Download Solution - Bryn Mawr College

Physics 102-1 Spring 2010 Final Exam Solution
Grading note: There are 12 problems. Point values are given with each problem.
They add up to 155 points. In multi-part problems, points are not necessarily evenly
divided between the parts.
1. [10 points] A person hums into the top of a well and finds that standing waves
are established at frequencies of 42, 70, and 98 Hz, but no frequencies in between
these. The frequency of 42 Hz is not necessarily the fundamental frequency. The
speed of sound is 343 ms . How deep is the well?
A well is open on the top and closed on the bottom. The fundamental frequency of
v
sounds waves in an open-closed tube is f1 = 4L
, where v is the speed of sound and
L is the length of the tube (the depth of the well). Other resonant frequencies are
fm = mf1 , where m is an odd integer, that is, they are
f1 , 3f1 , 5f1 , 7f1 , . . . .
Notice that each resonant frequency is 2f1 higher than the previous one.
We need to match the list of frequencies given in the problem to this pattern. Notice
that each frequency in the list given in the problem is 28 Hz higher than the previous
one: 42 Hz+28 Hz=70 Hz; 70 Hz+28 Hz=98 Hz. Thus
⇒
2f1 = 28 Hz
f1 = 14 Hz
From this value we can find the distance:
f1 =
v
4L
⇒
L=
343 ms
v
=
= 6.1 m .
4f1
(4)(14 Hz)
Another way to calculate this is to say that if one of the frequencies (42 Hz, say) is
the mth harmonic, then the next highest frequency (70 Hz) must be the (m+2)th
harmonic:
42 Hz
= mf1
70 Hz
=
(m + 2)f1 .
Subtracting the upper equation from the lower equation gives
70 Hz − 42 Hz
=
(m + 2)f1 − mf1
28 Hz
=
2f1 ,
from which f1 and L can be found as above. This same analysis works using 70 Hz
as the mth harmonic and 98 Hz as the (m+2)th harmonic.
1
2. [10 points] Two pulses travel down a string in opposite directions at speed 1 cm
.
s
The pulses are each half up and half down, as shown in the figure. The figure
shows the positions of the pulses at time t = 0. Draw the shape of the string
at t = 1 s, 2 s, 3 s, and 4 s.
1 cm/s
1
2
3
4
1 cm/s
5
6 7
x(cm)
8
9 10 11 12
To draw the shape at any given time, t, shift the left pulse rightward by t centimeters,
shift the right pulse leftward by t centimeters, as shown in the plots on the left below.
Then add the two waveforms together, as shown on the right.
Notice that at t = 2 the two pulses exactly cancel out.
t=0s
t=0s
1
2
3
4
5
6 7
x(cm)
8
9 10 11 12
t=1s
1
2
3
4
5
6 7
x(cm)
8
9 10 11 12
1
2
3
4
5
6 7
x(cm)
8
9 10 11 12
1
2
3
4
5
6 7
x(cm)
8
9 10 11 12
1
2
3
4
5
6 7
x(cm)
8
9 10 11 12
1
2
3
4
5
6 7
x(cm)
8
9 10 11 12
t=1s
1
2
3
4
5
6 7
x(cm)
8
9 10 11 12
t=2s
t=2s
1
2
3
4
5
6 7
x(cm)
8
9 10 11 12
t=3s
t=3s
1
2
3
4
5
6 7
x(cm)
8
9 10 11 12
t=4s
t=4s
1
2
3
4
5
6 7
x(cm)
8
9 10 11 12
2
3. [15 points] For each of the two situations given below, give (i) the type of image
formed (real or virtual), (ii) whether the image is upright or inverted, (iii) how
tall the image is, and (iv) where the image is located (say whether it is to the
right or the left of the lens, and how far from the lens it is located).
(a) A 20 cm tall flower is placed 30 cm to the left of a converging lens of focal
length 10 cm.
f = 10 cm, s = 30 cm, h = 20 cm.
1
1
1
+ =
s s0
f
M=
1
1
1
+ =
30 cm s0
10 cm
⇒
s0
h0
=−
h
s
⇒
h0
15 cm
=−
20 cm
30 cm
⇒
s0 = 15 cm
h0 = −10 cm
⇒
(i) Positive s0 implies real image.
