1. science
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Examination Cover Sheet
Princeton University Undergraduate Honor Committee
January 30, 2009
Course Name & Number: PHY 101
Professors: Galbiati, McDonald, Sondhi, Staggs
Date: January 20, 2009
Time: 7:30 PM
This examination is administered under the Princeton University Honor Code.
Students should sit one seat apart from each other, if possible, and refrain from
talking to other students during the exam. All suspected violations of the Honor
Code must be reported to the Honor Committee Chair at [email protected].
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that is not checked may not be used and should not be in your working space. Assume
items not on this list are not allowed for use on this examination. Please place items
you will not need out of view in your bag or under your working space at this time.
University policy does not allow the use of electronic devices such as cell phones,
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•
•
•
•
Course textbooks: NO
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Formula Sheet: YES, only the one available at the end of the exam
booklet
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functions or advanced functions, like equation solving
Students may only leave the examination room for a very brief period without the
explicit permission of the instructor. The exam may not be taken outside of the
examination room. During the examination, the Professor or a preceptor will be
available at the following location: outside the exam room.
This exam is a timed examination. You will have 3 hours and 0 minutes to
complete this exam.
Write your name in capital letters in the appropriate space in the next page. Also,
dont forget to write and sign the Honor Code pledge in the appropriate line on the
next page:
“I pledge my honor that I have not violated the Honor Code during this examination”
1
2
Your Name:
PHYSICS 101 FINAL EXAM
January 20, 2009
3 hours
Please Circle your section
1
9 am
Galbiati
2
10 am
3
11 am
McDonald
4 12:30 pm
5 12:30 pm Sondhi
Problem
Score
1
/15
2
/15
3
/20
4
/5
5
/5
6
/15
7
/15
8
/10
9
/16
10
/12
11
/3
12
/3
Total
/134
McDonald
Staggs
Instructions: When you are told to begin, check that this examination booklet
contains all the numbered pages from 2 through 27. The exam contains 9 problems.
Read each problem carefully. You must show your work. The grade you get depends
on your solution even when you write down the correct answer. BOX your final
answer. Do not panic or be discouraged if you cannot do every problem; there are
both easy and hard parts in this exam. If a part of a problem depends on
a previous answer you have not obtained, assume it and proceed. Keep
moving and finish as much as you can!
Possibly useful constants and equations are on the last page, which you
may want to tear off and keep handy.
Rewrite and sign the Honor Pledge: I pledge my honor that I have not violated the
Honor Code during this examination.
Signature
3
Problem 1
Circle all true statements. There may be none or more than one true statement
per group.
1. (3 points)
(a) In any perfectly inelastic collision, all the kinetic energy of the bodies
involved is lost.
(b) Mechanical energy is conserved in an elastic collision.
(c) After a perfectly inelastic collision, the bodies involved move with a
common velocity.
Statement a is false. The bodies, after a perfectly inelastic collision, move with the
velocity of the center of mass prior to the collision. Not all the kinetic energy is
necessarily lost.
Statement b is true.
Statement c is true.
2. (3 points)
(a) If the angular velocity of an object is zero at some instant, the net torque
on the object must be zero at that instant.
(b) If the net torque on an object is zero at some instant, the angular velocity
must be zero at that instant.
(c) The moment of inertia of an extended solid object depends on the location
of the axis of rotation.
Statement a is false. There is no such direct relationship between torque and angular
velocity, rather between torque and angular acceleration.
Statement b is false. There is no such direct relationship between torque and angular
velocity, rather between torque and angular acceleration.
Statement c is true.
4
3. (3 points) A bag of lead pellets (specific heat capacity C = 128 J kg−1 ◦ C−1 ) is
dropped from a height of 10 m. Upon impact, 50% of the kinetic energy of the
bag turns into thermal energy dissipated throughout the lead bag. Calculate
the change in temperature of the lead.
The energy balance is:
1
× mgh = mC∆T
2
Therefore:
∆T =
gh
(9.8 m/s2 )(10 m)
=
= 0.38 ◦ C
2C
2(128 J kg−1 ◦ C−1 )
4. (3 points) For this question, you MUST box your answer AND offer your
arguments as to why you are choosing that answer.
Imagine you are floating on a lake in a boat in which you have a large rock. You
throw the rock overboard. Does the level of the lake (relative to the shore):
• Stay the same?
• Increase?
• Decrease?
