Version 001 – Review 3: SHM and waves – tubman – (IBII201516)
This print-out should have 42 questions.
Multiple-choice questions may continue on
the next column or page – find all choices
before answering.
AP B 1993 MC 43
001 10.0 points
A particle oscillates up and down in simple
harmonic motion. Its height y as a function
of time t is shown in the diagram.
5
y (cm)
t (s)
1
2
3
4
5
5
At what time t in the period shown does
the particle achieve its maximum positive acceleration?
1. t = 3 s
2. None of these; the acceleration is constant.
3. t = 1 s correct
particle is slowing down. At t = 1 s, the particle is momentarily at rest and v = 0. Just
after t = 1 s , the velocity is positive since
the particle is speeding up. Remember that
∆v
a=
, acceleration is a positive maximum
∆t
because the velocity is changing from a negative to a positive value.
AP B 1998 MC 56
002 10.0 points
An object moves up and down the y-axis with
an acceleration given as a function of time t
by the expression a = A sin ω t, where A and
ω are constants.
What is the period of this motion?
1. T = ω 2 A
2. T =
ω
2π
3. T = 2 π ω
4. T = ω
2π
correct
ω
Explanation:
This motion is a simple harmonic motion.
5. T =
4. t = 4 s
5. t = 2 s
Explanation:
This oscillation is described by
πt
y(t) = − sin
,
2
dy
π
πt
v(t) =
= − cos
dt
2
2
2
d y
a(t) = 2
dt
π 2
πt
.
sin
=
2
2
Themaximum
acceleration will occur when
πt
sin
= 1, or at t = 1 s .
2
From a non-calculus perspective, the velocity is negative just before t = 1 s since the
1
ω = 2πf = 2π
T =
2π
.
ω
1
T
Piston in Harmonic Motion
003 (part 1 of 2) 10.0 points
A piston in an automobile engine is in simple
harmonic motion. Its amplitude of oscillation
from the equilibrium (centered) position is
±6.36 cm and its mass is 1.339 kg.
Find the maximum velocity of the piston
when the auto engine is running at the rate of
3040 rev/min.
Correct answer: 20.2469 m/s.
Explanation:
Let : A = 6.36 cm ,
Version 001 – Review 3: SHM and waves – tubman – (IBII201516)
m = 1.339 kg , and
f = 3040 rev/min .
ω = 2πf
= 2 π (3040 rev/min)
1 min
60 s
2
on a horizontal, frictionless track. The mass
is displaced 4 cm from the equilibrium point
and released from rest.
4 cm
2 N/m
x
621 g
= 318.348 rad/s .
The simple harmonic motion is described
by
x = A cos ω t ,
where ω is the frequency in rad/s if t is in
seconds.
The velocity is
v=
dx
= −A ω sin(ω t)
dt
meaning the maximum velocity is
vmax = A ω
= (0.0636 m) (318.348 rad/s)
x=0
Find the period of the motion.
Correct answer: 3.50115 s.
Explanation:
Let :
m = 621 g = 0.621 kg
k = 2 N/m .
and
This situation corresponds to the special
case
x(t) = A cos ωt , so
= 20.2469 m/s ,
since sine has a maximum value of 1.
004 (part 2 of 2) 10.0 points
Find the maximum acceleration of the piston
when the auto engine is running at this rate.
Correct answer: 6.44557 km/s2 .
Explanation:
The acceleration is
a=
d2 x
dv
=
= −A ω 2 cos(ω t) ,
2
dt
dt
so the maximum acceleration is
amax = A ω 2 = (0.0636 m) (318.348 rad/s)2
= 6.44557 km/s2 ,
since cosine has a maximum value of 1.
Oscillation on a Spring
005 (part 1 of 3) 10.0 points
A 621 g mass is connected to a light spring of
force constant 2 N/m that is free to oscillate
ω=
r
k
=
m
s
2 N/m
= 1.79461 s−1
0.621 kg
and the period is
T =
2π
2π
=
= 3.50115 s .
ω
1.79461 s−1
006 (part 2 of 3) 10.0 points
What is the maximum speed of the mass?
Correct answer: 0.0717843 m/s.
Explanation:
Let :
A = 4 cm .
The velocity as a function of time is
v(t) = −ω A sin(ω t) ,
so the maximum speed of the mass is
vmax = ω A = (1.79461 s−1 ) (0.04 m)
= 0.0717843 m/s .
