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Unformatted text preview: © Prep101 http://www.prep101.com/freestuff PRACTICE PHYSICS 101 EXAM
Problem 1
I. A student, holding a 500 Hz tuning fork, is walking at 1.50 m/s on a
direct line between two parallel smooth reflecting walls. What
frequency of sound are heard by the student?
Answer
⎛v⎞
At the wall the student approaches, f ' = f ⎜
⎜ v − v ⎟ because the wall is a
⎟
s⎠
⎝
stationary “observer” watching a moving source.
For the sound reflected by the wall, the wall is a stationary “source”
that reflects the same frequency it “hears”, and the student is a
moving observer, so if the student hears frequency f’’, where:
⎛ v + vo ⎞
⎛ v + vo ⎞
⎛ 341m / s + 1.5m / s ⎞
f '' = f '⎜
⎟= f⎜
⎟
⎜ v − v ⎟ = 500 Hz⎜ 341m / s − 1.5m / s ⎟ = 504.4 Hz
⎝
⎠
⎝v⎠
s⎠
⎝
Similarly, for receding wall, (the student is going away from the wall,
the velocities have inverse signs):
⎛ v + vo ⎞
⎛ 341m / s − 1.5m / s ⎞
f ''= f ⎜
⎟
⎜ v − v ⎟ = 500 Hz⎜ 341m / s + 1.5m / s ⎟ = 495.6 Hz
⎝
⎠
s⎠
⎝ II. A column of air in a tube is found to have standing waves at
frequencies of 380, 532 and 684 Hz. There are no standing wave
frequencies between the above frequencies.
a) What is the fundamental frequency of the tube?
Answer
Resonant frequency for sound waves in a tube are uniformly spaced.
Notice the frequency spacing here is 152 Hz because 684Hz532Hz=532Hz380Hz=152Hz. Keep subtracting multiples of 152Hz
until you reach a frequency <152Hz and you’ll reach the fundamental
frequency ffundamental=76Hz.
b) Is the tube open at both ends or open at one end and closed
at the other end?
Answer
For a tube open at both ends, the fundamental frequency is the same
as the frequency of spacing. For a tube closed at one end, the
fundamental frequency is on half of the frequency of spacing. The
latter is true, so the tube is closed at one end. Page 1 of 16 © Prep101 http://www.prep101.com/freestuff c) Draw a displacement curve for the 532Hz standing wave.
Answer
The fundamental frequency has one node on
the close end of the tube and no nodes
between the ends of the tube. The open end
of the tube has an antinode. For each
multiple of 152Hz we add to the fundamental
frequency to get to 532Hz wave, we add a
node.
532Hz=76Hz+152Hz×3, so we add 3 nodes between the ends of the
tube. d) the air in the tube is now replaced with carbon dioxide which
has a speed of sound of 280m/s. What are the new
frequencies corresponding to the given standing wave
frequencies?
Answer
Since geometry of the tube is unchanged, the resonant wavelength
vCO2
v
are unchanged. Since wavelength is unchanged, λ = air =
.
f air
f CO2
280m / s
= 0.8211 f air
v air
341m / s
so, the new frequencies are 312Hz, 436.8Hz and 561.6Hz.
f CO2 = f air × vCO2 = f air × Page 2 of 16 © Prep101 http://www.prep101.com/freestuff Problem 2
Two flat glass plates, each of length 6.0cm, are
separated at each end by two support of
thickness t1=0.00125 mm and t2=0.00575 mm,
as shown in the figure. The air between the
t1
glass plates forms a thin film. Green laser light
of wavelength 500 nm is incident from the top.
When viewed from above, an interference
pattern consisting of alternating bright and dark
fringes is observed.
a) What is the separation of the two dark fringes? t2 6.0 cm Answer
0
π
n>1
n=1 For dark fringes, we want the phase difference between
the two reflected waves to be π+2mπ for integer values
of m.
2nt
∆φ = π + 2mπ = π + 2 ×
×π λ 2nt=2t=mλ
(n=1 in air)
n>1
If the pieces of glass extended until they touched, this
formula would give the number of gaps between fringes, m, to the left
of wherever you choose to measure the thickness t. thus, you can
count the number of fringes along the 6 cm stretch by the following
formula:
2t − 2t1
2
m2 − m1 = 2
=
(5.75 × 10 −6 − 1.25 × 10 −6 )m = 18
−9
λ
500 × 10
The far left end of the plate has a dark fringe since m1=2t1λ=5 is an
integer.