The level decreases. In the initial situation, the water level is raised by the amount
necessary to make the rock float on the boat; i.e., the hull of the boat displaces a
volume of water whose mass is equal to the mass of the rock. In the final situation
the rock (now on the bottom of the lake) displaces a volume of water equal to its own
volume. The volume displace in the initial situation is larger than the rock’s own volume
by a factor given by the ratio of the rock density to the water density.
5
5. (3 points) Two disks of identical mass m but different radii (2r and r for disk
1 and 2 respectively) are spinning on frictionless bearings at the same angular
speed ω0 but in opposite directions. The two disks are brought slowly together.
The resulting frictional force between the surfaces eventually brings them to a
common angular velocity. What is the magnitude of that final angular velocity
in terms of ω0 ? Also specify: is the final system rotating in the initial sense of
disk 1 or disk 2?
Disk 1 Disk 2
The moments of inertia of the two disks are:
I1 = 2mr2
I2 =
1 2
mr
2
The angular momenta are:
L1 = 2mr2 ω0
1
L2 = − mr2 ω0
2
The total angular momentum in the initial condition is:
Ltot = L1 + L2 =
3 2
mr ω0
2
The angular momentum is conserved. Therefore, in the final condition:
Ltot = (I1 + I2 )ωf
where ωf is the final angular velocity:
ωf =
Ltot
=
I1 + I2
which is in the same direction as disk 1.
3
2
2 mr ω0
5
2
2 mr
=
3
ωf ,
5
6
Problem 2
A plastic tube of radius R=1.0 cm is placed at the bottom of a well on a hill. The
second end of the tube is located close to a large bucket (capacity 200 liters = 0.2 m3 )
located downhill (see the figure). Once the syphon is primed (the tube is filled with
water), the second end of the tube is inserted into the bucket. The surface of the
water in the well is y = 2.0 m above the level of the water in the bucket. The top of
the hill is h = 7.0 m above the level of the water in the well. Throughout the problem
you may take the atmospheric pressure to be patm = 105 Pa.
(top point
of water tube)
C
Water Tube
h
A
(water level
in the well)
y
(water level
in bucket)
B
7
1. (3 points) What is the velocity of the water as it leaves the tube and flows into
the bucket?
Let’s apply Bernoulli equation between point A (surface of the water in the well) and
point B (exit of the water tube):
1 2
1 2
pA + ρgy + ρvA
= pB + ρvB
2
2
Taking into account that pA = pB = patm and vA = 0, we get:
p
p
v = 2gy = 2(9.8 m s−2 )(2 m) = 6.3 m/s
2. (3 points) How much time ∆t is needed to fill the bucket (you can neglect the
variation of y during the time the bucket is filled)?
Let’s call with Q = AvB the volume flow rate at the exit of the pipe; A is the area of
the pipe. Then:
V = Qt = (AvB )t = πR2 vB t
Hence:
∆t =
V
0.2 m3
=
= 101 s .
πR2 vB
π(0.01 m)2 (6.3 m)
8
3. (5 points) What is the water pressure inside the tube at the top point (marked
as C in the associated figure)?
Let’s apply Bernoulli equation between point C and point B (exit of the water tube):
1 2
1 2
pC + ρg(h + y) + ρvC
= pB + ρvB
,
2
2
Taking into account that pB = patm and vC = vB , we get:
pC = pB − ρg(h + y) = 105 Pa − (1000 kg/m3 )(9.8 m s−2 )(9 m) = 11, 800 Pa .
4. (4 points) If the distance h gets increased (for example, by raising the top of
the tube C) at some point the syphon will stop working. For what value of h
does the syphon stop working?
The syphon stops working when the pressure in point C becomes equal to zero (can’t
have negative pressure). Applying the Bernoulli equation between points B and C, the
condition is:
ρg(h + y) = patm
Or:
h=
105 Pa
patm
− 2 m = 8.20 m .
−y =
ρg
(1000 kg/m3 )(9.8 m s−2 )
9
Problem 3
A closed cylinder is divided into two chambers by a massless disk which is connected
by a spring of spring constant k to one end of the cylinder, as shown in the figure.
(You may neglect any friction between the disk and the walls of the cylinder.) The
left chamber is filled with an ideal monatomic gas and the right chamber is completely
evacuated (i.e., its pressure is zero). Initially, the cylinder is in thermal equilibrium
with a thermal reservoir at temperature T1 =307 K, and the spring is compressed by
x1 =0.12 m. The spring constant is k=1.15×105 N/m.
