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Physics 2020
Spring 2009
Stephan LeBohec
COVER SHEET - EXAM 2
SEAT #
Constants:
k=
1
9
2
−2
=8.99×10 N⋅m ⋅C
4  0
−19
e=1.602×10
Waves:
=c⋅T =
c
1
as f =
f
T
In a gas c Sound =
Doppler effect: f o= f s
c String =

Q Proton =e & Q Electron=−e
C

m
F
with =
L

5
P
with = for a mono­atomic ideal gas
3

c±v o
c∓v s
Circular opening diffraction angle
sin =1.22 / D
Standing wave frequency (string or open pipe)
c
f n=n
2L
Electric force and field:
Force on a point charge in an electric field:
 =q E

F
Standing wave frequency in pipe open at one end
c
f n=n
4L
Electric field generated by a point charge:
 ∣=k q
∣E
r2
Force between two point charges:
q q
 ∣=k 1 2
∣F
2
r
Triangle geometry:
a
(ABD) and (ECD), similar triangles
AB DB AD
=
=
EC DC ED
c
µ
E
b
c 2=a 2b2
cos =b/ c
sin =a /c
tan =a/b
A
D
C
B
Physics 2020
Spring 2009
Stephan LeBohec
Name:_____________________________________
TA (circle one): Michael
A)
1
EXAM 2
Sarah
Student ID #:___________________________
Adam
Isaac
[36 points, 3 points per question] For each statement, circle the option you find appropriate.
You do not need to show your work.
1) Two sound waves with the same frequency are propagating in the same direction in a pipe. There might be
BEATS
A STANDING WAVE
INTERFERENCE
3) Two sound waves with different frequencies are propagating in opposite directions in a pipe. There might be
BEATS
A STANDING WAVE
4) Two sound waves with the same frequency are propagating in opposite directions in a pipe. There might be
BEATS
A STANDING WAVE
5) A wave described by y  x , t = A sin 2 f t 2  x / propagates toward
INCREASING X
DECREASING X
6) Two harmonic waves with the same frequency propagate on a string in the same direction. One has an
amplitude A1=2mm and the other A2=3mm . According to the superposition principle, the resulting
wave may have a maximal amplitude of
1mm
2mm
7) and a minimal amplitude of
1mm
2mm
3mm
4mm
5mm
6mm
3mm
4mm
5mm
6mm
8) A taut string has a fundamental frequency
25Hz
50Hz
75Hz
f 1 =25Hz . A standing wave with 4 anti-nodes has a frequency:
100Hz
125Hz
150Hz
150Hz
°
9) A loud speaker produces a f 1 =440Hz acoustic wave with a 25 diffraction angle. When the same
speaker produces a f 1 =880Hz acoustic wave the diffraction angle is
°
°
SMALLER THAN 25
LARGER THAN 25
10) A source moves toward and observer and the observer moves away from the source at the same
speed. Compared to the source frequency, the observed frequency is
SMALLER
EQUAL
GREATER
11) Two loud speakers are working in phase, producing waves with a wavelength of 1.44m. A listener is standing
half way between the two speakers. In order to get to a destructive interference point, he needs to move toward
one of the speakers by a distance of
0.18m
0.36m
0.72m
1.44m
2.88m
12) In order to double the fundamental frequency of a taut string, the tension must be multiplied by
1/ 2
1/  2
2
2
4
Physics 2020
Spring 2009
Stephan LeBohec
Name:_____________________________________
TA (circle one): Michael
B)
2
EXAM 2
Sarah
Adam
Student ID #:___________________________
Isaac
[15 points] A 0.80m long string under a 300N tension is observed to carry a standing wave with 5 antinodes and a 220Hz frequency . What is the mass of the string?
Since there are 5 anti-nodes, the standing wave is the harmonic n=5 for which
f n=n

c
with
2L
L=0.8m . With
T
where m is the mass of the string and m/ L the linear mass density. In total we
m/ L
T n 2
n T⋅L
have f n=
in which only m is unknown. Solving for m we get m= 
 . Numerically,
L 2f
2L m
2
300N
5
−2
m=

 =4.84×10 kg=48.4g . With N ≡kg⋅m⋅s−2 we check that the units workout.
0.8m 2×220Hz
T =300N we have c=

C)
[17 points] Two speakers are operating in phase at a frequency known to be between 100Hz and 160Hz .
A point located 25.8m from one speaker and 42.5m from the other is observed to be the site of destructive
interferences. The speed of sound is c=334m⋅s−1 What are the possible frequencies produced by the
speakers?
For the interference to be destructive, the difference ∣d 1−d 2∣=42.5m−25.8m=16.7m
between the distance to the
1
 so ∣d 1−d 2∣=n  with
2
−1
1 c
1
c
1 334m⋅s
1
=n 
=n 20Hz .
from which f = n 
n=0,1,2,3,⋯ or ∣d 1−d 2∣=n 
2 f
2 ∣d 1−d 2∣
2 16.7m
2
Staring with n=1 The frequencies are 30Hz, 50Hz, 70Hz, 90Hz, 110Hz, 130Hz, 150Hz, 170Hz, ⋯ . The only
frequencies in the range from 100Hz to 160Hz are 110Hz, 130Hz and 150Hz.
two speakers must be an multiple of the wavelength 
augmented by half
Physics 2020
Spring 2009
Stephan LeBohec
Name:_____________________________________
TA (circle one): Michael
D)
1)
3
EXAM 3
Sarah
Adam
Student ID #:___________________________
Isaac
[24 points] Two charges q 1=1.0 C , and q 2=−2.0 C
are positioned as indicated on the figure with w=0.60m and
h=0.80m . Calculate the magnitude and the direction (as
specified by the angle between the electric field and the x
direction) of the electric field at point M.
Staring with E1 ,
2
∣E1∣=k q 1 /d 1M
=9×10 9×1×10−6 /0.6 20.82 =0.9×104 N⋅C −1
w
E 1x =∣ E1∣
=0.9×10 4×0.6 /1.0=0.54×10 4 N⋅C−1
d 1M
h
E 1y=−∣E1∣
=−0.9×10 4×0.8 /1.0=−0.72×104 N⋅C −1
d 1M
E2
E 2x=−k q2 / w =−9×10 9×2×10−6 /0.62=−5.0×104 N⋅C −1
E 2y =0
E2
Θ
E1y E1
Then dealing with
2
EM
E1x
EM = E1 E2 so
E Mx =0.54×104 −5.0×104=−4.46×10 4 N⋅C−1
E My =−0.72×104 N⋅C −1
Then we have
and
∣EM∣= E 2Mx E 2My = 0.72×10 4 2 4.46×104 2=4.52×104 N⋅C −1
and
−1
−1
°
=−cos  E Mx /∣EM∣=−cos −4.46/ 4.52=−170
2)
[8 points] An electric charge q=−10 C is now placed at point M. What is the magnitude and
direction of the electric force acting on this charge?
 points
 ∣=∣q∣⋅∣E
M∣=10×10−6 ×4.52×104 =0.452N . Since q0 , F
∣F
°
°
°
making an angle ' =180=180 −170 =10 with the x direction.
The magnitude of the electric force is
in a direction opposite to
EM