PHY 113 A General Physics I 9-9:50 AM MWF Olin 101 Plan for Lecture 36: Review – Part I

PHY 113 A Fall 2012 -- Lecture 36 112/05/2012

PHY 113 A General Physics I9-9:50 AM MWF Olin 101

Plan for Lecture 36:Review – Part I

1. General advice about how to study2. Some comments on sound and waves3. Review of past exam questions

Review – Part II (Friday)4. Systematic review of PHY 113 topics



Comments on final exam for PHY 113

Date: Thursday, Dec. 13, 2012 at 9 AMPlace: Olin 101Format: Similar to previous exams; covering material in Lectures 1-37, Chapters 1-22 (no time pressure)Focus: Basic physics concepts; problem-solving techniquesBring:

1. Clear head2. Calculator3. Pencils, pens4. Up to 4 equation sheets


iclicker question:What is the purpose of the final exam in PHY 113

A. No purpose – just pain and sufferingB. To improve my grade in the courseC. It is a college tradition that must be

maintainedD. To check that I have actually learned the

materialE. To encourage students to review the course

material and solidify my learning


Comments on waves and sound1. Standing wave resonances for strings or

pipes2. Relationship between wave speed,

frequency, and wavelength3. Doppler effect


ctxytxyctxytxy leftright λ

π2sin),( λπ2sin),( 00

ftxytxtxy leftright π2cosλπ2sin2),(y),( 0

“Standing” wave: (lf=c)

Formation of standing waves; beautiful trigonometric identity: BABABA 2

12

1 cossin2sinsin


.....4,3,2,1 22

sin0

: wavestring ersefor transv shapes spatial Possible

nπxLnA)y(x,

...........4,3,2,1 2

2

2cos2sin :form waveStanding

nLncf

nL

πftλπxAy(x,t)

nnl


Musical scale (chromatic)

harmonic) 2( Hz 880 tension)adjusting(by m/s 440set must string, m 0.5 aon Hz 440 :Afor :Example

...........4,3,2,1 2

:string aon wavesstandingFor

2

1

nfcf

nLncfn


Standing waves in air c=343 m/s:

Hz 660 Hz 220 :A

m 39.0For

...........5,3,1 4

: wavesstanding closed-openFor

Hz 880 Hz 440 :A

m 39.0For

...........4,3,2,1 2

: wavesstandingopen -openFor

3

1

2

1

ffL

nLncf

ffL

nLncf

n

n


The “Doppler” effect v=sound velocity

observer stationary, source moving

vvvT

ftt

dTvTtvdvt

S

O

S

112

2

1

d

SSO vvvff

toward

awayS

OSO vv

vvff

:Summary


S

OSO vv

vvff

:effectDoppler sound ofSummary toward

away

R

RSO vv

vvff

: wavesneticelectromagfor effect Doppler

Relative velocity of source toward observer

Hz 104.4 Hz103/301103/301440 m/s103radar For

Hz 482 Hz )343/301/(440 m/s343 soundFor m/s30 and 0 suppose and Hz 440 :Example

5-8

88

SO

O

ROSS

ffv

fvvvvf


iclicker question:The previous calculation for “radar” Doppler was:

A. Encouraging me to speed because it is impossible to detect such a small frequency difference

B. Full of admiration that Doppler radar equipment can detect such a small frequency difference

C. Not relevant to actual “radar” Doppler -- still need to be careful not to speed

iclicker question:The fallacy in the previous analysis was

A. Incorrect value of fs

B. Incorrect value of v (speed of light)C. Calculator error


R

RSO vv

vvff

: wavesneticelectromagfor effect Doppler

Relative velocity of source toward observer

Hz 102 Hz103/301103/3011020 m/s103radar For

m/s30 suppose Hz; 1020 :sfrequencieradar Typical

38

898

9

SO

RS

ffv

vf




Review questions from Exam 4:

















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PHY 113 A General Physics I 9-9:50 AM MWF Olin 101 Plan for Lecture 36: Review – Part I