Download 1. (Chapter 16) Given the following travelling waves on different

y
1. (Chapter 16) Given the following travelling waves on different strings. Which of
the following corresponds to the correct order of phases?
(Note that the blue letter corresponds to the clicker button)
• A 1 → 2x − 4t
2 → 4x − 8t
3 → 8x − 16t
5
4
3
2
1
-0.2
-0.4
-0.6
-1
2
-2
6
(1)
-3
-4
-5
y
4
x
5
4
3
2
1
-0.2
-0.4
-0.6
-1
2
-2
6
(2)
-3
-4
-5
y
4
x
5
4
3
2
1
-0.2
-0.4
-0.6
-1
-2
-3
-4
-5
2
4
(3)
6
x
• B 1 → 2x − 4t
2 → 8x − 16t
3 → 4x − 8t
• C 1 → 4x − 8t
2 → 2x − 4t
3 → 8x − 16t
• D 1 → 4x − 8t
2 → 8x − 16t
3 → 2x − 4t
• E 1 → 8x − 16t
2 → 2x − 4t
3 → 4x − 8t
The phase is kx − ωt. The larger the value of the phase, the faster the
oscillations will occur.
• 8x − 16t – fastest
• 4x − 8t – middle
• 2x − 4t – slowest
So, A is the correct answer.
2. (Chapter 16) Two waves with equations
• y1 (x, t) = 5 sin (3x − 2t)
• y2 (x, t) = 5 sin (3x + 2t)
are travelling along the same string. What is the displacement, when x = 31 π
• A – y(x, t) = 5
• B – y(x, t) = −5
• C – y(x, t) = 10
• D – y(x, t) = 0
• E – y(x, t) = −10
Hint: The superposition principle gives:
y(x, t) = 2A sin (kx) cos (ωt)
This is not a travelling wave, because the phase no longer has both the
“kx” and the “ωt” terms. Instead this is a standing wave, with k = 3.
π
Hence, if x = , then kx = π, and sin (kx) = 0 such that y = 0. Note that
3
this is true for all times, which means that this is a node of the oscillation.
So, D is the correct answer.
3. (Chapter 17) Consider a merry-go-round of radius 10m which make a complete
revolution in the counter-clockwise direction every 5 seconds. Riding on the merrygo-round edge is a 1kHz sound source. To the right of the merry-go-round is a sound
detector. What are the frequencies that the detector will measure at the following
angles, θ:
θ
Clicker
y
ω
θ
x
Response
30◦
A
B
C
D
E
963.0 Hz
1037.0 Hz
1000.0 Hz
981.5 Hz
1018.5 Hz
0◦
1000.0
1000.0
1037.0
1000.0
1000.0
330◦
Hz 1037.0 Hz
Hz 963.0 Hz
Hz 963.0 Hz
Hz 1018.5 Hz
Hz 981.5 Hz
The equation for frequency shifting is:
v
±
v
D
f
f0 =
v ± vs
where v = 343m/s is the velocity of the wave in air, vD = 0 is the velocity
of the detector, and vS is the velocity of the source with respect to the
sound. Here the velocity of the source from the perspective of a distance
detector is:
2πr
2π10m
vS =
cos θ =
cos θ ≈ 12.6 cos θ m/s
τ
5s
For θ < π, the source is moving away from the detector, and hence the
frequency would be shifted down; for θ > π, the frequency is shifted up.
Ignoring, for the moment, the cosine factor, one has f 0 = 964.6 m/s for
3π
π
θ = and f 0 = 1038.0 m/s for θ = . However, the cosine factor will reduce
2
2
the relative velocity, so:
D is the correct answer
4. (Chapter 21) Consider the following four arrangements of static charges. In each
arrangement, the charges are at equal distances from the origin. Rank (in descending order) the arrangements according to the magnitude of net electric field at the
origin.