1. (4 points) How many moles of gas, n, are in the left chamber?
Let the area of the disk be A and the pressure of the gas be P1 . Then, by considering
the forces on the massless disk we see P1 A = kx1 , so that P1 = kx1 /A. Note that the
volume the gas occupies is given by V1 = Ax1 . Next we apply the ideal gas law:
kx1
P1 V1 = nRT1 =
(Ax1 ) = nRT1 .
A
This can be re-arranged to show:
n=
kx21
1.15 × 105 N m−1 · (0.12 m)2
=
= 0.65 mol .
RT1
8.315 J K −1 mol−1 · 307 K
10
2. (4 points) What is the internal energy Ugas of the gas?
Ugas =
3
3
3
nRT0 = kx21 = · 1.15 × 105 N m−1 · (0.12 m)2 = 2490 J .
2
2
2
3. (4 points) What is the mechanical energy Espring stored in the spring?
Espring =
1 2
1
kx = · 1.15 × 105 N m−1 · (0.12 m)2 = 830 J .
2 1
2
11
4. (4 points) Now the temperature is changed to 4T1 . The cylinder remains sealed.
How much work W is performed by the gas?
We can find the work W done by the gas by noting that the work is done on the
spring, increasing its potential energy from 12 kx21 to 12 kx22 . We can find x2 , the new
compression of the spring, by applying the ideal gas law again and using the result from
part (1):
P2 V2 = nR4T1 = 4kx21 .
We note that the volume is V2 = Ax2 and that P2 = kx2 /A so that we can write
P2 V2 − kx22 = 4kx21 ,
and thus we see x2 = 2x1 . So, putthing these things together we find:
W = 1/2k(2x1 )2 − 1/2kx21 = 3/2kx21 .
W =
3 2
3
kx = · 1.15 × 105 N m−1 · (0.12 m)2 = 2490 J .
2 1
2
5. (4 points) Find now the heat that entered the gas while raising the temperature
from T1 to the 4T1 . Mind the sign!
We know from the first law of thermodynamics that ∆Ugas = Q − W , where Q is the
heat that flows into the gas. We find
∆Ugas = U2 − U1 =
3
3
9
9
nR(4T1 ) − nRT1 = RT0 = kx21 ,
2
2
2
2
where in the last step we used the solution from part (1). Then:
Q = ∆Ugas +W =
9 2 3 2
kx − kx = 3kx21 = 3·1.15×105 N m−1 ·(0.12 m)2 = 4970 J .
2 1 2 1
12
Problem 4
(5 points) Circle the true statement(s) below.
1. A heat pump takes heat from a hot reservoir and uses it to increase the
temperature of a cold reservoir while work is done on the pump’s working
substance.
2. A Carnot heat pump operating operating between two reservoirs at
temperatures TH and TC is more efficient than any other heat pump using
those two reservoirs which does not use the Carnot process.
3. A Carnot cycle includes changing the volume of an ideal gas at constant
pressure, and changing the volume of the gas adiabatically.
4. A Carnot cycle includes an isothermal compression of an ideal gas and an
adiabatic expansion of the same gas.
5. The change of the entropy of the universe during one cycle of a Carnot heat
pump, operating between two reservoirs at TH and TC , is larger than the change
of entropy of the universe during one cycle of any other heat pump operating
between the same temperatures.
Statement a is false. A heat pump takes heat from a cold reservoir and uses it to
increase the temperature of a hot reservoir.
Statement b is false. A Carnot heat pump operating operating between two reservoirs
at temperatures TH and TC has the same efficiency of any reversible heat pump.
Statement c is false. A Carnot cycle two isothermal (not isocoric!) and two adiabatic
transformations.
Statement d is true.
Statement e is false. The change of the entropy of the universe during one cycle of
a Carnot heat pump is zero. The change of change of entropy of the universe during
one cycle of any other heat pump operating between the same temperatures is zero or
larger.
13
Problem 5
A Carnot heat pump is used to heat a building to a temperature of 21◦ C.
1. (3 points) How much work must the Carnot heat pump do to deliver 3350 J of
heat into the house on a day when the outside temperature is 0 ◦ C?
W = QH − QC = QH
1−
TC
TH
273 K
= 3350 J 1 −
= 239 J .
