AP Waves Review

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Transcript AP Waves Review

Do Now (2/20/14): • • What are the major topics and concepts in our unit on waves?

Write your group members’ names down on the piece of paper at your desk

AP Waves Review

Which of the following waves can propagate in a vacuum?

I) light waves II) sound waves III) microwaves A) I) only B) II) only C) I) and III) only D) III) only E) all

Which of the following is not an example of electromagnetic wave?

I) x-rays II) water waves III) radio waves A) II) only B) I) only C) III) only D) I) and III) only E) I) and II) only

Which of the following represents the relationship between the frequency f, velocity v and wavelength λ of a wave?

A) λ=f/v B) f=v/λ C) f=λ/v D) λ=f+v E) v ⋅ λ=f

If a wave propagate into different media, which of the following change?

I) speed II) frequency III) wavelength A) I) only B) II) only C) III) only D) I) and III) only E) II) and III) only

If the distance between two successive crest of a traveling wave is equal to 50 cm and its frequency is equal to 500 Hz, then its speed is A) 250 m/s B) 1000 m/s C) 0.1 m/s D) 10 m/s E) 25,000 m/s

As a wave propagate from medium (A) to medium (B) its speed doubles. Which of the following is true about the wave?

A) the period doubles B) the frequency halves C) the wavelength doubles D) the amplitude doubles E) no property of the wave changes

How many complete cycles would a pendulum make in 40 seconds if its frequency 0.5 Hz?

A) 80 B) 120 C) 100 D) 40 E) 20

Sound does not pass through (a) steel (b) diamond (c) nitrogen (d) water (e) vacuum

Sound requires a material medium for its propagation, so sound does not pass through vacuum.

Ultrasonic waves from a sonar undergoes refraction at the interface between water and air. Which one of the following characteristics of the wave remains unchanged?

(a) Wave length

The correct option is (c). The

(b) Speed

period (and of course frequency) of the wave

remains unchanged.

(d) Energy (e) None of the above

A stationary wave of frequency 30 Hz is set up in a string of length 1.5 m fixed at both ends. The string vibrates with 3 segments. The speed of the wave along the string is (a) 10 ms –1 (b) 20 ms –1 (c) 30 ms –1 (d) 60 ms –1 (e) 90 ms –1

The distance between consecutive nodes (or anti-nodes) in a stationary wave is λ/2 where λ is the wave length. Therefore we have (from the figure) λ/2 = 0.5 m so that λ = 1m. Since speed v = n λ where n is the frequency we have v = 30×1 = 30 ms –1

What is the fundamental frequency of vibration of the string in the above question? (A stationary wave of frequency 30 Hz is set up in a string of length 1.5 m fixed at both ends. The string vibrates with 3 segments.) (a) 5 Hz (b) 10 Hz (c) 15 Hz (d) 30 Hz •

(e) 60 Hz

The speed of waves in the string is unchanged since the tension is unchanged. Since speed v = n 1

1 fundamental frequency and λ 1 where n 1 is the is the wave length in the fundamental mode of vibration, we have n 1 = v/λ 1 In the fundamental mode of vibration, the entire length of the string forms a single segment (with anti-node at the middle and nodes at the ends). Therefore we have λ 1 /2 = length of string = 1.5 m so that λ 1 = 3 m.

Substituting, n 1 = v/λ 1 = 30/3 = 10 Hz.

[You can work out this problem in no time remembering that the fundamental frequency is one third of the frequency with which the string vibrates with three segments. If the string were originally vibrating with four segments, the fundamental frequency would be one fourth].

When two tuning forks are sounded together 4 beats ars heard per second. One tuning fork is of frequency 346 Hz. When its prong is loaded with a little wax, the number of beats is increased to 6 per second. The frequency of the other fork is (a) 352 Hz (b) 340 Hz (c) 342 Hz (d) 346 Hz (e) 350 Hz

The frequency of the unknown fork must be either 342 Hz or 350 Hz since 4 beats are produced initially. When the fork of frequency 346 Hz is loaded with wax, its frequency is reduced. The number of beats then increased since its frequency is lower than that of the unknown fork.

