information in mind, we'll need to use subtraction in both the numerator and denominator. Once we plug in the values given in the question stem, we have everything we need to solve for the answer.

### Example Question #3 : Electromagnetics, Waves, And Optics

On planet Borg, a male insect is flying toward his mate at while buzzing at . The female insect is stationary and is buzzing at . What is the speed of the sound on planet Borg?

**Possible Answers:**

**Correct answer:**

To begin, we'll start with the Doppler Effect equation:

Rearrange the equation by solving for to get:

Finally we plug in our known values and solve for :

### Example Question #1 : Electromagnetics, Waves, And Optics

Suppose that the fifth harmonic of a standing wave contained within a pipe closed at both ends has a wavelength of . What is the length of the pipe?

**Possible Answers:**

**Correct answer:**

For this question, we're told that a standing wave is contained within a pipe closed at both ends. We're also given the wavelength for the fifth harmonic, and are asked to find the length of the pipe.

The first step to solve this problem is to use the equation for a pipe closed at both ends.

Since we're told which harmonic the wave is on, as well as its wavelength, we have everything we need to solve for the length of the pipe.

### Example Question #71 : College Physics

A string under tension is oscillated at the 5th harmonic. What is the wavelength of this oscillation?

**Possible Answers:**

**Correct answer:**

Each harmonic has a wavelength that behaves according to the equation

, where is the wavelength and is the harmonic.

Since this is the 5th harmonic, and . Solving for wavelength gives .

### Example Question #3 : Electromagnetics, Waves, And Optics

A pipe has a length of . If the pipe is open at both ends, what is the frequency of the third harmonic assuming that the velocity in which sound moves through the air is .

**Possible Answers:**

**Correct answer:**

Since the pipe is open at both ends, we can use the equation where f is the frequency we are solving for, is the number of the harmonic, is the velocity at which sound moves through air, and L is the length of the pipe.

We know from the question that:

When we plug in these values into the frequency equation, we get as the answer.