y
y
−5q
2q
−5q
−3q
x
Clicker
4q
−2q
(i)
x
(ii)
y
y
−q
4q
−q
5q
(iii)
x
q
−4q
x
4q
(iv)
Order
Response
A
(i)
(ii) (iii) (iv)
B
(iv)
(i)
C
(iii) (ii)
(i)
(iv)
D
(ii) (iii) (iv)
(i)
E
(i)
(ii) (iii)
(iii) (ii) (iv)
The magnitude ofq
the electric field is the quadrature sum of the x and y
components: E = Ex2 + Ey2 . Let the distance to the origin be d, then:
s
2
5q
q 2
− 2 + − 2
E=
d
d
s 2
2
4q 2q
2q
2q
5q
5q
• (ii) Ex = 2 − 2 = 2 , Ey = − 2 ,
E=
+ − 2
d
d
d
d
d2
d
s 2
2
4q
q
3q
5q
5q
3q
+
• (iii) Ex = 2 − 2 = 2 , Ey = 2 ,
E=
2
d
d
d
d
d
d2
q
4q
3q
4q
q
3q
• (iv) Ex = 2 − 2 = − 2 , Ey = 2 − 2 = 2 ,
d
d
d
d
d
d s
2 2
3q
3q
− 2 +
.
E=
d
d2
q
5q
2q 3q
• (i) Ex = 2 − 2 = − 2 , Ey = − 2 ,
d
d
d
d
such that the order is: (iii), (ii), (i), (iv) .
C is the correct answer
√
26q
=
d2
√
29q
=
d2
√
36q
=
d2
√
18q
=
d2
5. (Chapter 23) Consider a cube of sides 2m. If the cube is placed in an electric field:
~ (3î − 4ĵ)N/C
E=
what is the electric flux through the right face of the cube?
Clicker
Response Order
z
A
B
C
y
D
x
E
The electric flux is given by:
ΦE =
Z
Nm2
-16
C
Nm2
-12
C
Nm2
12
C
Nm2
16
C
Nm2
20
C
~ · dA
~
E
~ is toward positive y, so ĵ. Substituting:
where the direction of A
Z
ΦE =
(3î − 4ĵ)N/C · ĵdA
!
Z
1
0
>
>
=
3(
î ·ĵ) − 4(
ĵ · ĵ) N/C dA
= −4
Z
dA
= −16 N m2 /C
since the area of the side is 4m2
A is the correct answer
.
6. (Chapter 23) Consider a butterfly net with radius, r = 0.5m, which is in a uniform
electric field and whose face is perpendicular to the direction of the field. If the
N m2
, what is the magnitude and unit vector of the electric
measured flux is 10
C
field?
Clicker
~
Response E
r
y
A
B
C
D
E
E
N
C
-12.7 î
-6.4 î
12.7 î
6.4 ĵ
12.7 ĵ
.
x
The electric field flux is given by the equation:
Z
~ · dA
~
ΦE = E
Note that the reason the integral is not over a closed surface is because
we are only interested in the flux through the butterfly net.
Since both the electric field and the area vector are parallel, then the
dot product gives cos 0◦ = 1, which means that the direction of the field
is î Furthermore, since the electric field is uniform, then it does not
depend on the position, and therefore the electric field can come out of
the integral:
Z
~
ΦE = |E|
dA
where the integral is simply the area of the butterfly net. Solving for the
electric field:
Finally:
ΦE
ΦE
10N m2/C
~
R
|E| =
=
=
= 12.7N/C
π(0.5m)2
dA πr2
C is the correct answer
~ = î12.7N/C
E
7. (Chapter 25) Consider the following circuit. Before the switch is closed, there is
each capacitor is without charge. What is the equivalent capacitance, C eq , of the
circuit, if C1 = 1µF , C2 = 2µF , C3 = 3µF , C4 = 4µF , and C5 = 5µF ?
Clicker
Response
Ceq
+
C1
C2
C3
−
C4
A
B
C
D
E
C5
19/12µF
201/47µF
47/201µF
37/12µF
12/19µF
.