Version 001 – Review 3: SHM and waves – tubman – (IBII201516)
007 (part 3 of 3) 10.0 points
What is the maximum acceleration of the
mass?
where θ is the phase of the oscillation. When
the object is at its maximum displacement
sin θ = 1
π
θ=
2
Correct answer: 0.128824 m/s2 .
Explanation:
The acceleration as a function of time is
a(t) = −ω 2 A cos(ω t) ,
so the maximum acceleration of the mass is
amax = ω 2 A = (1.79461 s−1 )2 (0.04 m)
= 0.128824 m/s2 .
AP B 1993 MC 9
008 10.0 points
When an object oscillating in simple harmonic
motion is at its maximum displacement from
the equilibrium position, which of the following is true of the values of its speed and the
magnitude of the restoring force?
Magnitude of
Speed
Restoring Force
1. Zero
so its speed is
v = ω A cos θ = ω A cos
π
=0
2
and the restoring force is
F = m A ω 2 sin θ = m A ω 2
π
= m A ω2 ,
2
at its maximum value.
AP M 1993 MC 24
009 10.0 points
Two identical massless springs are hung from
a horizontal support. A block of mass m is
suspended from the pair of springs, as shown.
k
k
Zero
1
3. maximum
2
1
maximum
2
1
maximum
2
4. Zero
Maximum correct
5. Maximum
Zero
2. Maximum
3
Explanation:
The maximum displacement occurs at the
turning points (where the velocity or speed
is zero). The magnitude of restoring force is
given by Hooke’s law
m
When the block is in equilibrium, each
spring is stretched an additional ∆x. Then
the block is set into oscillation with amplitude A; when it passes through its equilibrium
point it has a speed v.
In which of the following cases will the
block, when oscillating with amplitude A, also
have speed v when it passes through its equilibrium point? The acceleration of gravity is
g.
A) The block is hung from only one of the
two springs:
F = −k x ,
where k is the spring constant and x is the
displacement. Since x is a maximum, F is a
maximum.
From a different perspective, the displacement from the equilibrium position can be
written as
y = A sin θ ,
k
m
Version 001 – Review 3: SHM and waves – tubman – (IBII201516)
B) The block is hung from the same two
springs, but the springs are connected in
series rather than parallel:
4
Call the displacement of the mass x, and
choose the positive direction to be to the right.
For springs in parallel:
k1
k2
m
k
Hooke’s law is
F = −k x
k
and the frequency of oscillation is
m
f≡
C) An additional mass of m is attached to
the block:
k
ω
.
2π
The forces from the springs on the mass m
are to the left: F1 = −k1 x , F2 = −k2 x, and
F = F1 + F2 so that force equilibrium is
−k1 x − k2 x = m a = m
k
d2 x
.
dt2
This is a differential equation for x(t)
k1 + k2
d2 x
x = 0,
+
dt2
m
2m
which has a sine solution of the form
1. A, B, and C
x(t) = A sin(ω t + δ) ,
2. A only
where the angular frequency ω is
r
r
k1 + k2
2k
ω=
=
.
m
m
3. C only
4. B and C only
Note:
5. B only
kparallel = k1 + k2 .
6. A and C only
(1)
For springs in series:
k1
k2
7. None of these correct
m
8. A and B only
Explanation:
Let
k1 = k ,
k2 = k .
and
Consider the forces from a spring’s point of
view. The oscillating mass exerts the same
force F (at some instant in time) on each
spring, so
Version 001 – Review 3: SHM and waves – tubman – (IBII201516)
F
k1
F
.
F = k 2 x2 ⇒ x2 =
k2
Now consider the effective spring constant
kseries , where x = x1 + x2 .
F = k 1 x1 ⇒ x1 =
F
F
=
x
x1 + x2
F
k1 k2
k1 k2
=
·
=
,
F
F k1 k2
k2 + k1
+
k
k2
r1
r
2
k
k
=
.
=
2km
2m
kseries =
ωseries
5
AP B 1998 MC 38
010 10.0 points
A block of mass 3 kg is hung from a spring,
causing it to stretch 18 cm at equilibrium, as
shown below.
