Lecture 30
Goals:
• Wrap up chapter 20.
• Review for the final.
• Final exam on Friday, Dec 23, at 10:05 am in BASCOM 272.
•
Final will cover Chapters 1-20 (excluding Chapter 13).
Semi-cumulative.
HW 12 due tomorrow night.
Physics 207: Lecture 30, Pg 1
The Doppler effect
The frequency of the wave that is observed depends on the
relative speed between the observer and the source.
vs
observer
Physics 207: Lecture 30, Pg 2
The Doppler effect
Approaching source:
f=f0/(1-vs/v)
Receding source:
f=f0/(1+vs/v)
Physics 207: Lecture 30, Pg 3
Waves
The figure shows a snapshot graph D(x, t = 2 s) taken at
t = 2 s of a pulse traveling to the left along a string at a
speed of 2.0 m/s. Draw the history graph D(x = −2 m, t) of
the wave at the position x = −2 m.
Physics 207: Lecture 30, Pg 4
History Graph:
D(x = −2 m, t)
2
-2
2
3
4
6
5
time (sec)
7
Physics 207: Lecture 30, Pg 5
A concert loudspeaker emits 35 W of sound power. A small
microphone with an area of 1 cm2 is 50 m away from the
speaker.
What is the sound intensity at the position of the
microphone?
How much sound energy impinges on the microphone
each second?
Physics 207: Lecture 30, Pg 6
R
Psource
I=
2
4pR
Psource=35 W
R=50 m
The power hitting the microphone is:
Pmicrophone= I Amicrophone
Physics 207: Lecture 30, Pg 7
Intensity of sounds
If we were asked to calculate the intensity level in
decibels:
æIö
b = 10log10 ç ÷
è I0 ø
I0: threshold of human hearing
I0=10-12 W/m2
Physics 207: Lecture 30, Pg 8
Suppose that we measure intensity of a sound wave
at two places and found them to be different by 3 dB.
By which factor, do the intensities differ?
æ I1 ö
æ I2 ö
b1 = 10log10 ç ÷ b 2 = 10log10 ç ÷
è I0 ø
è I0 ø
æ I1 ö
b1 - b 2 = 10log10 ç ÷ = 3
è I2 ø
I1
0.3
= 10 = 2
I2
Physics 207: Lecture 30, Pg 9
Engines
For the engine shown below, find, Wout, QH and
the thermal efficiency. Assume ideal monatomic
gas.
4Pi
Q=90J
Pi
Q=25J
Vi
2Vi
Physics 207: Lecture 30, Pg 10
First, use the ideal gas law to find temperatures
4Pi
4Ti
8Ti
Q=90J
Pi
Ti
Vi
2Ti
Q=25J
2Vi
From the right branch, we have:
nCVΔT=90 J
n(3R/2)6Ti=90J
nRTi=10J
Physics 207: Lecture 30, Pg 11
Work output is the area enclosed by the curve:
4Pi
4Ti
8Ti
Q=90J
Pi
Ti
Vi
2Ti
Q=25J
2Vi
Wout=area=3PiVi=3nRTi=30J
Physics 207: Lecture 30, Pg 12
From energy conservation:
Wout=QH-QC
Wout=30J
QC=115J
QH=145J
The thermal efficiency is:
η=0.2
Physics 207: Lecture 30, Pg 13
The Carnot Engine
Carnot showed that the
thermal efficiency of a
Carnot engine is:
Tcold
Carnotcycle  1
Thot
All real engines are less efficient than the Carnot
engine because they operate irreversibly due to the
path and friction as they complete a cycle in a brief
time period.
Physics 207: Lecture 30, Pg 14
For which reservoir temperatures would you expect
to construct a more efficient engine?
A) Tcold=10o C, Thot=20o C
B) Tcold=10o C, Thot=800o C
C) Tcold=750o C, Thot=800o C
Physics 207: Lecture 30, Pg 15
Kinetic theory
A monatomic gas is compressed isothermally to 1/8 of its
original volume.
Do each of the following quantities change? If so, does the
quantity increase or decrease, and by what factor? If not, why
not?
a. The temperature
b. The rms speed vrms
c. The mean free path
d. The molar heat capacity CV
Physics 207: Lecture 30, Pg 16
The average translational kinetic energy is:
εavg=(1/2) mvrms2=(3/2) kBT
Mean free path is the average distance particle moves
between collisions:
1
l=
4 2p ( N /V )r 2
N/V: particles per unit volume
The specific heat for a monatomic gas is:
CV=3R/2 (monatomic gas)
Physics 207: Lecture 30, Pg 17
Simple Harmonic Motion
A Hooke’s Law spring is on a horizontal frictionless surface is
stretched 2.0 m from its equilibrium position. An object with
mass m is initially attached to the spring however, at
equilibrium position a lump of clay with mass 2m is dropped
onto the object. The clay sticks.
What is the new amplitude?
2m
k
m
2m
m
k
-2
0(≡Xeq)
2
Physics 207: Lecture 30, Pg 18
The speed when the mass reaches the equilibrium position:
½ k A2=½ m vmax2
vmax=ωA
The speed after clay sticks can be found using momentum
conservation:
m vmax=(m+2m)vnew
vnew=vmax/3
The new amplitude can be found using energy conservation:
½ (m+2m)vnew2=½ k Anew2
Anew=A/√3
Physics 207: Lecture 30, Pg 19
Fluids
What happens with two fluids?
Consider a U tube containing liquids of
density r1 and r2 as shown:
r2
r1
Compare the densities of the liquids:
(A) r1 < r2
(B) r1 = r2
(C) r1 > r2
Physics 207: Lecture 30, Pg 20
Fluids
What happens with two fluids??
Consider a U tube containing liquids of
density r1 and r2 as shown:
r2
x r1
At the red arrow the pressure must be the
same on either side. r1 x = r2 y
 Compare the densities of the liquids:
(A) r1 < r2
(B) r1 = r2
y
(C) r1 > r2
Physics 207: Lecture 30, Pg 21