The frequency of the unknown fork must therefore be 350 Hz

For a wave on the ocean, the amplitude is: a. the distance between crests.

b. the height difference between a crest and a trough.

c. one half the height difference between a crest and a trough.

d. how far the wave goes up on the beach.

e. one half the distance between crests.

Consider two identical and symmetrical wave pulses on a string. Suppose the first pulse reaches the fixed end of the string and is reflected back and then meets the second pulse. When the two pulses overlap exactly, the superposition principle predicts that the amplitude of the resultant pulses, at that moment, will be what factor times the amplitude of one of the original pulses?

a. 0 b. 1 c. 2 d. 4 e. 8

Equal wavelength waves of amplitude 0.25 m and 0.15 m interfere with one another. What is the resulting minimum amplitude that can result?

a. 0.15 m b. 0.10 m c. 0 m d. -0.40 m e. -0.60 m

Bats can detect small objects such as insects that are of a size approximately that of one wavelength. If bats emit a chirp at a frequency of 60 kHz, and the speed of sound waves in air is 330 m/s, what is the smallest size insect they can detect?

a. 1.5 mm b. 3.5 mm c. 5.5 mm d. 7.5 mm e. 9.5 mm

In an aquarium, light traveling through water (n = 1.3) is incident upon the glass container (n = 1.5) at an angle of 36° from the normal. What is the angle of transmission in the glass?

A. The light will not enter the glass because of total internal reflection.

B. 31° c. 36° d. 41° e. 52°

B—If you had a calculator, you could use Snell's law, calling the water medium "1" and the glass medium "2": 1.3·sin 36° = 1.5·sin θ 2 . You would find that the angle of transmission is 31°. But, you don't have a calculator … so look at the choices. The light must bend toward the normal when traveling into a material with higher index of refraction, and choice B is the only angle smaller than the angle of incidence. Choice A is silly because total internal reflection can only occur when light goes from high to low index of refraction.

Which of the following optical instruments can produce a virtual image with magnification 0.5?

I. convex mirror II. concave mirror III. convex lens IV. concave lens a. I and IV only b. II and IV only c. I and II only d. III and IV only e. I, II, III, and IV

A—The converging optical instruments— convex lens and concave mirror—only produce virtual images if the object is inside the focal point. But when that happens, the virtual image is larger than the object, as when you look at yourself in a spoon or a shaving mirror. But a diverging optical instrument—a convex mirror and a concave lens—always produces a smaller, upright image, as when you look at yourself reflected in a Christmas tree ornament .

Light waves traveling through air strike the surface of water at an angle. Which of the following statements about the light's wave properties upon entering the water is correct?

A. The light's speed, frequency, and wavelength all stay the same.

B. The light's speed, frequency, and wavelength all change.

C. The light's speed and frequency change, but the wavelength stays the same.

D. The light's wavelength and frequency change, but the light's speed stays the same.

E. The light's wavelength and speed change, but the frequency stays the same.

E—The speed of light (or any wave) depends upon the material through which the wave travels; by moving into the water, the light's speed slows down. But the frequency of a wave does not change, even when the wave changes material. This is why tree leaves still look green under water—color is determined by frequency, and the frequency of light under water is the same as in air. So, if speed changes and frequency stays the same, by v = λf, the wavelength must also change.

An object is placed at the center of a concave spherical mirror. What kind of image is formed, and where is that image located?

A. A real image is formed at the focal point of the mirror.

B. A real image is formed at the center of the mirror.

C. A real image is formed one focal length beyond the center of the mirror.

D. A virtual image is formed one focal length behind the mirror.

E. A virtual image is formed one radius behind the mirror.

B—You could approximate the answer by making a ray diagram, but the mirror equation works, too: Because the radius of a spherical mirror is twice the focal length, and we have placed the object at the center, the object distance is equal to 2f. Solve the mirror equation for d

by finding a common denominator: This works out to ( f )(2f )/(2f – f ) which is just 2f. The image distance is twice the focal length, and at the center point. This is a real image because d

is positive.