The 3 kg block is then replaced by a 4 kg
block, and the new block is released from
the position shown below, at which point the
spring is unstretched.
so
Using Eq. 2, the velocity is
v=
dx
= ω A cos(ω t + δ) .
dt
At the equilibrium point
r
v = ωA =
ωparallel =
k1 + k2
=
m
r
ymax
3 kg
equilbrium
k
A.
m
Therefore the angular velocity ω as presented in the question should be the same in
cases A, B, and/or C.
The question presents the springs in parallel (Eq. 3), so
r
18 cm
2k
.
m
Case A: Only one spring is present:
r
r
k
2k
6=
.
ωsingle =
m
m
Case B: Eq. 4, the springs are in series:
r
r
k
2k
6=
.
ωseries =
2m
m
Case C: Eq. 1, but the mass is doubled:
r
r
r
2k
k
k
ω2 m =
=
6=
.
2m
m
2m
Consequently, none of the choices is the
correct answer.
4 kg
How far will the 4 kg block fall before its
direction is reversed? For the new block case,
the total distance of fall is twice the amplitude
of the oscillation. The acceleration due to
gravity is 9.8 m/s2 .
Correct answer: 48 cm.
Explanation:
Let :
m1
m2
y1
g
= 3 kg ,
= 4 kg ,
= 18 cm , and
= 9.8 m/s2 .
Applying Hook’s law,
F = ky
so
F1
m1 g
k y1
=
=
F2
m2 g
k y2
m2 y1
y2 =
m1
Version 001 – Review 3: SHM and waves – tubman – (IBII201516)
(18 cm) (3 kg)
4 kg
y2 = 24 cm
6
K
Kmax
y2 =
1.
is the amplitude, which is half the vertical
motion, so the block will drop
−xmax
K
x
+xmax
Kmax
ymax = 2 (24 cm) = 48 cm
before its direction is reversed.
Alternate Solution: The equilibrium position for the 3 kg is 18 cm, as given. The
maxium stretched position for 4 kg after released from rest is ymax.
m1 g
.
The spring constant is k =
x1
Conservation of energy implies that
1
2
k ymax
so
2
m2
2 m2 g
=2
x1
=
k
m1
4 kg
=2
(18 cm) = 48 cm .
3 kg
2.
−xmax
K
x
+xmax
Kmax
3.
−xmax
x
+xmax
−xmax
x
+xmax
K
Kmax
m2 g ymax =
ymax
correct
4.
K
Kmax
AP B 1998 MC 8 v2
011 10.0 points
The graph below represents the potential energy U as a function of displacement x for
an object on the end of a spring (F = −k x)
oscillating in simple harmonic motion with
amplitude xmax .
U
Umax
x
−xmax
+xmax
Which graph represents the kinetic energy
K of the object as a function of displacement
x?
5.
x
−xmax
+xmax
Explanation:
At the equilibrium point (x = 0), the velocity is maximum and the kinetic energy is Umax
due to conservation of energy. At the maximum displacement points +xmax and −xmax
the velocity is zero and the kinetic energy is
zero.
From a different perspective, in simple harmonic motion of an object on the end of a
spring, the total energy is conserved. At the
maximum displacement xmax , the kinetic energy is 0, so E = U (xmax ) = Umax .
U + K = E = Umax
K(x) = Umax − U (x) .
Thus, K(x) looks like an upsidedown U (x).
Version 001 – Review 3: SHM and waves – tubman – (IBII201516)
7
Correct answer: 0.576 J.
AP M 1998 MC 35
012 10.0 points
An ideal massless spring is fixed to the wall
at one end, as shown below. A block of mass
M attached to the other end of the spring
oscillates with amplitude A on a frictionless,
horizontal surface. The maximum speed of
the block is vm .
k
vm
m
−A
0
Explanation:
Let : x = 0.8 m and
k = 1.8 N/m .
The potential energy U is
1
k x2
2
1
= (1.8 N/m) (0.8 m)2
2
= 0.576 J .
U=
+A
What is the force constant k of the spring?
2
m vm
2A
m g vm
2. k =
2A
2
m vm
correct
3. k =
A2
2
m vm
4. k =
2 A2
mg
5. k =
A
Explanation:
For the ideal harmonic oscillation of the
spring system, the kinetic energy maximum is
equal to the potential energy maximum which
is also the total energy of the system, so we
obtain
AM Radio Signals
014 (part 1 of 4) 10.0 points
AM radio signals are broadcast at frequencies
between 550 kHz and 1600 kHz and travel
2.99792 × 108 m/s.
