Version 001 – Waves extra practice - optional – tubman – (20131B)
This print-out should have 30 questions.
Multiple-choice questions may continue on
the next column or page – find all choices
before answering.
Conceptual 14 03
001 10.0 points
Andrea asked her brother to take a 4 ft floating raft out of the water near the wave-swept
shore. Using this raft as a measuring tool,
she estimated that the wavelengths of these
particular ocean waves were about 15 ft.
How fast are these surface ocean waves if
1
the frequency remains Hz?
4
Correct answer: 3.75 ft/s.
1
Holt SF 12Rev 51
003 (part 1 of 2) 10.0 points
The notes produced by a violin range in frequency from approximately 196 Hz to 2637
Hz.
Find the maximum wavelength in air produced by this instrument when the speed of
sound in air is 345 m/s.
Correct answer: 1.7602 m.
Explanation:
Let : fmin = 196 Hz and
v = 345 m/s .
Explanation:
Let : λ = 15 ft and
1
f = Hz .
4
1
Hz (15 ft) = 3.75 ft/s .
v=fλ=
4
v
λmax =
fmin
=
345 m/s
= 1.7602 m .
196 Hz
004 (part 2 of 2) 10.0 points
Find the minimum wavelength produced by
this instrument.
Correct answer: 0.13083 m.
keywords:
Explanation:
Concept 19 05
002 10.0 points
Radio waves travel at the speed of light:
3 × 105 km/s.
What is the wavelength of radio waves received at 94.9 MHz on your FM radio dial?
Correct answer: 3.16122 m.
Explanation:
Let :
v = 3 × 105 km/s
= 300 million m/s and
f = 94.9 MHz
= 94.9 million Hz .
Let :
λmin =
v
fmax
fmax = 2637 Hz .
=
345 m/s
= 0.13083 m .
2637 Hz
Longitudinal Wave Propagation
005 10.0 points
Longitudinal sound waves cannot propagate
through
1. solids
2. steam
3. gases
4. liquids
v
300 million m/s
λ= =
= 3.16122 m .
f
94.9 million Hz
5. a vacuum correct
Version 001 – Waves extra practice - optional – tubman – (20131B)
This sound travels to the object and returns, so
6. wet sand
Explanation:
A longitudinal sound wave is a pressure
wave which can propagate through any media
whose pressure may vary, including gases, liquids, solids or even mixtures such as wet sand
or steam (tiny water droplets in air / water
vapor mixture). The one ‘medium’ through
which a pressure wave cannot propagate is a
vacuum.
AP B 1993 MC 30
006 10.0 points
Sound in air can best be described as which
of the following type of wave?
2 ∆x = v∆t
v ∆t
(341 m/s)(3.7 s)
∆x =
=
2
2
= 630.85 m .
Tipler PSE5 16 24
008 10.0 points
Two sound sources oscillate in phase with the
same amplitude A. They are separated in
5λ
space by
.
3
What is the amplitude (as a multiple of A)
of the resultant wave formed from the two
sources at a point that is on the line that
passes through the sources but is not between
the sources?
Correct answer: 1 A.
1. Transverse
Explanation:
The phase shift in the waves generated by
these two sources is due to their separation of
5λ
∆x =
. The phase difference due to the
3
path difference is
2. Polarized
3. Electromagnetic
4. Longitudinal correct
∆x
λ
∆x
5
5π
δ
=π
=π =
,
2
λ
3
3
δ = 2π
5. Torsional
Explanation:
A sound wave in the air is propagated by
the oscillation of air molecules. It is best
described as longitudinal wave.
Holt SF 12Rev 50
007 10.0 points
A sound wave traveling at 341 m/s is emitted
by the foghorn of a tugboat. An echo is heard
3.70 s later.
How far away is the reflecting object?
Correct answer: 630.85 m.
Explanation:
Let : v = 341 m/s
∆t = 3.70 s .
2
and
so the amplitude of the resultant wave is
5π
δ
Ares = 2 y0 cos = 2 A cos 2
3
= 1A .
keywords:
Tipler PSE5 16 33
009 (part 1 of 2) 10.0 points
Sound source A is located at x = 0, y = 0, and
sound source B is place at x = 0, y = 2.5 m.
The two sources radiate coherently in phase.
