PHY 113 A General Physics I
9-9:50 AM MWF Olin 101
Plan for Lecture 36:
Review – Part I
1. General advice about how to study
2. Some comments on sound and waves
3. Review of past exam questions
Review – Part II (Friday)
1. Systematic review of PHY 113 topics
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Comments on final exam for PHY 113
Date: Thursday, Dec. 13, 2012 at 9 AM
Place: Olin 101
Format: Similar to previous exams; covering material
in Lectures 1-37, Chapters 1-22 (no time pressure)
Focus: Basic physics concepts; problem-solving
techniques
Bring:
1. Clear head
2. Calculator
3. Pencils, pens
4. Up to 4 equation sheets
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iclicker question:
What is the purpose of the final exam in PHY 113
A. No purpose – just pain and suffering
B. To improve my grade in the course
C. It is a college tradition that must be
maintained
D. To check that I have actually learned the
material
E. To encourage students to review the course
material and solidify my learning
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Comments on waves and sound
1. Standing wave resonances for strings or
pipes
2. Relationship between wave speed,
frequency, and wavelength
3. Doppler effect
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Formation of standing waves; beautiful trigonometric
identity:
1
1
sin A  sin B  2 sin
 2π

yright ( x, t )  y0 sin   x  ct 
 λ

“Standing” wave:
2
 A  B cos 2  A  B 
 2π

yleft ( x, t )  y0 sin   x  ct 
 λ

(lf=c)
 2 πx 
yright ( x, t )  y left ( x, t )  2 y0 sin 
 cos2 πft 
 λ 
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Possible spatial shapes for transverse string wave :
 n

y(x,0 )  A sin 
2πx 
 2L

n  1,2,3,4.....
 2πx 
Standing wave form : y(x,t)  A sin 
 cos2πft 
 λ 
2L
nc
 ln 
fn 
n  1,2,3,4...........
n
2L
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Musical scale (chromatic)
For standing waves on a string :
nc
fn 
n  1,2,3,4...........
2L
Example : for A : f1  440 Hz on a 0.5 m string,
must set c  440 m/s (by adjusting tension)
f 2  880 Hz (n  2 harmonic)
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Standing waves in air c=343 m/s:
For open - open standing waves :
nc
fn 
n  1,2,3,4...........
2L
For L  0.39 m
A : f1  440 Hz
f 2  880 Hz
For open - closed standing waves :
nc
fn 
n  1,3,5...........
4L
For L  0.39 m
A : f1  220 Hz
f 3  660 Hz
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The “Doppler” effect
v=sound velocity
observer stationary, source moving
vt1  d
vt 2  T   vS T  d
v  vS
1
t 2  t1 
T
fO
v
fO  f S
d
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Summary :
v  vO
fO  f S
v  vS
PHY 113 A Fall 2012 -- Lecture 36
v
v  vS
toward
away
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Summary of sound Doppler effect :
v  vO
fO  f S
v  vS
toward
away
Doppler effect for electromagnetic waves :
fO  f S
v  vR
v  vR
Relative velocity of source
toward observer
Example : f S  440 Hz and suppose vS  0 and vO  vR  30m/s
For sound v  343m/s
For radar v  3 108 m/s
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f O  440 /(1  30 / 343) Hz  482 Hz
1  30 / 3 108
-5
f O  f S  440
Hz

4.4

10
Hz
8
1  30 / 3 10
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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
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Doppler effect for electromagnetic waves :
fO  f S
v  vR
v  vR
Relative velocity of source
toward observer
Typical radar frequencies : f S  20 109 Hz; suppose vR  30m/s
For radar v  3 108 m/s
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f O  f S  20 109
1  30 / 3 108
3
Hz

2

10
Hz
8
1  30 / 3 10
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Review questions from Exam 4:
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Review questions from Exam 4:
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Review questions from Exam 4:
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Review questions from Exam 4:
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Review questions from Exam 3:
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Review questions from Exam 3:
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Review questions from Exam 2:
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Review questions from Exam 2:
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Review questions from Exam 2:
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