What is the shortest AM wavelength?
1
1
2
k A2 = m vm
2
2
2
m vm
k=
.
A2
c = f λ.
1. k =
Harmonic Motion of a Spring 02
013 10.0 points
A 0.35 kg mass at the end of a horizontal
spring is displaced 1.8 m and released, then
moves in SHM at the end of the spring of force
constant 1.8 N/m.
Find the potential energy of the system
when the spring is stretched 0.8 m. The
acceleration of gravity is 9.8 m/s2 .
Correct answer: 187.37 m.
Explanation:
Let :
flargest = 1600 kHz ,
and
c = 2.99792 × 108 m/s .
The shortest wavelength corresponds to
largest frequency, and vice versa since
λshortest =
c
flargest
2.99792 × 108 m/s 1 kHz
· 3
1600 kHz
10 Hz
= 187.37 m .
=
015 (part 2 of 4) 10.0 points
What is the longest AM wavelength?
Correct answer: 545.077 m.
Explanation:
Version 001 – Review 3: SHM and waves – tubman – (IBII201516)
Let : fsmallest = 550 kHz ,
λlongest =
c
fsmallest
2.99792 × 108 m/s 1 kHz
=
· 3
550 kHz
10 Hz
= 545.077 m .
016 (part 3 of 4) 10.0 points
FM frequencies range between 88 MHz and
108 MHz and travel at the same speed.
What is the shortest FM wavelength?
8
Sound in air can best be described as which
of the following type of wave?
1. Longitudinal correct
2. Electromagnetic
3. Polarized
4. Torsional
5. Transverse
Explanation:
A sound wave in the air is propagated by
the oscillation of air molecules. It is best
described as longitudinal wave.
Correct answer: 2.77586 m.
Explanation:
Let : flargest = 108 MHz ,
λshortest =
c
flargest
2.99792 × 108 m/s 1 MHz
=
· 6
108 MHz
10 Hz
= 2.77586 m .
Waves in a Pond
019 10.0 points
A rock dropped into a pond produces a wave
that takes 10.9 s to reach the opposite shore,
35 m away. The distance between consecutive
crests of the wave is 5.6 m.
What is the frequency of the wave?
Correct answer: 0.573394 Hz.
Explanation:
Let : t = 10.9 s ,
d = 35 m , and
λ = 5.6 m .
017 (part 4 of 4) 10.0 points
What is the longest FM wavelength?
Correct answer: 3.40673 m.
Explanation:
Let :
λlongest =
fsmallest = 88 MHz ,
c
fsmallest
2.99792 × 108 m/s 1 MHz
· 6
=
88 MHz
10 Hz
= 3.40673 m .
AP B 1993 MC 30
018 10.0 points
The wavelength λ is the distance d1 between consecutive crests, so
v=fλ=
f=
d
t
(35 m)
d
=
= 0.573394 Hz .
tλ
(10.9 s) (5.6 m)
keywords:
AP B 1993 MC 59
020 10.0 points
The figure shows two wave pulses that are
approaching each other.
Version 001 – Review 3: SHM and waves – tubman – (IBII201516)
9
P
Q
P
Q
Which of the following best shows the shape
of the resultant pulse when the centers of the
pulses, points P and Q, coincide?
P +Q
1.
Conceptual 14 Q03
021 (part 1 of 4) 10.0 points
Consider two waves traveling through the
same medium in the same time frame.
A
B
correct
2.
3.
Compare the wavelengths.
4.
1. A has the longer wavelength. correct
2. B has the longer wavelength.
3. Cannot be determined
5.
4. A and B have the same wavelength.
Explanation:
Notice that the two pulses have the same
width and amplitude.
Choosing the the point P (the same as point
Q when the two pulses coincide) as the origin,
the two pulses can be described as:
P :
Q:
y1 =
y2 =
A , −d ≤ x ≤ d
(
A
−A
Explanation:
A exhibits three complete wavelengths in
the same time that B exhibits five complete
wavelengths, so A has a longer wavelength.
022 (part 2 of 4) 10.0 points
Compare the amplitudes.
, −d ≤ x < 0
,
0<x<d
Using the principle of superposition, the resultant pulse is
2 A , −d ≤ x < 0
y = y1 + y2 =
0 , 0<x<d
1. Cannot be determined
2. A has the smaller amplitude.
3. B has the smaller amplitude. correct
4. A and B have the same amplitude.
Version 001 – Review 3: SHM and waves – tubman – (IBII201516)
Explanation:
The vertical distance between the troughs
and peaks of A is greater, so it has the larger
amplitude.