An observer at x = 80 m, y = 0 notes that
Version 001 – Waves extra practice - optional – tubman – (20131B)
as she takes a few steps in either the positive
or negative y direction away from y = 0, the
sound intensity diminishes.
What is the lowest frequency of the sources
that can account for that observation?
Correct answer: 8706.12 Hz.
Explanation:
Let : rA = 80 m ,
rAB = 2.5 m , and
vsound = 340 m/s .
She observes constructive interference initially, so her distance from source B is
q
2 ,
rB = rA 2 + rAB
and the path difference is
p
∆r = rB − rA = rA 2 + rAB 2 − rA
q
= (80 m)2 + (2.5 m)2 − 80 m
= 0.039053 m .
For constructive interference at (80 m, 0)
∆r = nλ, n = 1, 2, 3, · · ·
∆r
, so
λ=
n
vsound = f λ
v
n · vsound
fn = sound =
.
λ
∆r
and the lowest frequency is
fn=1 =
1(340 m/s)
= 8706.12 Hz .
0.039053 m
The next higher frequency is
fn=2 =
2(340 m/s)
= 17412.2 Hz .
0.039053 m
keywords:
Two Waves
011 (part 1 of 2) 10.0 points
Consider two waves traveling through the
same medium in the same time frame.
A
B
Compare the wavelengths.
1. Cannot be determined
2. A has the longer wavelength. correct
3. A and B have the same wavelength.
4. B has the longer wavelength.
Explanation:
A exhibits three complete wavelengths in
the same time that B exhibits five complete
wavelengths, so A has a longer wavelength.
012 (part 2 of 2) 10.0 points
Compare the frequencies.
1. A and B have the same frequency.
2. A has the higher frequency.
3. B has the higher frequency. correct
010 (part 2 of 2) 10.0 points
What is the next higher frequency of the
sources that can account for that observation?
Correct answer: 17412.2 Hz.
Explanation:
3
4. Cannot be determined
Explanation:
v = λf
v
f= .
λ
Version 001 – Waves extra practice - optional – tubman – (20131B)
Since the speeds are the same and A has a
longer wavelength λ, then A must have the
lower frequency.
Fixed End Pulse Reflection
013 10.0 points
A pulse moves on a string at 1 m/s, traveling
to the right. At point A, the string is tightly
clamped and cannot move.
1 m/s
4
Explanation:
The pulse travels to the right and is reflected back from the fixed point. The wave
cannot propagate to the right beyond the
point A. Reflected pulses from fixed end
points are inverted.
Reflection at a barrier
014 10.0 points
A
1m
Which of the following shows how the string
would look soon after 2 seconds?
1.
2.
3.
4.
5.
6.
A
A
A
A
A
A
1. has a larger speed than the original
wave.
correct
7.
8.
Wave Reflected From Fixed End
015 10.0 points
A wave on a string is reflected from a fixed
end.
The reflected wave
A
A
2. cannot be transverse.
3. is 180◦ out of phase with the original wave
at the end. correct
4. has a larger amplitude than the original
Version 001 – Waves extra practice - optional – tubman – (20131B)
5
wave.
5. is in phase with the original wave at the
end.
Explanation:
At a fixed end, the incident and reflected
wave must always add to zero (since the end is
fixed and cannot move), so they must always
be 180◦ out of phase with each other at that
end.
Wave Action at a Boundary
016 10.0 points
A wave is turned back when it meets the
boundary of the medium in which it is traveling.
The wave is said to have undergone
4. Jamie hears a much lower amplitude whistle than Dustin. correct
5. Jamie hears a moderately lower amplitude
whistle than Dustin.
Explanation:
Destructive interference between the two
whistles is greatest when the distance to the
two speakers differs by half a wavelength.
Constructive interference between the two
whistles is greatest when the distance to the
two speakers differs by a full wavelength.
The distance from Jamie to the two speakers differs by half a wavelength, and the distance from Dustin to the two speakers differs
by a full wavelength, so Jamie hears a much
lower amplitude whistle than Dustin.
1. interference.
2. reflection. correct
3. diffraction.
4. refraction.
Explanation:
A wave that is turned back at a boundary
has undergone reflection.
1. Jamie hears a moderately higher amplitude whistle than Dustin.
2. Jamie hears a much higher amplitude
whistle than Dustin.
