AP Waves Review

Report
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
period (and of course
(b) Speed
frequency) of the wave
(c) Period
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
The distance between consecutive
(a) 10 ms–1
nodes (or anti-nodes) in a
stationary wave is λ/2 where λ is
(b) 20 ms–1
the wave length. Therefore we
–1
(c) 30 ms
have (from the figure) λ/2 = 0.5 m
so that λ = 1m. Since speed v = n
(d) 60 ms–1
λ where n is the frequency we have
–1
(e) 90 ms
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
• B.
(d) 30 Hz
(e) 60 Hz
The speed of waves in the string is unchanged since the
tension is unchanged. Since speed v = n1λ1 where n1 is the
fundamental frequency and λ1 is the wave length in the
fundamental mode of vibration, we have n1 = v/λ1
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 λ1/2 = length of
string = 1.5 m so that λ1 = 3 m.
Substituting, n1 = v/λ1 = 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 θ2. 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 di 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 di 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 11/2 "humps," so its length equals threequarters 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 × 10–7m, also known
as 600 nm.
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.
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?
B.
• This is a geometry problem. tan 25° = (40 cm)/L;
solve for L to get 86 cm, or about 3 feet.
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.
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.)

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