023 (part 3 of 4) 10.0 points
Compare the frequencies.
1. A and B have the same frequency.
2. B has the higher frequency. correct
3. A has the higher frequency.
4. Cannot be determined
Explanation:
10
v
What is/are correct concerning the reflected pulse?
A) The reflected pulse forms a valley.
B) The reflected pulse forms a crest.
C) The reflected pulse has a greater speed
than that of the incident pulse.
D) The reflected pulse has a greater amplitude than that of the incident pulse.
1. D only
v = λf
v
f= .
λ
Since the speeds are the same and A has a
longer wavelength λ, then A must have the
lower frequency.
024 (part 4 of 4) 10.0 points
Compare the periods.
1. Cannot be determined.
2. A and B have the same period.
3. A has the shorter period.
4. B has the shorter period. correct
Explanation:
1
T = . Since A has a lower frequency, its
f
period must be larger.
AP B 1998 MC 29 v2
025 10.0 points
One end of a horizontal string is fixed to a
wall. A transverse wave pulse in the form of a
crest is generated at the other end and moves
toward the wall, as shown. Consider the pulse
after it has been reflected by the wall.
2. C only
3. A only correct
4. B only
5. None of these
Explanation:
The speed of the pulse is determined by the
tension of the string and the mass per unit
length, and is the same for the incident and
reflected pulses.
The reflected pulse cannot have an amplitude greater than that of the incident pulse,
since no energy is added to the pulse at the
wall.
There is a phase shift of π at the wall,
because the end fixed at the wall cannot move,
so the reflected pulse is on the opposite side
of the string from the incident pulse, forming
a valley.
Incident crests reflect as valleys, and vice
versa.
AP B 1993 MC 58
026 10.0 points
Consider the following:
I. the loudness of the sound;
II. the speed of the observer;
III. the speed of the source.
In the Doppler effect for sound waves, which
Version 001 – Review 3: SHM and waves – tubman – (IBII201516)
factors affect the frequency that the observer
hears?
2. A
1. I and III only
3. It will be the same for all four points.
2. I only
4. D
3. None is true.
5. B correct
Explanation:
The Doppler effect for stationary observers
is
vsound
f′ =
f0 .
vsound − vsource
Since the source is moving directly toward
point B, the measured frequency at point B
by a stationary observer will be greatest.
4. I and II only
5. All are true.
6. II only
7. II and III only correct
8. III only
Explanation:
In the Doppler effect of sound waves, both
the speed of the source and the speed of the
observer affect the frequency that the observer
hears. (The directions of the movement are
factors, too.)
On the other hand, the loudness of the
sound (the intensity of the sound wave) is
irrelevant.
AP B 1998 MC 49
027 10.0 points
A small vibrating object on the surface of a
ripple tank is the source of waves of frequency
20 Hz and speed 60 cm/s. The source S is
moving to the right with speed 20 cm/s, as
shown.
1. 1.1 kHz correct
2. 1.5 kHz
3. 1.05 kHz
4. 0.95 kHz
Explanation:
S
B
A
At which of the labeled points will the frequency measured by a stationary observer be
greatest?
1. C
Doppler Shift of an Echo
028 10.0 points
Suppose you are at the bottom of a canyon.
You are driving toward the canyon wall at
35 m/s . A stationary siren behind you produces a sound also at the bottom of the canyon
at a frequency of 1.00 kHz.
You will pick up an echo at what approximate frequency? The speed of sound is
350 m/s .
5. 0.9 kHz
C
D
11
Let :
v = 35 m/s ,
cs = 350 m/s , and
f = 1 Hz .
The Doppler shift in frequency according
to an observer (the driver) moving toward a
stationary source (the echo from the canyon
wall) is
v
′
f =f 1+
cs
Version 001 – Review 3: SHM and waves – tubman – (IBII201516)
35 m/s
= (1 Hz) 1 +
= 1.1 Hz .
Correct answer: 3782 Hz.
350 m/s
Falling Sky Diver
029 (part 1 of 2) 10.0 points
In order to be able to determine her speed,
a skydiver carries a tone generator. A friend
on the ground at the landing site has equipment for receiving and analyzing sound waves.