(ii) Negative y 0 implies inverted image.
(iii) Size of y 0 is 10 cm. .
(iv) Positive s0 gives location of image as the opposite side of the lens as
the object, that is, 15 cm to the right of the lens.
(b) A 20 cm tall flower is placed 30 cm to the left of a diverging lens of focal
length 10 cm.
f = −10 cm, s0 = 30 cm, y0 = 20 cm.
1
1
1
+ 0 =
s s
f
M=
y0
s0
=−
y
s
1
1
1
+ 0 =
30 cm s
−10 cm
⇒
⇒
y0
−7.5 cm
=−
20 cm
30 cm
⇒
⇒
s0 = −7.5 cm
y 0 = 5 cm
(i) Negative s0 implies virtual image.
(ii) Positive y 0 implies upright image.
(iii) Size of y 0 is 5 cm. .
(iv) Negative s0 gives location of image as the same side of the lens as the
object, that is, 7.5 cm to the left of the lens.
3
4. [10 points] A prism is made of glass with index of refraction 1.50. The angles
as of the prism are shown in the diagram. A beam of light enters the prism
perpendicular to one side as shown in the diagram. At what angle, relative to
the horizontal, does the beam of light leave the prism?
30o
o
o
θ2
90
60
θ1=30o
incoming
light beam
90o
60o
See the annotated diagram above.
Coming into the prism, the beam is perpendicular to the surface and parallel to
the normal line. Its incoming angle is therefore θ = 0 (not explicitly shown above).
Plug sin 0 = 0 for the incoming angle into Snell’s law to find that the outgoing angle
is also 0. In other words, because the beam is perpendicular to the first surface, it
does not bend there.
When the beam hits the far side of the prism, the angle relative to the normal line
is 30◦ . (See diagram above. The angle of 30◦ may be found by considering the
angles in the 30-60-90 triangle above the light beam in the diagram.)
Use Snell’s law, with n1 = 1.50, n2 = 1.00, and θ1 = 30◦ to find θ2 :
n1 sin θ1
sin θ2
θ2
= n2 sin θ2
n1
1.5
=
sin θ1 =
sin 30◦ = 0.75
n2
1.0
= 48.6◦
Angle θ2 is relative to the normal line. The problem asks for the angle relative to
the horizontal, which is 30◦ less than θ2 , that is, 18.6◦ . See diagram below.
48.6
30.0
48.6−30.0=18.6
4
5. [20 points] A velocity selector is used in experiments where all the charged
particles in a beam are required to have the same velocity (for example,
when entering a mass spectrometer). A velocity selector has a region of
uniform electric and magnetic fields that are perpendicular to each other and
perpendicular to the motion of the charged particles. The electric field and
magnetic fields each exert a force on the particles. If a particle has precisely
the right velocity, the two forces exactly cancel and the particle is not deflected.
Equating the electric and magnetic forces gives the following equation, which
we can solve for velocity:
⇒
qE = qvB
v=
E
B
A particle moving at this velocity will pass through the region of uniform fields
with no deflection, as shown in the figure. For higher or lower velocities than
this, the particles will feel a net force and will be deflected. A slit at the end of
the region allows only the particles with the correct velocity to pass.
~ down parallel to the page, and B
~ going into the page. A
The figure shows E
particle of mass m and charge q follows the path marked ~v .
B
v
particle
source
E
(a) Assuming that the particle in the figure is positively charged, what is the
direction of the force due to the electric field?
~ the direction of the force is the same as the direction of the electric
From F~ = q E,
field for a positively charged particle. In this situation, it is down.
(b) Assuming that the particle in the figure is positively charged, what is the
direction of the force due to the magnetic field?
One way to solve it. The fields have been arranged so that the magnetic force is
opposite the electric force, so it must be up.
~ use the right hand rule to find the
Another way to solve it. From F~ = q~v × B,
~ into the page, F~ is up.
direction of the magnetic field. For ~v to the right and B
This problem is continued on the next page....
5
(c) How does the kinetic energy of the particle shown in the figure change as it
traverses the velocity selector: does it increase, decrease, or stay the same?
There is no net force on the particle. Its speed does not change, and its kinetic
energy stays the same.
