Physics 207: Lecture 30, Pg 1

Lecture 30Goals:Goals:

• Wrap up chapter 20. Wrap up chapter 20.

• Review for the final.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 2

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 3

The Doppler effect

f=f0/(1-vs/v)

Approaching source:

Receding source:

f=f0/(1+vs/v)

Physics 207: Lecture 30, Pg 4

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.

Waves

Physics 207: Lecture 30, Pg 5

History Graph:

2 3 64 7time (sec)

2

-25

D(x = −2 m, t)

Physics 207: Lecture 30, Pg 6

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 7

R

Psource=35 WR=50 m

The power hitting the microphone is:

Pmicrophone= I Amicrophone

Physics 207: Lecture 30, Pg 8

Intensity of sounds

If we were asked to calculate the intensity level in decibels:

I0: threshold of human hearing

I0=10-12 W/m2

Physics 207: Lecture 30, Pg 9

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?

Physics 207: Lecture 30, Pg 10

Engines

For the engine shown below, find, Wout, QH and the thermal efficiency. Assume ideal monatomic gas.

Pi

4Pi

Vi 2Vi

Q=90J

Q=25J

Physics 207: Lecture 30, Pg 11

First, use the ideal gas law to find temperatures

Pi

4Pi

Vi2Vi

Q=90J

Q=25JTi

4Ti8Ti

2Ti

From the right branch, we have:

nCVΔT=90 J

n(3R/2)6Ti=90JnRTi=10J

Physics 207: Lecture 30, Pg 12

Work output is the area enclosed by the curve:

Pi

4Pi

Vi2Vi

Q=90J

Q=25JTi

4Ti8Ti

2Ti

Wout=area=3PiVi=3nRTi=30J

Physics 207: Lecture 30, Pg 13

From energy conservation:

Wout=QH-QC

Wout=30JQC=115J

η=0.2

The thermal efficiency is:

QH=145J

Physics 207: Lecture 30, Pg 14

The Carnot Engine

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.

Carnot showed that the Carnot showed that the thermal efficiency of a thermal efficiency of a Carnot engine is:Carnot engine is:

hot

coldcycleCarnot T

T1

Physics 207: Lecture 30, Pg 15

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 16

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 17

N/V: particles per unit volume

Mean free path is the average distance particle moves between collisions:

The average translational kinetic energy is:

εavg=(1/2) mvrms2=(3/2) kBT

CV=3R/2 (monatomic gas)

The specific heat for a monatomic gas is:

Physics 207: Lecture 30, Pg 18

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?

km

km

-2 20(≡Xeq)

2m

2m

Physics 207: Lecture 30, Pg 19

½ k A2=½ m vmax2

vmax=ωA

The speed when the mass reaches the equilibrium position:

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 Anew

2

Anew=A/√3

Physics 207: Lecture 30, Pg 20

Fluids

What happens with two fluids?

Consider a U tube containing liquids of density 1 and 2 as shown:

Compare the densities of the liquids:

(A) 1 < 2 (B) 1 = 2 (C) 1 > 2

1

2

Physics 207: Lecture 30, Pg 21

Fluids

What happens with two fluids??

Consider a U tube containing liquids of density 1 and 2 as shown:

At the red arrow the pressure must be the same on either side. 1 x = 2 y

Compare the densities of the liquids:

(A) 1 < 2 (B) 1 = 2 (C) 1 > 2

1

2

x

y