While the skydiver is falling at terminal speed,
her tone generator emits a steady tone of
918 Hz.
If her friend on the ground (directly beneath the skydiver) receives waves of frequency 2350 Hz, what is the skydiver’s speed
of descent? Assume the air is calm and the velocity of sound in air is 343 m/s, independent
of altitude.
Correct answer: 209.011 m/s.
Explanation:
Let :
fe = 918 Hz ,
fg = 2350 Hz ,
v = 343 m/s .
12
Explanation:
When the waves are reflected, and the skydiver is moving toward them,
v + vdiver
frec = fg
v
= (2350 Hz)
343 m/s + 209.011 m/s
×
343 m/s
= 3782 Hz .
AP B 1993 MC 27 28num
031 (part 1 of 2) 10.0 points
A standing wave of frequency 3.42 Hz is set
up on a string 1.4 m long with nodes at both
ends and in the center, as shown.
1.4 m
and
Since the diver (the source of the sound
waves) is moving toward her friend on the
ground (the receiver of the waves) the Doppler
formula takes the form
v
fg = fe
v − vdiver
vdiver
fe
=1−
fg
v
fe
vdiver = v 1 −
fg
918 Hz
= (343 m/s) 1 −
2350 Hz
Find the speed at which waves propagate
on the string.
Correct answer: 4.788 m/s.
Explanation:
Let : f = 3.42 Hz
L = 1.4 m .
and
The wavelength λ = L = 1.4 m, so the wave
speed is
v = f λ = (3.42 Hz) (1.4 m) = 4.788 m/s .
= 209.011 m/s .
030 (part 2 of 2) 10.0 points
If the skydiver were also carrying soundreceiving equipment sensitive enough to detect waves reflected from the ground, what
frequency would she receive?
032 (part 2 of 2) 10.0 points
Find the fundamental frequency of vibration
of the string.
Correct answer: 1.71 Hz.
Explanation:
Version 001 – Review 3: SHM and waves – tubman – (IBII201516)
1.4 m
13
4λ
L
a
3λ
L
10. y3 =
2a
Explanation:
The third minimum occurs at β = 6 π,
which corresponds to a path difference between two end rays:
9. y3 =
The fundamental wave has only two nodes
at the ends, so its wave length is
λ1 = 2 L = 2.8 m ,
b3 =
and the fundamental frequency is
6π
2π
λ
= 3λ
b3
θ=
a
y3
=
L
b3
L
y3 =
a
3λ
L.
=
a
=
4.788 m/s
v
=
= 1.71 Hz .
f1 =
λ1
2.8 m
S2
L
viewing
screen
a
×10
θ
y3
Exam Single Aperture
033 10.0 points
Consider the setup of a single slit experiment.
Hint: Use a small angle approximation; e.g.,
sin θ = tan θ .
S1
Determine the height y3 , where the third
minimum occurs.
1. y3 =
2. y3 =
3. y3 =
4. y3 =
5. y3 =
6. y3 =
7. y3 =
8. y3 =
2λ
L
a
λ
L
2a
5λ
L
2a
7λ
L
2a
9λ
L
2a
3λ
L correct
a
λ
L
a
5λ
L
a
β
k
Eye Resolution
034 10.0 points
Assume a lens can act like a one-dimensional
single slit, with the diameter of the lens equivalent to the slit width. The resolution of the
lens is then equivalent to the distance from
the middle of the central bright band to the
first-order dark band. Suppose the image
formed on the retina of the eye shows the effect of diffraction. The diameter of the iris
opening in bright light is 2.992 mm and the
distance from the iris to the retina is 2.41 cm.
Find the resolution of the eye for light of
wavelength 595.4 nm. Assume the index of
refraction of the interior of the eye is 1.31 .
Correct answer: 0.000366094 cm.
Explanation:
Let :
a = 2.992 mm = 0.002992 m ,
L = 2.41 cm ,
λ = 595.4 nm = 5.954 × 10−7 m ,
neye = 1.31 .
and
Version 001 – Review 3: SHM and waves – tubman – (IBII201516)
(645 nm) (1.44 mi)
5.99 mm
(6.45 × 10−7 m) (2316.96 m)
= 1.22
0.00599 m
= 0.304377 m .