(d) Suppose a particle with twice the velocity of the particle in the figure enters
the velocity selector. How will the path of the particle curve: upward
(toward the top of the page), downward (toward the bottom of the page),
out of the page, into the page, or not at all?
~ Doubling v doubles the (upward) magnetic force.
Magnetic force: F~ = q~v × B.
~ Doubling v does not change the (downward) electric force.
Electric force: F~ = q E.
Since the upward force is increased while the downward force in unchanged, the
path of the particle will curve upward.
(e) Suppose a particle with the same mass and velocity of the particle in the
figure but twice as much charge, 2q, enters the velocity selector. How
will the path of the particle curve: upward (toward the top of the page),
downward (toward the bottom of the page), out of the page, into the page,
or not at all?
Both the electric and magnetic forces are proportional to charge. Doubling the
charge would double each force, so the two forces would still cancel out. The
particle would curve not at all.
6
6. [10 points] A resistor has a conical shape, as shown in the figure. The top of
the resistor has a smaller radius than the bottom of the resistor. It is hooked
to a 10 V battery, so that there is a 10 V drop from the top of the resistor to
the bottom.
Consider the voltage difference between the middle of the resistor (halfway
between top and bottom) and the bottom. Is this difference greater than, equal
to, or less than, 5 V? Briefly explain your reasoning.
10V
10V
V=?
Think of cutting the resistor into two parts at the midpoint, as suggested by the
figure below, so that each part has the same length, L. Because the cross-sectional
area A of each part varies over the length of the part, we cannot precisely calculate
the resistance of each part using R = ρL
A . However, we can make the qualitative
statement that A throughout the lower part is greater than A throughout the upper
part. Thus the resistance of the lower part must be smaller than that of the upper
part.
The two parts are in series, they have the came current. From V = IR, because the
lower part has less resistance, it has less voltage. The two voltages must add up to
10 V. In order for the lower part to have less voltage, it must be less than 5V.
Rbigger
10V
10V
Rsmaller
7
7. [20 points] Two resistors, two batteries, two switches, and a capacitor are wired
up as shown. Initially the switches are open (as shown) and the capacitor has
no charge.
R1=5Ω
Switch 1
Switch 2
10V
5V
R2=5Ω
C=0.1F
(a) At time t = 0, both switches are closed. Give a formula for the voltage
across the capacitor as a function of time, VC (t), and give a rough sketch
this function. Your formula should be a numerical expression with t as the
only variable.
R1=5Ω
10V
See the redrawn circuit at left. There is always a 5 V
difference across the series combination of resistor R2
and the capacitor, since this combination is parallel
to the 5 V battery. The presence of the 10 V battery
and R1 do not affect the behavior of R2 and C
because their total voltage is fixed at 5 V. Thus this
situation is exactly the same as a capacitor charging
up through a resistor attached to a single 5 V battery.
The final voltage is the battery voltage, Vf = 5 V.
The time constant is τ = (R2)(C) = 0.5 s. Thus
VC (t) = 5 V 1 − e−t/0.5 s .
5V
10V
R2=5Ω
5V
5V
C=0.1F
(b) If resistors R1 and R2 are light bulbs, describe qualitatively their behavior
over time, starting at t = 0.
See the diagram in the answer to part (a). Resistor R1 always has 5 V across it, so
it is steadily illuminated.
Resistor R2 starts off with 5 V (because the capacitor is
initially uncharged), but its voltage drops off as the capacitor is charged up.
Thus
they start off equally bright; R1 does not change; R2 fades out with time constant 0.5 s.
This problem is continued on the next page....
8
(c) At some much later time, switch 2 is opened. Switch 1 remains closed. Give
a rough sketch of the voltage of the capacitor over time, VC (t), starting
when switch 2 is opened. What is the characteristic time scale on which
the capacitor changes? (You are not required to give a formula for VC (t).
Just draw the sketch and calculate the characteristic time scale.)
See the circuit drawing at the bottom of the page. Once the switch is opened, current
cannot flow through the 5 V battery, so it has no effect on the behavior of the circuit.
In effect, the circuit is a 10 V battery with a series combination of R1 and R2 (whose
equivalent resistance is 5Ω+5Ω=10Ω) and the capacitor.