The far right edge of the plate has a dark fringe since m2=2t2λ=23 is n
integer. If there are 18 gaps between fringes, then the spacing
between fringes is 6cm/18=0.33cm. Page 3 of 16 © Prep101 http://www.prep101.com/freestuff b) What must be the index of refraction of a 150 nm thick
coating (with index of refraction less than 1.5) on a glass
surface (n=1.5) if it is to eliminate reflection of 650 nm light
incident on the coating from air?
Answer
We want destructive interference.
Light rays reflected off the coating
and light rays reflected off the glass
both receive π phase shifts.
Therefore, we only need to consider
phase shift due to the path
difference. We want 2nt=λ/2 for the
thinnest coating that blocks
λ=650nm light.
Thus:
650nm
λ
n=
=
= 1.08
4t 4 × 150nm π
π
n=1
t Page 4 of 16 air 1<n<1.5 coating n=1.5 glass © Prep101 http://www.prep101.com/freestuff Problem 3
I. A hydrometer consists of a spherical bulb with a cylindrical stem.
The crosssectional area of the stem
is 4.0 cm2. The total volume of bulb
and stem is 16.8 cm2. When
immersed in water, the hydrometer
floats with 1.7 cm of the stem
above the water surface. In alcohol,
1.0 cm of the stem is above the
water surface. Find the density of
alcohol.
Answer
Volume under water=Vwater=VhydrometerVabove water
=16.8cm3(1.7 cm×4cm2)=10 cm3
Mass of hydrometer=m=ρwaterVwater=1g/cm3×10cm3=10g
Volume under alcohol=Valcohol=VhydrometerVabove alcohol
=16.8cm3(4cm2×1cm)=12.8cm3
m
10 g
ρ alcohol =
=
= 0.78125 g / cm 3
3
Valcohol 12.8cm
II. Water stands to a height of 2.50 m
in a large tank which contains
compressed air maintained at an
absolute pressure of 1.100×105
N/m2. The horizontal outlet of the
pipe has crosssectional areas of
0.0300 m2 and 0.0100 m2 at the
larger and smaller sections. 1
3
2.5m
h
B 2 a) What is the velocity of the fluid as it leaves the outlet into the
atmosphere?
Answer
Use Bernoulli’s equation at point 1 and 2 as seen in the diagram.
For a large tank, v1≈0. Define h2=0.
1
1
2
2
P1 + ρv1 + ρgh1 = P2 + ρv 2 + ρgh2
2
2
2
v2 =
( P1 − P2 + ρgh1 ) ρ = 2m 3
(1.1 × 10 5 N / m 2 − 1.013 × 10 5 N / m 2 + 10 3 kg / m 3 × 9.81N / kg × 2.5m)
3
10 kg Page 5 of 16 © Prep101 http://www.prep101.com/freestuff = 66.45m 2 / s 2 = 8.15m / s b) What is the velocity of the fluid at the point B, as shown in
the figure?
Answer
A2 v 2 = AB v B vB = v1 A1 8.15m / s × 0.01m 2
=
= 2.7m / s
AB
0.03m 2 c) To what height h does water stand in the open ended pipe?
Answer
Use Bernoulli’s equation between points 1 and B. Again, use v1≈0 for a
large tank and define hB=0.
1
1
2
2
P1 + ρgh1 = PB + ρv 2
PB = P1 + ρgh1 − ρv 2
2
2
1
PB = 1.1 × 10 5 N / m 2 + 10 3 kg / m 3 × 9.81N / kg × 2.5m − × 10 3 kg / m 3 × (2.7m / s ) 2
2
2
PB=130880 N/m
The height h is only affected by the static pressure at point B, so
define P3=101.3kPa (atmospheric pressure) and since
P − P3 (130880 − 101300) N / m 2
PB + ρghB = P3 + ρgh
=
= 3.015m
h= B
ρg
1000kg / m 3 × 9.81N / kg Page 6 of 16 © Prep101 http://www.prep101.com/freestuff Problem 4
A simple harmonic oscillator, consisting of a mass, m, on a frictionless
horizontal surface connected to a spring, has a period of 0.8s and an
amplitude of 8.0cm.
a) Write an equation for displacement, x, as a cosine function of
time if at time t=0s the mass is at x=0.30cm from the
equilibrium position.
Answer
We seek an equation of the form x(t)=Acos(ωt+φ).
Since f=1/T=1.25 s1, then ω=2πf=2.5π s1.
We can write an equation x(t)=0.08 m × cos(2.5π s1 × t +φ).
We are given x(0)=0.03m so:
x(0)=0.03=0.08×cosφ
cosφ=0.267
φ=105.5°
Therefore, the answer is: (t)=0.08 m × cos(2.5π s1 × t +105.5°). b) Draw a diagram of displacement as a function of time,
starting at time t=0 with a positive velocity.