The wavelength of the light in the eye is
λeye =
= 1.22
λ
5.954 × 10−7 m
=
neye
1.31
= 4.54504 × 10−7 m .
x
w
L
For single slit interference,
λeye
y
=
L
a
L λeye
y=
a
(2.41 cm) (4.54504 × 10−7 m)
=
0.002992 m
= 0.000366094 cm .
Paul Revere and Resolution
035 10.0 points
On the night of April 18, 1775, a signal was
to be sent from the Old North Church steeple
to Paul Revere, who was 1.44 mi away: “One
if by land, two if by sea.” Assume that Paul
Revere’s pupils had a diameter of 5.99 mm
at night, and that the lantern light had a
predominant wavelength of 645 nm.
At what minimum separation did the sexton have to set the lanterns so that Revere
could receive the correct message? One mile
is approximately equal to 1.609 km.
Three Polarizers 02
036 (part 1 of 4) 10.0 points
Consider 3 polarizers #1, #2, and #3 ordered
sequentially. The incident light is unpolarized with intensity I0 . The intensities after
the light passes through the subsequent polarizers are labeled as I1 , I2 , and I3 respectively.
See the sketch.
I0
I1
θmin = 1.22
λL
D
#2
I3
#3
Polarizers #1 and #3 are “crossed” such
that their transmission axes are perpendicular to each other. Polarizer #2 is placed
between the polarizers #1 and #3 with its
transmission axis at 60 ◦ with respect to the
transmission axis of the polarizer #1 (see the
sketch).
#1
60◦
#2
#3
After passing through the first polarizer,
the intensity of the light I1 is
1. I1 =
2. I1 =
3. I1 =
4. I1 =
Therefore
d = 1.22
d
λ
= .
D
L
I2
#1
Correct answer: 0.304377 m.
Explanation:
The angle of resolution for the Paul Revere’s pupils is
14
5. I1 =
I0
8
I0
12
I0
correct
2
3 I0
8
3 I0
32
Version 001 – Review 3: SHM and waves – tubman – (IBII201516)
I0
4
3 I0
7. I1 =
4
I0
8. I1 =
16
I0
9. I1 =
32
3 I0
10. I1 =
16
Explanation:
Basic Concepts: Malus’ law states
6. I1 =
I ′ = I0 cos2 θ ,
where θ is the angle between the polarization
of the light (before the light hits the polarizer)
and the transmission axis of the polarizer.
Solution: When unpolarized light falls on a
polarizer, no matter what the orientation of
the polarizer is, one half of the light is transmitted, and the polarization of the transmitted light is along the direction of the transmission axis of the polarizer (vertical in the
present case). Thus the answer is
I1 =
1
I0 .
2
037 (part 2 of 4) 10.0 points
After passing through polarizer #2 the intensity I2 (in terms of I1 ) is
1. I2 =
2. I2 =
3. I2 =
4. I2 =
5. I2 =
6. I2 =
7. I2 =
I1
8
I1
correct
4
I1
32
3 I1
32
I1
16
3 I1
8
I1
12
15
3 I1
16
3 I1
9. I2 =
4
I1
10. I2 =
2
8. I2 =
Explanation:
When polarized light passes through a polarizer, the transmitted intensity is I2 =
I1 cos2 θ, where θ is the angle between the
polarization of the light (of I1 ) and the orientation of the polarizer #2 and I1 is the result
we got from Part 1. When the light passes
through the polarizer #1 it is polarized vertically. Thus the angle between its polarization
and the orientation of polarizer #2 is θ = 60◦ .
Thus the transmitted intensity is
I2 = I1 cos2 θ
= I1 cos2 (60◦ )
=
1
1
I1 = I0 .
4
8
038 (part 3 of 4) 10.0 points
After passing through polarizer #3 the intensity I3 (in terms of I2 ) is
1. I3 =
2. I3 =
3. I3 =
4. I3 =
5. I3 =
6. I3 =
7. I3 =
8. I3 =
9. I3 =
I2
8
3 I2
32
I2
2
3 I2
16
I2
12
3 I2
correct
4
I2
4
I2
16
3 I2
8
Version 001 – Review 3: SHM and waves – tubman – (IBII201516)
3
I2
4 3
1
=
I1
4
4
1
1
3
I0
=
4
4
2
=
I2
10. I3 =
32
Explanation:
After the polarizer #1
I1 =
I0
2
=
After the polarizer #2
I2 = I1 cos2 (60◦ ) =
I3 = I2 cos2 (90◦ − 60◦ )
3
I2 .