Before the switch was opened, the capacitor voltage was 5 V. (This was the “final
voltage” in part (a).) After it is opened, the capacitor charges up, asymptotically
approaching 10 V. The time scale is τ = RC = (10)(0.1) = 1 s.
(d) If resistors R1 and R2 are light bulbs, describe qualitatively their behavior
over time. Specifically, state what changes (if any) happen at the moment
switch 2 is opened, and describe what (if anything) happens after that.
Before the switch was opened, as described in (b), bulb R1 has 5 V and bulb R2 has
0 V; thus R1 is lit up and R2 is dim.
Immediately upon opening, the 10 V battery voltage is divided between the capacitor
and the two resistors. The capacitor gets 5 V (see part (c)), so the remaining 5 V is
divided between the resistors, that is, they each get 2.5 V. This makes them dimmer
than R1 was before the switch was opened.
As the capacitor charges up, the voltage across the bulbs falls off with time constant
1 s, as calculated in (c). Writing in terms of brightness (power), we have the following.
When switch 2 is opened, bulb 1, which had been bright, gets dimmer; bulb 2,
which had been off, turns on with the same intensity as bulb 1. They both then
fade out with time constant 1 s.
R1=5Ω
10V
Switch 2
R1=5Ω
R2=5Ω
10V
R2=5Ω
5V
C=0.1F
9
C=0.1F
8. [15 points] Asteroids are small bodies, typically a few kilometers in size, orbiting
the Sun. In 1993, the space probe Galileo flew near asteroid 243 Ida and
discovered that it was orbited by a moon. The moon was named Dactyl.
The orbit of Dactyl isn’t precisely known. For this problem, assume that it
moves around Ida with speed 10 ms at a distance 90 km from the center of Ida.
Assume that both Dactyl and Ida are bright (i.e., they reflect enough light to
be seen) and that they both can readily reflect radar signals.
(a) The Hubble space telescope has diameter 2.4 m and uses optical light.
(Take λ = 500 nm for this problem.) How close to the Earth would 243 Ida
have to be in order for Hubble space telescope to resolve the 243 Ida and
Dactyl (i.e., to tell that they are two separate objects)?
λ
Telescope resolution (small angle formula, θ in radians): θ ≈ D
Geometry (see figure; small angle formula, θ in radians): θ ≈ tan θ =
x
d
3
(90×10 m)(2.4 m)
λ
Equate these formulas: D
= xd
⇒ d = xD
= 4 × 1011 m.
λ =
(500×10−9 m)
Depending on the moon’s position in its orbit, its apparent separation when observed from Earth
might be less than 90 km (for example, if the moon were “in front of” the asteroid rather than
“next to” it), in which case smaller distance would be needed.
Dactyl
Hubble
x
θ
d
243 Ida
(b) Imagine a space probe with a radar transmitter and receiver located near
the asteroid and moon. At some moment, the asteroid and moon are
approaching the space probe, with the asteroid having speed 100 ms relative
to the probe. The space probe emits a radar signal toward the asteroid
and moon with frequency 8.0000000 GHz. Describe the reflected signal
detected by the space probe. Depending on how you approach this, your
answer may be a mix of quantitative and qualitative parts.
First consider the asteroid. It is moving at 100 ms toward the wave source (the space probe), so
vr = 100 ms . The wave speed is v = c = 3 × 1010 ms , and the source frequency is fs = 8.00000000 Hz.
Plug these numbers in to the Doppler formula for reflected waves:
100 ms
vr
(8.0000000 GHz) = 0.0000053 GHz = 5300 Hz.
∆f = 2 fs = (2)
v
3 × 108 ms
The reflector (asteroid) is moving toward the source, so the reflected wave is observed at a higher
frequency, fo = fs + ∆f = 8.0000053 GHz.
What about the moon? It is moving at 10 ms relative to the asteroid, but we don’t know the
direction. It could be moving as fast as 110 ms toward us or as slowly as 90 ms toward us. Plug
this range of velocities into the reflected Doppler equation to find ∆f between 0.0000048 GHz
and 0.0000059 GHz. Thus the observed signal is between 8.0000048 GHz and 8.0000059 GHz.