294K
2. (2 points) How much work must the Carnot heat pump do to deliver the same
3350 J of heat into the house when the outside temperature is −21 ◦ C?
W = QH − QC = QH
TC
1−
TH
252 K
= 3350 J 1 −
294K
= 479 J .
14
Problem 6
Two monkeys of mass m=10 kg each are hanging from a spring of spring constant
k=196 N/m and unstretched length l0 =2.1 m. The spring is attached to the ceiling
whose height is h=5.0 m above the floor. You may consider the monkeys to be point
masses, i.e., ignore their size.
1. (4 points) What is the total length l of the spring when both monkeys hang from
it in equilibrium and at rest? Include both the original, unstretched length l0
and the displacement x in the answer.
The spring is stretched by an amount determined by 2mg = k(l − l0 ) ≡ kx. Hence
x=
2 × 10 kg × 9.8 m/s2
= 1.0 m .
196 N/m
Further:
l = l0 + x = 3.1 m .
2. (3 points) What is the elastic potential energy stored in the spring?
E=
1 2
kx = 98 J .
2
15
3. (4 points) One of the monkeys drops off, gently. As a result, the remaining
one starts to rise. To what maximum height above his initial position does the
second one rise?
With one monkey off, the equilibrium displacement of the spring is now:
x0 =
10kg × 9.8m/s2
= 0.50 m,
196N/m
Therefore the second monkey, which starts at an amplitude A=0.50 m below the new
equilibrium position, will rise to a height of 0.50 m above it and thus to a total height
of:
2A = 1.0 m ,
above his initial position.
4. (4 points) On his way back down, the second monkey lets go of the spring when
his velocity downwards is greatest. How long after the first monkey does he
reach the ground?
The first monkey drops
p a distance d1 = 5.0 m − 3.1 m = 1.9 m starting from rest which
takes a time t1 = 2d1 /g = 0.62 s. The second monkey moves up with the spring and
drops off at the equilibrium position
on his way down. It take him 3/4 of a time period
p
to get there which is ta2 = 3π
m/k
= 1.06 s. At this point his downward velocity is
2
p
v0 = Aω = A k/m = 2.2m/s and his height is d2 = 1.9 m + 0.5 m = 2.4 m above
ground. Thus he covers this distance in a time tb2 which solves the quadratic equation:
2
1
d2 = v0 tb2 + g tb2 ,
2
which can be written as:
2
1
2.4 m = 2.2 m/s · tb2 + 9.8 m/s2 · tb2 .
2
This yields tb2 = 0.51s and hence the second monkey lands a time:
ta2 + tb2 − t1 = 1.06 s + 0.51 s − 0.62 = 0.95 s ,
after the first one.
16
Problem 7
A bowler releases a bowling ball of mass m and radius R straight down a lane at
a bowling alley. The ball has initial velocity vo and the coefficient of kinetic friction
between the ball and the surface of the lane is µk . Initially the ball slides (or slips) but
starts to spin as well. Some time later it starts to roll without slipping and eventually
it strikes the pins.
1. (3 points) For the first part of the motion, find the velocity (of the center of
mass) as a function of time. Express your answer as a function of v0 , µk , g, and
the time t.
The translational motion of the ball is affected by the horizontal frictional force while
it slides. We will identify the frictional force as Ffr = µk mg. Hence:
v(t) = v0 − at = v0 −
Ffr
t = v0 − µk gt .
m
2. (4 points) What is the torque due to friction about the horizontal axis,
perpendicular to the direction of motion, which passes through the center of
the ball? Express your answer as a function of m, µk , g, and R.
The frictional force has the magnitude:
τ = |Ffr × R| = µk mgR .
17
3. (4 points) Again for the first part of the motion, find the angular velocity about
the axis in part 2 as a function of time. Express your answer as a function of
µk , g, R, and t.
The torque calculated above produces an angular acceleration α = τ /I = (5/2)(µk g/R)
where I = (2/5)mR2 is the moment of inertia of the ball. Initially, the angular velocity
is zero. Hence:
ω(t) = αt = (5/2)(µk g/R)t .
4. (4 points) The ball stops slipping when the velocity of the ball becomes equal
to the product of the angular velocity and the radius. How long does the ball
slide before it starts to roll without slipping? Express your answer as a function
of v0 , µk , and g.