Answer
In an x vs. t graph, there is positive velocity if v=dx/dt>0. Hence, we
want the slope of such a graph to be positive at t=0s. Also, the graph
should have amplitude A=8cm and period T=0.8s. Page 7 of 16 © Prep101 http://www.prep101.com/freestuff Distance x (m)
0.08 0.04 0.00 0.04 0.08
0.2 0.0
Time t (s) 0.2 0.4 0.6 c) It is found that a force of 5.0 N stretches the spring 3.0 cm
further than does a force of 3.0 N. Find the value of the mass
m.
Answer
Use the equation F=kx. Let x1 and x2 be the distances that the mass
has been stretched. From the information given, we know 3N=kx1,
5N=kx2 and x2x1=3cm=0.03m. So:
5N3N=kx2kx1
2N=k(x2x1)=k(3 cm)
2N
k=
= 666.7 N / m
0.03m
k
666.7 N / m
m= 2 =
= 10.81kg
ω
(2.5π s −1 ) 2 d) Find an expression for the acceleration, a, in terms of the
angular velocity, ω, when the displacement is 4.0 cm.
Answer
d 2x d 2
a = 2 = 2 [ A cos(ωt + φ )] = − Aω 2 cos(ωt + φ ) = −ω 2 x
dt
dt Page 8 of 16 http://www.prep101.com/freestuff © Prep101 When the displacemement is x=4 cm=0.04m, the acceleration is:
A=0.04ω2. e) At what displacement is the potential energy equal to the
kinetic energy?
Answer 12
1
mv = U = kx 2 . Usinf the general form
2
2
of the harmonic oscillator equation x(t)=Acos(ωt+φ), one can find:
1
1
V=dx/dt=ωAsin(ωt+φ)
E k = mv 2 = mω 2 A 2 sin 2 (ωt + φ )
2
2
A judicious substitutions of sin2θ=1cos2θ and m=k/ω2 gives us:
1⎛ k ⎞
1
E k = ⎜ 2 ⎟ω 2 A 2 1 − cos 2 (ωt + φ ) = k A 2 − x 2
2⎝ω ⎠
2
Substituting into the equation Ek=U gives:
1
1
12
E k = k A 2 − x 2 = U = kx 2
kA = kx 2
2
2
2
2
2
A
(0.08m)
x=
=
= 0.056568m
2
2
We must find x such that E k = [ [ [ Page 9 of 16 © Prep101 http://www.prep101.com/freestuff Problem 5
I. The graph shows a plot at t=0.50s of the displacement of a 3.0 Hz
wave traveling in the –x direction along a string. 8.00 B 6.00
4.00
2.00 D (cm) A 0.00
0.05 0.10 0.15 0.20 0.25 2.00
4.00
6.00
8.00 x (m) a) What is the speed of the wave?
Answer
We are given the frequency f=3Hz. To get the wavelength λ, we can
measure the distance between two peaks on the graph. We get
λ=0.165m0.065m=0.1m. Thus, v=λf=0.1m×3Hz=0.3m/s. Page 10 of 16 © Prep101 http://www.prep101.com/freestuff b) Write an equation describing the wave with all the constants
evaluated.
Answer
We can fit the equation to a function of the form:
D(x,t)=Asin(kx+ωt+φ).
We use this form instead of D(x,t)=Asin(kxωt+φ) because the wave is
moving in the –x direction. A cosine function can be done too.
The amplitude A can be read from the graph: A=0.08m.
We also have:
2π
2π
k=
=
= 20πm −1 ; ω = 2πf = 2π × 3Hz = 6π Hz
λ 0.1m
D( x, t ) = 0.08m × sin(20π × x + 6π × t + φ ) .
We know that D(x,t)=A at x=0.065m and t=0.1s, so sin(kx+ωt+φ)=1
kx+ωt+φ=π/2+nπ, where n is an integer. You can use any
number n. Let’s arbitrarily pick n=2 to get a nice small number for φ.
Thus:
π
5π
φ = + 2π − kx − ωt =
− 20π × 0.065 − 6π × 0.1 = −1.8π = −5.65 rad (= −324°)
2
2
Therefore D( x, t ) = 0.08m × sin(20π x + 10π t − 1.8π ) c) What is the transverse speed of the string a t=0.10s for
points A and B?
Answer
Since B is at maximum of the displacement, vB=0.
Since A is at equilibrium position, va=vmax=Aω=0.08m×6π s1
va=1.5m/s. Page 11 of 16 © Prep101 http://www.prep101.com/freestuff d) On the graph, sketch the displacement at t=0.