Summary.
A signal at 8.0000053 GHz is detected from the asteroid.
A signal between 8.0000048 GHz and 8.0000059 GHz is detected from the moon.
10
9. [10 points] A student studying the motion of dancers has a device which
measures forces the dancers’ feet impose on the floor. It produces a signal
whose voltage is proportional to the force. The motions made by the dancers
which she is interested in measuring take place over time scales of 0.1 seconds
and longer. Unfortunately, in addition to detecting the dancers’ motions, the
device vibrates on its own at a frequency of 100 Hz; she does not want this
100 Hz signal corrupting her measurements of the dancers.
She wants to use a resistor and a capacitor in a circuit to remove the 100 Hz
vibration signal while retaining signals from the dancers’ movements.
(Note: this problem may have a range of possible answers.)
(a) Which should she make: a low-pass filter or a high-pass filter?
(b) She will use a 10 kΩ resistor. What value of capacitor could she use?
(c) Sketch a diagram of what her circuit should look like. Indicate the input
(the signal from the force measurement device) as Vin . Indicate the output
(which she will record with a computer) as Vout .
She wants to detect signals with periods of 0.1 s or longer, which corresponds to
frequencies of f = 1/T = 10 Hz or smaller. She wants to eliminate signals at
100 Hz. Thus she wants a (a) low pass filter.
The characteristic frequency of the filter should be between 10 Hz and 100 Hz.
Anything in between is acceptable. We’ll pick 30 Hz. (We haven’t discussed this in
class, but usually one picks an in-between frequency using a logarithmic scale rather
than a geometric scale. That puts the half-way-in-between frequency at 30 Hz rather
than 55 Hz.) This gives characteristic time scale τ = 1/(30 Hz) = 0.033 s. Solving
τ = RC for capacitance gives C = τ /R = 0.033/10000 = 3.3 × 10−6 F = 3.3 µF.
In a lower-pass RC filter, the output voltage is across the capacitor, as in the drawing
below. The order of the resistor and capacitor doesn’t matter; what’s important is
that Vout is parallel to the capacitor.
R
Vin
C
11
Vout
10. [10 points] An infinitely long, straight wire carries a current, I1 , which steadily
increases, going from 0 to 6 Amp in 0.2 seconds. A conducting coil is 0.1 m away
from the wire. The coil consists of 1000 turns with radius 0.01 m. It is oriented
as shown, with the coils parallel to the plane of the paper. The ends of the coil
are attached to a resistor of resistance R = 1Ω.
What is the current through the resistor? Give both the magnitude of the
current and its direction, “up” (toward the top of the page) or “down” (toward
the bottom of the page).
Note. Because the radius of the coil is much smaller than its distance from
the wire, you may assume that, at any instant, the magnetic field is the same
everywhere in and near the coil.
I1
R
0.1 m
Direction. According to the right hand rule, the upward current in the wire, I1 ,
produces a magnetic field into the page at the location of the coil. The current,
I1 , is increasing, so the magnetic field through the coil, and hence its flux, into
the page, is increasing. A current is induced in the coil to oppose this field, so
the induced current will produce a field within the coil which points out of the
page. According to the right hand rule, this requires a counterclockwise current
around the coil. Following the geometry of this wire, such a current would flow
up, toward the top of the page when going through the resistor.
Magnitude. Let d be the distance from the wire to the coil and r be the radius of
the coil. The area of the coil is A = πr2 .
B
ΦB
E
I
=
µ0 I1
2πd
µ0 I1 πr2
µ0 I1 r2
=
2πd
2d
∆Φ
N µ0 r2 ∆I1
=
= N
∆t
2d∆t
E
N µ0 r2 ∆I1
(1000)(4π × 10−7 )(0.01)2 (6)
=
=
=
= 1.9 × 10−5 Amp
R
2dR∆t
2(0.1)(1)(0.2)
= BA =
12
11. [10 points]
(a) You have a rectangular block of dielectric material with dimensions
10 cm×20 cm×40 cm. You can put metal plates on any two opposite sides
of the block to form a capacitor. Which sides of the block should you put
the plates on to maximize the capacitance? (Answer “A”, “B”, or “C”,
based on the figure; obviously, one of the plates will be on a side of the
block not visible on the figure.)