The ball starts rolling without slipping when its velocity and angular velocity satisfy the
relation v(t) = Rω(t). Using our results from parts 1 and 3 we obtain the equation
that fixes the time when this happens:
v0 − µk gt = (5/2)µk gt,
which is solved by:
t=
2 v0
.
7 µk g
18
Problem 8
Lake Carnegie is a strip of water 70 m wide and 5.2 km long. In the middle of
January 2009, lake Carnegie has a 33 cm thick surface layer of ice on a night when the
air above it is at a temperature of -15 ◦ C. The water below the ice is at a temperature
of 0 ◦ C. Useful constants: the thermal conductivity of ice is κice =2.2 J/(m s ◦ C), the
latent heat of fusion of water is Lfusion =33.5×104 J/kg, and the density of water is
ρwater =1,000 kg/m3 .
1. (4 points) How much heat does the entire lake lose to the atmosphere in one
hour by conduction of heat through the ice?
The surface area of the lake is:
A = 70 m × 5.2 × 103 m = 3.64 × 105 m2 ,
which is then the area of the sheet of ice of thickness 0.33m which conducts heat from
the water beneath it to the air above it. The heat lost by the lake is
Q=
κA(Twater − Tair )t
2.2 J/(m s ◦ C) · 3.64 × 105 m2 · 15 ◦ C · 3600 s
=
= 1.31×1011 J
L
0.33 m
2. (2 points) If the heat is all lost from the layer of water just below the ice, by
how much does the thickness of the ice layer grow in one hour?
The amount of heat lost by the lake in an hour equals the latent heat loss of
the water which is transformed into ice, whose thickness is identified as d. We
neglect the difference between the density of water and ice, which is also taken to
be ρice = ρwater =1,000 kg/m3 . Therefore:
Q = Adρice Lfusion ,
which implies:
d=
Q
1.31 × 1011 J
= 1.07 mm .
=
Aρice Lfusion
3.64 × 105 m2 · 103 kg/m3 · 33.5 × 104 J/kg
19
3. (4 points) Imagine that the layer of ice is magically replaced by vacuum. How
much heat does the entire lake lose by heat radiation in one hour to the
atmosphere above the lake? You can assume the emissivities of both air and
water to be both 1.
By the Stefan-Boltzmann law, the net energy Erad radiated from the water to the air
equals:
4
4
Erad = Pnet · t = eσA(Twater
− Tair
) · t,
J
Erad = 1·5.67×10−8
·3.64×105 m2 · 2734 − 2584 K4 ·3600s = 8.35 × 1010 J .
s m2 K4
20
Problem 9
On the website www.rubegoldberg.com we learn that once upon a time Rube
Goldberg mistook a lot of broken glass for bath salts and when they pulled him
out of the tub he mumbled an idea for dodging bill collectors:
As Tailor (A) fits Customer (B) and calls out measurements, College Boy (CB)
mistakes them for football signals and makes a flying tackle at clothing dummy (D).
Dummy bumps head against paddle (E) causing it to pull hook (F) and throw bottle
(G) on end of folding hat rack (H) which spreads and pushes head of cabbage (I) into
net (J). Weight of cabbage pulls cord (K) causing shears (L) to cut string (M). Bag
of sand (N) drops on scale (O) and pushes broom (P) against pail of whitewash (Q)
which upsets all over you (R) causing you to look like a marble statue and making it
impossible for you to be recognized by bill collectors. Don’t worry about posing as
any particular historical statue because bill collectors don’t know much about art.
21
1. (4 points) The College Boy (CB) has mass M =97 kg and the clothing dummy
(D) has mass m=9.7 kg. The CB has velocity vCB =2.2 m/s when he collides
inelastically with the dummy. What is the final velocity vfinal of the CB +
dummy, and how much kinetic energy is lost during the collision?
Conservation of momentum implies M vCB = (M + m)vfinal , so:
vfinal =
∆KE =
M
2.2 m/s · 97 kg
vCB =
= 2.0 m/s .
M +m
(97 + 9.7) kg
2
97 kg · (2.2 m/s)2 (97 + 9.7) kg · (2 m/s)2
M v 2 (M + m)vfinal
−
=
−
= 21 J .