In the figure below, the original circuit is equivalent to the left circuit by
adding the parallel capacitors C3, C4 , and C5 together as C345. Then, this
circuit is equivalent to the circuit to the right by adding the three series
capacitors C1, C2 , and C345 to form Ceq . Where:
+
C1
C2
+
C345
−
Ceq
−
C345 = C3 + C4 + C5 = (3 + 4 + 5) µF = 12µF
Then:
−1
−1
−1 12
1
1
1
12
6
1
1
1 1
µF =
µF = µF
Ceq =
=
+
+
+ +
+
+
C1 C2 C345
1 2 12
12 12 12
19
or
Ceq =
E is the correct answer
19
12
−1
µF =
12
µF
19
8. (Chapter 26) Consider the following circuit. What is the magnitude and the
direction of the current in the lower right corner of the circuit?
1A
2A
5A
3A
1 2A
3
4
3A
2A
2
4A
2A
6A
8A
5
Clicker
Response
A
B
C
D
E
i
8A
6A
6A
8A
3A
i=?
direction
left
right
left
right
right
.
Modifying the drawing and excersizing the approach of divide and conquer, I am using the numbers in circles to label the junctions, and the
currents and arrows in blue to indicate the current along that leg of the
circuit. At each junction, I use Kirchhoff ’s junction law, which is based
on the principle of conservation of charge: the sum of currents entering
a junction is equal to the sum of currents exiting that junction. So:
• At junction 1: 2A and 3A enters, so 5A exits
• At junction 2: 1A and 2A exist, so 3A enters
• At junction 3: 5A enters and 3A exist, so 2A exits
• At junction 4: 2A and 4A enters, so 6A exits
• At junction 5: 6A and 2A enters, so 8A exits
D is the correct answer
9. (Chapter 26) Consider the following circuit. Before the switch is closed, there is
no current in the circuit. What is the equivalent resistance, R eq , of the circuit, if
R1 = 1Ω, R2 = 2Ω, R3 = 3Ω, R4 = 4Ω, and R5 = 5Ω?
Clicker
Response
Req
R1
+
R2
R3
−
R4
A
B
C
D
E
R5
19/12Ω
201/47Ω
47/201Ω
37/12Ω
12/19Ω
.
In the figure below, the original circuit is equivalent to the left circuit by
adding the parallel capacitors R3, R4 , and R5 together as R345. Then, this
circuit is equivalent to the circuit to the right by adding the three series
capacitors R1, R2, and R345 to form Req . Where:
+
R1
−
R345 =
or:
1
1
1
+
+
R3 R4 R5
R2
+
R345
−1
=
R345 =
Then:
Req = R1 + R2 + R345 =
B is the correct answer
Req
−
1 1 1
+ +
3 4 5
47
60
−1
−1
Ω=
Ω=
20 15 12
+
+
60 60 60
−1
60
Ω
47
60
141 60
201
1+2+
+
Ω
Ω=
Ω=
47
47
47
47
Ω
10. (Chapter 28) In the figure, a wire is situated along the x-axis with the current, i,
~ such that a
flowing in the −î direction. The wire is in a uniform magnetic field B,
magnetic force, F~B , is along the +k̂ direction. The field is oriented so that the force
is maximum. In which direction is the field?
y
Clicker
Response
i
x
z
FB
~ direction
B
A
+î
B
−î
C
+ĵ
D
−ĵ
E
−k̂
.
Using the right hand rule with the thumb pointed along the force F~B in
the +k̂ direction, and pointing the fingers in the −î direction with the
current, then one must curl the fingers in the −ĵ direction for the field,
~
and that must be the direction of B.
D is the correct answer
11. (Chapter 29) In the figure, the three wires each have the same magnitude current,
i, flowing in the direction indicated. HThe four loops are Amperian loops. Rank the
~ ·d~s along each, in decreasing order.
loops according to the magnitude of B
(a)
Clicker
Order
Response
(b)
(c)
.