40
cm
C
B
10 cm
A
20 cm
Use C = A
d . To maximize C, you need to maximize A while minimizing d.
This is done by putting the plates on the 20 cm×40 cm sides, i.e., side C .
(b) Consider the two capacitors shown below. The first one, in the upper
figure, has plate separation d. The second one, in the lower figure, has
plate separation 2d but also has a metal slab of thickness d in the middle.
Is the capacitance of the second capacitor greater than, less than, or the
same as the capacitance of the first capacitor. conductor?
Hint. One way to start this problem is to compare what happens if you
put charge Q on each capacitor.
d
2d
d/2
d
d/2
∆V1
∆V1 /2
∆V1 /2
Consider charge Q on the original capacitor, as in the
figure. This produces electric field E = 0QA inside the
capacitor. To find the voltage, rewrite E = ∆V
∆x as
∆V1 = E∆x = Ed = Qd
.
0A
Now consider the same charge Q on the second
capacitor. Charges −Q and +Q are induced on the
center conductor. In the gaps between plates and
conductors, the electric field is E = 0QA , just as the
original capacitor. In each gap, the distance is half the
original distance, d2 , so the voltage is half the original
Qd
2
1
voltage: 0 A
= 2Qd
= ∆V
2 . The two gaps are in series,
0A
∆V1
1
so their voltages add up, ∆V
2 + 2 = ∆V1 . Thus the
voltage difference between the top and bottom plate is
the same as in the original capacitor.
Capacitance is defined through Q = CV . Since, for a
given Q, both capacitors give the same voltage, ∆V1 ,
they must have the same capacitance.
Note: you can think of this as two capacitors in
series. However, unlike resistors, you can’t just add the
capacitances of two capacitors in series. See text for
details.
13
12. [15 points] A person consumes 2000 calories in a day. This means that he
whatever food is needed for his body’s metabolic process to generate 2000
calories of energy a day by digesting the food.
His mass is 75 kg and the surface area of his body is 2.0 m2 .
A “food calorie” is really a kilocalorie, which is equal to approximately 4186 J.
To convert from Celsius to Kelvin: T (Kelvins) = T (◦ C) + 273.15
Latent heat of fusion of water
Latent heat of vaporization of water
Specific heat of ice
Specific heat of liquid water
Specific heat of steam (at constant pressure)
J
Lf = 334, 000 kg
J
Lv = 2, 260, 000 kg
cice = 2090 kgJ◦ C
cwater = 4186 kgJ◦ C
csteam = 2010 kgJ◦ C
(a) If he retained all of the food energy in his body, by how much would his
temperature increase in one day? (Assume that he is made entirely of
water.)
Total energy input is Q=(2000 kcal)(4186 J/kcal)=8.4 × 106 J.
Q = mc∆T
∆T
=
Q
mc
=
8.4 × 106 J
(75 kg) 4186 kgJ◦ C
=
27◦ C
This problem is continued on the next page....
14
(b) In order to maintain temperature equilibrium, he must get rid of 2000
calories of energy a day. One way to do this is by sweating, in which
the energy is used to evaporate water. A model for this evaporation is
that energy from the body is used to bring the water from human skin
temperature, 34◦ C, up to the boiling point of water, 100◦ C, and then to
vaporize the water.
(Note. Although water evaporated from human skin is really evaporated at
a lower temperature than 100◦ C, the heat-then-vaporize model suggested
here gives reasonably accurate numerical results.)
Assume that sweating is the only mechanism by which he loses energy.
What mass of water he evaporate every day in order to maintain a steady
temperature (in other words, how much must he sweat)?
As in part (a), total energy input is Q=(2000 kcal)(4186 J/kcal)=8.4 × 106 J.
The water must be increased from 34◦ C to 100◦ C, a difference of ∆T = 66◦ C, and
then vaporized.
Q = mc∆T + mLv
Q = m(c∆T + Lv )
Q
m =
c∆T + Lv
8.4 × 106 J
= J
4186 kgJ◦ C (66◦ C) + 2, 260, 000 kg
=
8.4 × 106 J
J
2, 540, 000 kg
=
3.3 kg
15