2
2
2
2
2. (4 points) As the clothing dummy (D) strikes the paddle (E), the bottle (G)
takes on an initial horizontal velocity of 0.90·vfinal and falls height h=0.85 m onto
the folding hat rack (H). This contraption has the same effect on the cabbage
head (I) as an elastic collision with a bottle whose velocity equals the actual
vertical velocity of the bottle when it hits the hat rack. If the bottle and the
cabbage have the same mass m0 =1 kg, what is the velocity of the cabbage after
the collision? (If you did not get part 1, assume vfinal =1 m/s.)
In elastic collisions between objects of equal mass, the velocities of the two objects are
interchanged. Hence, the final velocity of the cabbage equals the initial, vertical velocity
of the bottle.
p
p
vcabbage = vbottle,y = 2gh = 2 · 9.8 m/s2 · 0.85 m = 4.1 m/s .
22
3. (4 points) If the bag of sand (N) has a mass mN equal to 7 times the mass mP
of the broom (P), and the effect of the scale (O) is to transfer all of the energy
of the falling bag to the broom, to which height l would the broom rise above its
initial position if the pail (Q) weren’t it its way? The bag falls height d=0.66 m
before it hits the scale, which is basically a beam that pivots about a horizontal
axis through its center.
The bag of sand gains energy EN = mN gd when it falls height d. Therefore, the initial
(kinetic) energy of the broom, just after the bag strikes the scale, is EP = EN = mN gd.
Hence, the broom would rise to height l given by:
mP gl = EP ,
therefore:
l=
mN gd
EP
=
= 7 · d = 7 · 0.66 m = 4.6 m .
mP g
mP g
4. (4 points) Suppose that instead of falling onto the scale (O), the bag of sand
(N) is just placed into one of its pans, such that the broom (P) pushes up
against the pail (Q) of whitewash, whose mass (including the whitewash) is
mQ =8.3 kg. What is the least mass mN of the sand bag such that the broom
of mass mP =0.71 kg could (slowly) tip the pail of whitewash onto you (R)? For
the whitewash to (begin to) spill out, the center of mass of the pail + whitewash
must rise by h0 =4.0 cm, during which time the point of contact of the broom
with the pail must rise by h1 =9.0,cm.
To raise the center of mass of the pail of whitewash by h0 , work WQ = mQ gh0 must be
done on it by the broom. The tension in the hook from which the pail is suspended
does no work.
If F is the vertical force of the broom on the pail, the work done by this force as the
point of contact moves by height h1 is:
F h1 = WQ = mQ gh0 .
Thus, we derive:
F =
mQ gh0
.
h1
Meanwhile, the effect of the scale is that the weight mb g of the broom plus force Fb
balance the weight of the sand bag, ms g (if the motion is slow). That is:
mN g = m P g + F = mP g +
mQ gh0
,
h1
so that:
m N = mP g +
8.3 kg · 0.04 m
mQ gh0
= 0.71 kg +
= 4.4 kg .
h1
9m
23
Problem 10
The engine of a jet airplane operates at 12500 rpm (revolutions per minute) when
it flies at Mach 1.25 (equivalent to 1.25 times the speed of sound). You may assume
in parts 1 and 2 that the jet is flying very low, near to the ground on which the
observers are standing.
1. (4 points) What frequency would you hear ass the jet approaches you?
Because the jet is flying faster than the speed of sound,
you hear nothing as the jet approaches you.
2. (4 points) What frequency would you hear later as the jet flies away from you?
Doppler’s formula for source s moving away from fixed observer o is:
fo =
fs
,
1 + vvs
where fs is the frequency emitted by the source, fo is the frequency perceived by the
observer, vs is the frequency of the source relative to the medium carrying the sound
waves, and and v is the speed of sound (for sound in air, v=343 m/s.) In the present
case:
12, 500 min−1
fs = 12500 rpm =
= 208 Hz,
60 s min−1
and:
vs
= 1.25
v
So:
fs
208 Hz
fo =
=
= 94 Hz .
1 + vvs
1 + 1.25
24
3. (4 points) When the jet is a distance d=500 m past you, the sound level of its
engine is 110 dB (as heard by you). How much time ∆t does it take for you to
hear the sound level to drop to 90 dB?
For the sound level to decrease by 20 dB, the sound intensity must drop by a factor of
100. Since the intensity of sound falls off as 1/r2 , the jet must be 10 times as far away,
i.e., 5 km from you. The time delay between your hearing 110 dB and 90 dB is the sum
of the time for the jet to travel the extra distance (l = 4.5 km), and for the sound to
return that distance (i.e., 4.5 km and not 5 km).