4
039 (part 4 of 4) 10.0 points
After passing through polarizer #3 the intensity I3 (in terms of I0 ) is
AP B 1998 MC 51 A
040 10.0 points
Plane sound waves of wavelength 0.17 m are
incident on two narrow slits in a box with
nonreflecting walls, as shown in the figure
below.
At a perpendicular distance of 7 m from the
center of slits, a first order maximum occurs
at a point which is 5 m from the central
maximum.
sound
5m
I0
1. I3 =
16
I0
2. I3 =
12
I0
3. I3 =
2
3 I0
4. I3 =
4
I0
5. I3 =
32
3 I0
6. I3 =
8
I0
7. I3 =
4
I0
8. I3 =
8
3 I0
9. I3 =
correct
32
3 I0
10. I3 =
16
Explanation:
After the polarizer #3
2
3
I0 .
32
1
I1 .
4
After the polarizer #3
=
16
wavelength
7m
0.17 m
What is the distance between the two slits?
Correct answer: 0.292498 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 ,
◦
◦
I3 = I2 cos (90 − 60 )
where n = 1 for the first maximum.
Solution: Double Slit interference.
(1)
Version 001 – Review 3: SHM and waves – tubman – (IBII201516)
d1
⊗
y−
d2
d
2 y
17
n λ [d2 + d1 ]
2y
(1) (0.17 m)
=
2 (5 m)
× [(8.67203 m) + (8.53371 m)]
d=
y+
d
d
2
⊗
= 0.292498 m .
L
Let : λ = 0.17 m ,
L = 7 m , and
y = 5 m,
δ
The approximation sin θ =
requires
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
s
2
d
2
d1 = L + y −
2
and the wave from the lower slit travels a
distance
s
2
d
d2 = L 2 + y +
2
[d2 +d1 ] [d2 − d1 ] = d22 − d21 , so
[d2 +d1 ] n λ
2
2
d
d
2
2
=L + y+
−L − y−
2
2
d2
d2
= y2 + y d +
− y2 + y d −
4
4
= 2yd.
Since
s
2
0.238
m
d1 = (7 m)2 + 5 m −
2
=s
8.53371 m and
2
0.238
m
d2 = (7 m)2 + 5 m +
2
= 8.67203 m ,
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
d=
nλ
sin θ
nλ
h
y i
sin arctan
L
(1) (0.17 m)
=
5m
sin arctan
7m
=
= 0.292479 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≈
= 0.238 m
y
sin θ ≈
with percent error
0.292498 m − 0.238 m
100%
0.292498 m
= 18.6318 % .
% Error =
Difraction Third Null
Version 001 – Review 3: SHM and waves – tubman – (IBII201516)
041 10.0 points
In a diffraction slit, light at a frequence f
has its second null directed at 45◦ from the
horizontal direction of the beam.
Will there be a distinct third null present
in the diffraction pattern?
1. No correct
Here we have taken into account that high
reflectivity is achieved for constructive interference. The phase changes at both the “aircryolite” and the “cryolite-glass” surfaces is
φ = 180◦ (nair = 1 < nc < ng ).
Note: Two phase changes of 180◦ .
For minimum thickness m = 1
λ1
2 nc
524 nm
=
2 × 1.37
= 191.241 nm .
t1 =
2. Yes
3. Not enough information
Explanation:
The angular position of a diffraction null is
given by
a sin θ = m λ =
mc
,
f
with a the slit width and m an integer.
The wavelength of the second null satisfies
a sin θ1
λ =
while if there were a distinct
2
third null, it would satify
3 2λ
3λ
=
a
2 a
3
= sin 45◦ = 1.06066 > 1 .
2
sin θ2 =
Thus there is no distinct third null in the
diffraction pattern.
Coating on a Camera Lens 2
042 10.0 points
A thin film of cryolite ( nc = 1.37 ) is applied
to a camera lens ( ng = 1.54 ). The coating
is designed to reflect wavelengths at the blue
end of the spectrum and transmit wavelengths
in the near infrared.
What minimum non-zero thickness gives
high reflectivity at 524 nm?
Correct answer: 191.241 nm.
Explanation:
For camera lens coating of cryolite (nc =
1.37) over glass (ng = 1.54), high reflectivity
is achieved for
2 nc t1 = m λ1 .
18