Answer
Substitutiong t=0 in our equation from b) is like shifting the original
graph by a phase of –3π radians.
To see this, note that at t=0.1s, ωt=2π×3Hz×0.1s=3π. With t=0, we
must remove this phase contribution due to time by subtracting π from
the phase. This has the effect of shifting the graph half a wavelength
to the right. e) Write the equation of the wave that, when added, will
produce a standing wave.
Answer
D( x, t ) = 0.08m × sin(−20π x + 6π t − 1.8π ) will work.
So will D( x, t ) = 0.08m × sin( 20π x − 6πt − 1.8π ) .
Basically, any sinusoidal wave of form
D( x, t ) = 0.08m × sin(±20π x ± 10πt − 1.8π ) will work. Page 12 of 16 © Prep101 http://www.prep101.com/freestuff II. A jet plane emits 8.0×105 J of sound energy per second.
a) What is the sound level in decibels 40.0m away?
Answer
P
P
8 × 10 5 W
I= =
=
= 39.79W / m 2
2
2
A 4πr
4π × (40.0m) β (dB) = 10 log10 39.79W / m 2
dB = 135.99dB
10 −12 b) Air absorbs sound at a rate of 7.0 dB/km. What will the sound
level 1.5 km away from the plane be, taking into account air
absorption?
Answer
Let position A be at 10m away from the plane. Let position B be at
1000 m away from the plane. Without considering the air absorption,
r
β 1 − β 2 = 20 log10 2 dB = 31.48dB . Now, also taking absorption into account,
r1
β 2 = β1 − 30.46dB − β absorbed = 135.99 − 31.48 − 7 × 1.5 = 94dB Page 13 of 16 © Prep101 http://www.prep101.com/freestuff Problem 6
a) A steel tape measure gives the length of a brass rod to be 300.00
cm when both the tape measure and the rod are at 25°C. What would
the tape measure read when both are at 60°C?
[αbrass=19×106 K1 ; αsteel=11×106 K1 ] Answer
The temperature difference is ∆T=60°C25°C=35°C=35 K (since it’s a
difference of temperature, ∆T is the same in °C or in K).
The steel tape measure will expend by:
∆Lsteel = L0 _ steel α steel ∆T = 300cm × 11 × 10 −6 K −1 × 35 = 0.1155cm
The brass rod will expend by:
∆Lsteel = L0 _ steel α steel ∆T = 300cm × 19 × 10 −6 K −1 × 35 = 0.1995cm
The tape is therefore shorter than the rob by 0.084cm. The tape will
therefore measure that the rod is 300.084 cm at 40°C. Page 14 of 16 © Prep101 http://www.prep101.com/freestuff b) The average rate at which heat flows through Mars’ surface is 70
W/m2 and the average thermal conductivity of Mars’ crust is k=4.1
W/m/K.
What is the temperature at the base of the crust if its thickness is
20km and the surface temperature is 210°C?
Answer
The heat flow for conduction is given by: H = kA ∆T
L Therefore ∆T = Tbase − Tsurface = LH L
20km
= × heat flow =
× 70W / m 2 = 341.5K (= 341.5°C )
kA k
4.1 W/m/K H 1 dQ
=
= heat flow through a surface
A A dt
(since it’s a difference of temperature, ∆T is the same in °C or in K)
Since Tsurface=210°C, we have:
Tbase=Tsurface+∆T=210°C+341.5°C=551.5°C
where Page 15 of 16 © Prep101 http://www.prep101.com/freestuff c) How long would it take in space to turn a 1.4 kg sphere of water
with surface area 0.06 m2 into ice in an environment with temperature
Text=5K? (assume ε=1.0, Lsol_w=3.33 and that the sphere is “floating”
in space)
Answer
The change of temperature of the ice will come from radiation. The
heat current between the ice (Tice=0°C) and the outside is therefore:
dQ
H=
= εσAT 4 = 1 × 5.67 × 10 −8 W / m 2 / K 4 × 0.06m 2 × 273 4 = 18.9 W
dt
To become ice, the water needs to “loose”:
Q = mL = 1.4kg × 3.33 × 10 5 J / kg = 466,200 J
It will therefore take:
466,200 J
Q
t=
=
= 24,666.67 s = 6h 51 min 6.7 s
dQ / dt 18.9 J / s Page 16 of 16 ...
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This note was uploaded on 01/06/2011 for the course PHYS 101 taught by Professor Bates during the Fall '08 term at The University of British Columbia.
 Fall '08
 BATES
 Physics

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