∆t =
d
d
4500 m
4500 m
+ =
+
= 23.6 s.
vs
v
1.25 · 343 m/s 343 m/s
25
Problem 11
By flexing a stretched string, whose far end is tied to a wall, you generate a pulse
that moves towards the wall as shown below.
1. (3 points) Sketch on the lower figure how the pulse looks after it reflects from
the wall.
The reflected pulse can be thought of as coming in from the right, beyond the wall.
The sum of the initial pulse and the reflected pulse must be zero at the wall, where
the string is tied. Hence, the reflected pulse must be inverted, as sketched below.
26
Problem 12
(3 points) A drainpipe outside your (second floor) window has a length L=5.13 m
and is open at both ends.
1. (3 points) When the wind blows hard, what are the highest and lowest
frequencies you hear from standing waves in the pipe, assuming that you can
hear frequencies between 50 and 15,000 Hz?
Standing waves in an open pipe of length L must have antinodes at both ends, which
implies that the wavelength λ of the wave obeys:
L=
nv
nλ
=
,
2
2f
for any positive integer n. The possible frequencies are therefore:
fn = nv/2L.
The lowest frequency is:
f1 =
343 m/s
v
=
= 33.43 Hz.
2L
2 · 5.13 m
However, you do not hear this because it’s too low for the human ear. The lowest
frequency that you hear is:
f2 = 2 · f1 = 2 · 33.43 Hz = 67 Hz .
The highest frequency that you (barely) hear has index:
15, 000 Hz
n = int
= 448,
33.43 Hz
and the corresponding frequency is:
f448 = 448 · f1 = 448 · 33.43 Hz = 14, 977 Hz .
27
POSSIBLY USEFUL CONSTANTS AND EQUATIONS
You may want to tear this out to keep at your side
x = x0 + v0 t + 12 at2
F = µFN
s = rθ
p = mv
ω = ω0 + αt
KE = 21 Iω 2
W = F s cos θ
2
ac = vr
τ = F ` sin θ
Ifull cylinder = 21 mr2
Ifull sphere = 52 mr2
F = Y A ∆L
L0
P V = nRT
Q = mL
V
Wisotherm = nRT ln Vfi
Q
= kA
∆T
t
L
Q = σT 4 At
v = λf
β = (10 dB) log II0
f0 = fs / 1 ∓ vvs
sin θ = Dλ
Q = kA∆T t/L
v = v0 + at
F = −kx
v = rω
F ∆t = ∆p
P E = 12 kx2
KE = 12 mv 2
L = Iω
ω 2 =p
ω02 + 2α∆θ
ω = k/m√
2πr3/2 = T GM
Ihollow cylinder = mr2
∆L = αL0 ∆T
P1 V1γ = P2 V2γ
P V = nkT
Q = cm∆T
γ =q
CP /CV
v=
γkT
p m
v = Bad /ρ v
1± 0
f0 = fs 1∓ vvs
v
v
fn = n 2L
, n=1,2,3, ...
sin θ = 1.22 λd
v2 = v02 + 2ax
F = −G Mr2m
a = rα
P
xcm = M1tot i xi mi
P E = mgh
Wnc = ∆KE + ∆P E
P
τ = Iα
∆θ =pω0 t + 12 αt2
ω = g/l
P
I = mi ri2
Q = πR4 ∆P/8ηL
Q = Av
P + ρgh + 21 ρv 2 = const
∆S = ∆Q
T
∆U = Q − W
U = 32 nRT
q
F
v = m/L
√
v = Yρ
f0 = fs (1 ± vv0 )
v
fn = n 4L
, n=1,3,5, ...
Pnet = eσA (T 4 − T04 )
Monatomic:
Diatomic:
CV = 3R/2
CV = 5R/2
CP = 5R/2
CP = 7R/2
R = 8.315 J/K/mol
u = 1.66 × 10−27 kg
σ = 5.67 × 10−8 W m−2 K−4 k = 1.38 × 10−23 J K−1
NA = 6.022 × 1023 mol−1
Mearth = 5.98 × 1024 kg
Mmoon = 7.35 × 1022 kg
Rearth = 6.36 × 106 m
Rmoon = 1.74 × 106 m
GNewton = 6.67 × 10−11 N m2 kg−2
0 ◦ C = 273.15 K
1 kcal = 4186 J
vsound,air = 343 m/s
Asphere = 4πR2
Vsphere = 43 πR3