3. Jamie hears the whistle at the same amplitude as Dustin.
10 m
center line
intensity
Interference From Two Speakers
017 10.0 points
Two electronic speakers, 1 and 2, are generating a fixed frequency whistle at the same amplitude. Jamie is sitting one wavelength away
from speaker A and 1.5 wavelengths from
speaker B . Dustin is sitting one wavelength
away from speaker A and 2 wavelengths away
from speaker B .
Which statement below is correct?
Serway CP 14 32 A new
018 (part 1 of 2) 10.0 points
Two loudspeakers are placed a distance of 4 m
above and below one another and driven by
the same source at a frequency of 737.85 Hz .
A detector is positioned a perpendicular distance of 10 m from the two speakers and a
vertical height 3.2 m from the lower speaker.
This results in constructive interference at the
first order maximum.
3.2 m
4m
What minimum distance directly toward or
away from the detector should the top speaker
be moved in order to create destructive interference between the two speakers at the location of the detector? The velocity of sound is
345 m/s .
Correct answer: 0.233787 m.
Explanation:
Version 001 – Waves extra practice - optional – tubman – (20131B)
The two waves will be in phase and constructively interfere at the detector as long
as the speakers’ distances from the detector
only differ by an integer multiple of the wavelength. For destructive interference, the distances must differ by a half-integer multiple
of the wavelength. Therefore the smallest distance the top speaker could be moved either
directly toward or away from the detector
would be half a wavelength.
We can find the wavelength using the formula
v = f λ.
We are given the speed of sound and also the
frequency of the waves, so the wavelength is
λ=
v
345 m/s
=
= 0.467575 m.
f
737.85 Hz
Half of a wavelength is therefore
λ
0.467575 m
=
= 0.233787 m.
2
2
6
Which of the labeled points are...
...on nodal lines?
1. C, E, F, H
2. A, B, D, G, I, J correct
3. A, B, C, E, F, H
4. C, E, G, I, J
5. A, C, D, G, H, J
Explanation:
Nodal lines, in this context, are where crests
overlap with troughs, so the correct points are
A, B, D, G, I, and J.
021 (part 2 of 4) 10.0 points
...on antinodal lines?
1. A, B, C, E, F, H
019 (part 2 of 2) 10.0 points
If the top speaker is moved by twice the distance calculated in the previous part, what
will the interference be?
2. C, E, F, H correct
3. C, E, G, I, J
1. Unable to determine
4. A, C, D, G, H, J
2. Constructive correct
5. A, B, D, G, I, J
3. Destructive
Explanation:
The speakers will now be separated by a
full wavelength and constructive interference
will occur again.
nodallines
020 (part 1 of 4) 10.0 points
Explanation:
Antinodal lines are where crests overlap
with crests, or troughs with troughs, so the
correct points are C, E, F, and H.
022 (part 3 of 4) 10.0 points
...formed as the result of constructive interference?
1. A, C, D, G, H, J
2. A, B, D, G, I, J
3. C, E, G, I, J
4. A, B, C, E, F, H
5. C, E, F, H correct
Version 001 – Waves extra practice - optional – tubman – (20131B)
Explanation:
Constructive interference occurs along
antinodal lines, so the answer here is the same
as for part 2.
023 (part 4 of 4) 10.0 points
...formed as the result of destructive interference?
7
pointsources
025 10.0 points
Two point sources are generating periodic
waves in phase. The wavelength is 4.0 cm.
A point on the second antinodal line is 30.0
cm from the nearest source. How far is this
point from the farthest source? Begin by constructing a sketch of the physical situation.
1. C, E, F, H
Correct answer: 38 cm.
2. A, C, D, G, H, J
3. A, B, C, E, F, H
4. A, B, D, G, I, J correct
5. C, E, G, I, J
P.D. = mλ,
Explanation:
Destructive interference occurs along nodal
lines, so the answer here is the same as for
part 1.
Standing Waves 19
024 10.0 points
The distance between two nearest nodes of
a standing wave is 21.7 cm . Hand generated
pulses move up and down through a complete
cycle five times every nine seconds.
Find the velocity of the wave.
Correct answer: 0.241111 m/s.