A talk show host inhales helium; as a result, the pitch of his voice rises. What happens to the standing waves in his vocal cords to cause this effect?

A. The wavelength of these waves increases.

B. The wavelength of these waves decreases.

C. The speed of these waves increases.

D. The speed of these waves decreases.

E. The frequency of these waves decreases.

C—The frequency of these waves must go up, because the pitch of a sound is determined by its frequency. The wavelength of the waves in the host's voice box doesn't change, though, because the wavelength is dependent on the physical structure of the host's body. Thus, by v = λf, the speed of the waves in his vocal cords must go up. You can even look it up—the speed of sound in helium is faster than the speed of sound in normal air.

In a pipe closed at one end and filled with air, a 384-Hz tuning fork resonates when the pipe is 22-cm long; this tuning fork does not resonate for any smaller pipes. For which of these closed pipe lengths will this tuning fork also resonate?

A. 11 cm b. 44 cm c. 66 cm d. 88 cm e. 384 cm

C—A wave in a pipe closed at one end has a node at one end and an antinode at the other. 22 cm is the length of the pipe for the fundamental oscillation, which looks like this: You can see that 1 / 2 of a "hump" is contained in the pipe, so the total wavelength (two full "humps") must be 88 cm. The next harmonic oscillation occurs when there is again a node at one end of the pipe and an antinode at the other, like this: This pipe contains 1 1 / 2 "humps," so its length equals three quarters of the total wavelength. The pipe length is thus 66 cm.

Monochromatic light passed through a double slit produces an interference pattern on a screen a distance 2.0 m away. The third-order maximum is located 1.5 cm away from the central maximum. Which of the following adjustments would cause the third-order maximum instead to be located 3.0 cm from the central maximum?

A. doubling the distance between slits b. tripling the wavelength of the light c. using a screen 1.0 m away from the slits d. using a screen 3.0 m away from the slits e. halving the distance between slits

E—Use the equation Here m = 3 because we are dealing with the third-order maximum. We want to double the distance to this third-order maximum, which means we want to double x in the equation. To do this, halve the denominator; d in the denominator represents the distance between slits.

The two wave pulses shown are moving toward each other along a string. When the two pulses interfere, what is the maximum amplitude of the resultant pulse?

A. (1/2)A b. A c. (3/2)A d. 2A e. (5/2)A

D—When wave pulses interfere, their amplitudes add algebraically. The question asks for the maximum amplitude, so the widths of the pulses are irrelevant. When both pulses are right on top of one another, each pulse will have amplitude A; these amplitudes will add to a resultant of 2A.

Free Response:

Laser light is passed through a diffraction grating with 7000 lines per centimeter. Light is projected onto a screen far away. An observer by the diffraction grating observes the first order maximum 25° away from the central maximum.

A. What is the wavelength of the laser?

A. (a) Use d sin θ = mλ. Here d is not 7000! d represents the distance between slits. Because there are 7000 lines per centimeter, there's 1/7000 centimeter per line; thus, the distance between lines is 1.4 × 10–4 cm, or 1.4 × 10–6 m. θ is 25° for the first-order maximum, where m = 1. Plugging in, you get a wavelength of just about 6 × 10 –7 m, also known as 600 nm.

B. If the first order maximum is 40 cm away from the central maximum on the screen, how far away is the screen from the diffraction grating?

• This is a geometry problem. tan 25° = (40 cm)/L; solve for L to get 86 cm, or about 3 feet.

C. far, measured along the screen, from the central maximum will the second-order maximum be?

C. (c) Use d sin θ = mλ; solve for θ using m = 2, and convert everything to meters. We get sin θ = 2(6.0 × 10 –7 m)/(1.4 × 10–6 m). The angle will be 59°. Now, use the same geometry from part (b) to find the distance along the screen: tan 59° = x/(0.86 m), so x= 143 cm. (Your answer will be counted correct if you rounded differently and just came close to 143 cm.)