Ampere’s Law is:
(d)
I
A
B
C
D
E
(a)
(d)
(d)
(c)
(d)
(b)
(b)
(a)
(a)
(c)
~ · d~s = µ0 itot
B
where the current i is the total current in the loop. So loop:
(a) itot = −i + i + i = i
(b) itot = −i + i = 0
(c) itot = −i
(d) itot = i + i
such that in decreasing order, the magnitude of the loops is:
(d) (a) (c) (b)
C is the correct answer
(c)
(c)
(c)
(b)
(b)
(d)
(a)
(b)
(d)
(a)
12. (Chapter 30) In the figure, the graph gives the magnitude of B(t) of a uniform
magnetic field that exists throughout a conducting loop, with the direction of the
field perpendicular to the plane of the loop. Rank the five regions of the graph
according to the magnitude of the EMF induced in the loop in descending order.
Clicker
Response
B(t)
.
a
b
c
d
A
B
C
D
E
t
e
Faraday’s law states that:
E=
where:
ΦB =
Z
Order
(a)
(e)
(d)
(a)
(b)
(b)
(d)
(b)
(d)
(d)
(c)
(c)
(a)
(b)
(e)
(d)
(b)
(c)
(e)
(a)
(e)
(a)
(e)
(c)
(c)
d ΦB
dt
~ · dA.
~
B
~ and A
~ are parallel, so that the field is maximum through the
Here, B
loop. However, since B = B(t), the flux ΦB will change with time, and so
induce an EMF. Here, the change in flux is:
(a)
(b)
(c)
(d)
(e)
So,
∆B
0
= = 0 such that E = 0
∆t
1
∆B
2
= = 2 such that |E| = 2 units
∆t
1
0
∆B
= = 0 such that E = 0
∆t
1
−1
∆B
=
= −1 such that |E| = 1 unit
∆t
1
∆B
−1
=
= −1 such that |E| = 1 unit
∆t
1
in decreasing order of magnitude:
(b) (d)=(e) (a)=(c)
E is the correct answer
13. (Chapter 30) In the figure, the induced EMF in the inductor is shown. Which of
the following describes the current in the coil?
εL
Clicker
Response
A
B
C
D
E
.
Order
constant and to the left
constant and to the right
increasing and to the left
increasing and to the right
decreasing and to the left
The EMF induced in an inductor is given by:
E = −L
di
dt
where the inductance L is fixed by the inductor and depends only on the
physical characteristics of the coil, and the negative sign indicates that
the EMF is opposing the change in current. Note that if di/dt = 0, then
there is no induced EMF; hence the first two answers cannot be right.
Using the right hand rule with the thumb following the current, if the
current is going to the right and increasing, then the magnetic field produced by the current goes to the right inside the inductor. Since the
EMF opposes this change in current, then the EMF would be going to
the left. This is also true for decreasing current to the left; the decreasing
magnetic field would go to the left, and so the EMF would go to the left
to oppose the change.
On the other hand, using the right hand rule with the thumb following
the current, if the current is going to the left, then the magnetic field
produced by the current goes to the left inside the inductor. Since the
EMF opposes this change in current, then the EMF would be going to
the right.
C is the correct answer
14. (Chapter 31) In the figure, the three plots show the oscillations of voltage, v(t)
and current, i(t) as a function of the argument of the sine function. The three plots
correspond to the three simple circuits with a single of the following components
and an alternating EMF source with:
• “R” a series resistor
• “C” a series capacitor
1
Amplitude
Amplitude
Amplitude
• “L” a series inductor
1
1
0
0
0
-1
-1
-1
2π
2π
phase
2π
phase
phase
Match the circuits with the graphs:
Clicker
Response
A
B
C
D
E
Order
R
L
C
L
C
L
R
R
C
L
C
C
L
R
R
.
For each circuit:
• R: iR (t) and vR (t) are in phase and follow each other =⇒ 3rd graph
• C: iC (t) leads vC (t) by φ = π/2 =⇒ 1rst graph
• L: iL (t) lags vL(t) by φ = π/2 =⇒ 2nd graph
E is the correct answer