Explanation:
Let :
d = 21.7 cm ,
n = 5 , and
t = 9 s.
The distance between two nodes is half the
wavelength, so λ = 2 d . The frequency is the
n
number of cycles per unit time, so f = and
t
the velocity is
2nd
2 (5) (21.7 cm)
=
t
9s
= 0.241111 m/s .
v=fλ=
Explanation:
This question is really asking us to find the
path difference (P.D.) between the near and
far sources to the point in question. The
equation here is just
where m is the order of the line and λ is the
wavelength. Since the point in question is
on the second antinodal line, we know that
m = 2. We also know that λ = 4.0 cm. Then
the path difference is
P.D. = (2)(4.0 cm) = 8.0 cm,
meaning the point in question must be 38 cm
from the farther source.
pointsources3
026 10.0 points
Two point sources are generating periodic
waves in phase. A point on the fourth nodal
line is 25 cm from one source and 39 cm from
the farthest source. Construct a sketch of the
physical situation and determine the wavelength.
Correct answer: 4 cm.
Explanation:
The equation to use here is just
P.D. = mλ,
where P.D. is the path difference, m is the
order of the nodal line and λ is the wavelength.
The path difference between the two sources
to the point is 14 cm, and we know m = 3.5
Version 001 – Waves extra practice - optional – tubman – (20131B)
since we are on the fourth nodal line, so we
have
14 cm = 3.5 λ ⇒ λ = 4 cm.
AP B 1998 MC 51 A
027 10.0 points
Plane sound waves of wavelength 0.11 m are
incident on two narrow slits in a box with
nonreflecting walls, as shown in the figure
below.
At a perpendicular distance of 10 m from
the center of slits, a first order maximum
occurs at a point which is 6 m from the central
maximum.
6m
sound
δ
requires
The approximation sin θ =
d
L ≫ d, which does NOT apply here; the signals are NOT traveling nearly parallel to each
other. We must go back to the definitions and
basic concepts of constructive and destructive interference. From the picture and using
the Pythagorean Theorem, the wave from the
upper slit travels a distance
d1 =
s
L2
d
+ y−
2
2
and the wave from the lower slit travels a
distance
s
2
d
2
d2 = L + y +
2
10 m
wavelength
0.11 m
What is the distance between the two slits?
Correct answer: 0.213806 m.
Explanation:
Basic Concept: The rules for determining
interference maximum or minimum are the
same for sound waves and light waves.
Thus, the path length difference is
δ = d2 − d1 = n λ ,
(1)
where n = 1 for the first maximum.
Solution: Double Slit interference.
d1
⊗
8
y−
d2
d
⊗
d
2 y
y+
d
2
[d2 +d1 ] [d2 − d1 ] = d22 − d21 , so
[d2 +d1 ] n λ
2
2
d
d
2
2
−L − y−
=L + y+
2
2
2
2
d
d
= y2 + y d +
− y2 + y d −
4
4
= 2yd.
Since
s
2
0.183333
m
d1 = (10 m)2 + 6 m −
2
=s
11.615 m and
2
0.183333 m
2
d2 = (10 m) + 6 m +
2
= 11.7093 m ,
n λ [d2 + d1 ]
2y
(1) (0.11 m)
=
2 (6 m)
× [(11.7093 m) + (11.615 m)]
d=
L
Let : λ = 0.11 m ,
L = 10 m , and
y = 6 m,
= 0.213806 m .
Version 001 – Waves extra practice - optional – tubman – (20131B)
d=
nλ
sin θ
nλ
h
y i
sin arctan
L
(1) (0.11 m)
=
6m
sin arctan
10 m
=
= 0.213802 m .
Such an estimate, if L > d, is usually fairly
good, but may not be close enough to give one
percent accuracy.
Small Angle Approximation:
The
small angle approximation assumes that
sin θ = tan θ = θ, where θ is in radians. Thus
y
δ
≈ , which gives
d
L
δL
d≈
= 0.183333 m
y
sin θ ≈
with percent error
0.213806 m − 0.183333 m
100%
0.213806 m
= 14.2527 % .
% Error =
AP B 1998 MC 51 B
028 (part 1 of 2) 10.0 points
Plane sound waves of wavelength 0.14 m are
incident on two narrow slits in a box with
nonreflecting walls, as shown in the figure.
At a perpendicular distance of 10 m from
the center of slits, a first order maximum
occurs at point which is 8 m from the central
maximum.
sound
8m
Alternative Approximate Solution:
Since the receiver is at the first maximum,
y
n = 1. From trigonometry, tan θ ≡ . For
L
constructive interference (using the approximation), n λ = δ = d sin θ. This approximation assumes that L ≫ d, which is only good
to a few percent in this case. Solving for d, we
have
9
10 m
wavelength
0.14 m
What is the distance between the two slits?
Correct answer: 0.224113 m.
Explanation:
The path length difference is
δ = d2 − d1 = n λ ,
(1)
where n = 1 for the first maximum.
d1
⊗
y−
d2
d
2 y
y+
d
d
2
⊗
L
Let :
λ = 0.14 m ,
L = 10 m ,
y = 8 m , and
n = 1.
δ
The approximation sin θ =
requires
d
L ≫ d, which does NOT apply here. Using the Pythagorean Theorem, the wave from
the upper slit travels a distance
s
2
d
d1 = L 2 + y −
2
and the wave from the lower slit a distance
s
2
d
.
d2 = L2 + y +
2
d22 −d21 = (d2 +d1 ) (d2 −d1 ) = (d2 +d1 ) n λ ,
and
Version 001 – Waves extra practice - optional – tubman – (20131B)
d22
− d21
2
2
d
d
2
=L + y+
−L − y−
2
2
2
2
d
d
− y2 + y d −
= y2 + y d +
4
4
= 2yd.
2
10
Double Slit Fringe Distance 01
030 10.0 points
A screen is illuminated by 613 nm light as
shown in the figure. The distance from the
slits to the screen is 3.8 m .
Since
d22 − d21 = (d2 + d1 ) (d2 − d1 )
2 y d = (d2 + d1 ) n λ
n λ (d2 + d1 )
d=
2y
(1) (0.14 m)
=
2(8 m)
× (12.8611 m + 12.7518 m)
= 0.224113 m .
029 (part 2 of 2) 10.0 points
What is the percent error in the determination
of the distance between the slits when using
the small angle approximation?
Correct answer: 21.9142 %.
Explanation:
The small angle approximation assumes
that sin θ = tan θ = θ, where θ is in radians. Thus
δ
y
sin θ ≈ ≈
d
L
nλL
δL
≈
d≈
y
y
(1) (0.14 m) (10 m)
≈
≈ 0.175 m .
8m
The percent error in is thus
0.224113 m − 0.175 m
× 100%
% Error =
0.224113 m
= 21.9142 % .
y
S1
viewing
screen
θ
S2
3.8 m
How far apart y are the central bright region
and the second bright fringe (in cm)?
Correct answer: 1.22601 cm.
Explanation:
Basic Concepts: For bright fringes, we
have
d sin θ = m λ ,
and for dark fringes, we have
1
d sin θ = m +
λ,
2
where m = 0 , ±1 , ±2 , ±3 , · · · .
From geometry, we have y = L tan θ .
Let : y = 1.22601 cm = 0.0122601 m ,
L = 3.8 m , and
d = 0.38 mm = 0.00038 m .
6
r1
S2
−1
an
r2 θ = t
S1
d
P
S2 Q S1 ≈ 90◦
y
L
Q
δ ≈ d sin θ ≈ r2 − r1
L
viewing
screen
2
0.175 m
+ 8m−
d1 = (10
2
=s
12.7518 m and
2
0.175 m
2
d2 = (10 m) + 8 m +
2
= 12.8611 m , then
m)2
0.38 mm
s
O
y
Version 001 – Waves extra practice - optional – tubman – (20131B)
r1
S1
−1
θ = t an
d
θ
S2
y
L
r2
Q
◦
0
≈9
1
r1
S
Q
r2 −
6 S2
≈
nθ
d si
≈
δ
Solution: For the second bright fringe,
m = 2, so
λ
θ = arcsin m
d
(6.13 × 10−7 m)
= arcsin (2)
(0.00038 m)
◦
= 0.184855 .
The distance from the central to the second
bright fringe is
y = L tan θ
= (3.8 m) tan 0.184855◦
= 0.0122601 m = 1.22601 cm .
11