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Waves
y(x, t) = Acos(kx – ωt) | A = amplitude| k = constant | ω = angular velocityt) | A = amplitude| k = constant | ωt) | A = amplitude| k = constant | ω = angular velocity = angular velocity
ωt) | A = amplitude| k = constant | ω = angular velocity =
2 π
T
| T = Period
T =
1
f
| f = frequency
ωt) | A = amplitude| k = constant | ω = angular velocity = 2 πf
k =
2 π
λ
v = λf
| λ = wavelength
v=
T
μ
| T = Tension |
μ
= Linear Density (
mass
length
)
T = v
2
μ | T = Tension
Wave Speed - Use eqtns. and algebra
Transverse Velocity – Use derivative of wave eqtn.
f beat
=f 2
-f 1
Doppler Effect – Source Receding
o f observed
=
v + v
observer
v + v
source
∗ f
source
Doppler Effect – Source Approaching
o f observed
=
v + v
observer
v − v
source
∗ f
source
In Doppler Effect, v = c = speed of light = 343 m/s
Open Pipes, Closed Pipes, and Strings Tied at Both Ends
o n = 1, 2, 3, etc.
o f =
nv
2 L
| L = length of pipe
Open-Closed Pipes and Strings Tied at One End
o n = 1, 3, 5, etc.
o f =
nv
4 L
| L = length of pipe
Sound Intensity
o I = 10 ∗log
I
I
0
| I = Intensity that can be detected
o
I ∝
I
0
r
2
I 0
= Intensity that is made
Electricity
Electric field points in direction that protons move
They point towards electrons, away from protons
k = Coulomb’s constant =
1
4 π ϵ
0
F =
Gm
1
m
2
r
2
| G = 6.011 x 10
E ∗ dA =¿
q
enc
ϵ
0
E =
Gm
r
2
| For point charge
E =
λ
2 π ϵ
0
r
| For infinite line | λ=linear charge density
λ=
charge
length
E =
σ
2 ϵ
0
| For infinite sheet
σ
= charge density =
Q
A
Flux = φ = EA(cos(θ)) = )) =
E ∗ dA
| THERE IS NOT FLUX INSIDE A CHARGED
SHAPE
φtotal =
Q
ϵ
0
Problems
One of the harmonics of a column of air open at one end and closed at the other has a
frequency of 448 Hz and the next higher resonance
has a frequency of 576 Hz. What is the
fundamental frequency of the air column?
Since it is CLOSED at one end…
n = 1, 3, 5, etc. and f =
nv
4 L
SO
f=
Δ nv
4 L
=
2 v
4 L
=
v
2 L
=
Δ f
2
=
128
2
= 64 Hz.
Three point charges q, q and –q are placed at the corners of a square of side d as shown.
What is the magnitude of the net electric field at the center of the square?
q
First, find the distance to the center. If the side is length d,
q -q r
2
= (
1
2
d)
2
1
2
d)
2
| r =
1
2
d
2
=
d
The bottom left field is: E =
kq
r
2
=
2 kq
d
2
The bottom right and top left field work together to create: E = 2
kq
r
2
=
4 kq
d
2
To find the net electric field, we need to get the magnitude by using Pyth. Theorem:
E =
√
(
2 kq
d
2
)
2
+¿ ¿
=
√
4 k
2
q
2
d
4
16 k
2
q
2
d
4
=
20 ∗ kq
d
2
The figure shows two tiny 5.0 gram spheres suspended from
two very thin 1.0 m long threads. The spheres repel each other
after being charged and hang at rest as shown at angle
θ=0.006π rad. What is the charge on each sphere?
F
down
= mg =(.005 kg ) ( 9.8)=.049 N
F
side
= F
side
∗tan( .006 π )=9.24∗ 10
− 4
N
r = (1m)sin(.006π)2 )
F
side
=
kqq
r
2
=
k q
2
r
2
=
( 9 ∗ 10
9
)( q )
2
.
2
q =
√
F
side
∗ r
2
k
=
√
9.24∗ 10
− 4
∗.
2
9 ∗ 10
9
= 12 nC
A line of uniform charge is extended from x = -5.0m to x = -4.0m, and has a linear charge
density of 7.0 x 10
-
C/m. What is the magnitude of the resulting electric field at x = 0m?
E =
kdq
r
= kλ
x 1
x 2
dr
r
2
= kλ
(
1
x
1
−
1
x
2
)
9
− 6
(
1
4
−
1
5
)
A traveling wave on a horizontal vibrating string has wavelength 2.0m and vertical
amplitude 5.0cm. The string has a wave speed of 5.0m/s. A light insect with mass 6.0mg
holds tight to the top side of the string. What is the maximal normal force on the insect
exerted upwards by the string (do not ignore gravity)?
A = .05m | m = .006g | λ = 2m | v = 5m/s
F = ma = F N
d
2
y
d t
2
=− ω
2
Asin ( kx − ωt )
a
max
= ω
2
A =( 2 πf )
2
∗ A =
(
2 π
v
λ
)
2
∗ A
F = m
( a + g
) =.
(
4 π
2
∗ 5
2
4
∗
( .
)
)
=.13 N
Raptors, having a much more advanced ability to hear, can detect sound with an
intensity level as low as -4.0 dB. When you walk, the resulting sound has an intensity
level of 40 dB at a distance of 0.5 m. How close to the raptor can you walk?
10 log
I
1
I
2
=− 40 − 4 =− 44
| log
I
1
I
2
=−4.
Then raise both sides to the 10
th
power and make ratio of distances:
I
1
I
2
= 10
−4.
=(
R
2
R
1
)
2
R 2
= .5m |
.
R
1
= √
10
−4.
|
R
1
=
.
√
10
−4.
= 79 m
A standard demonstration charges a conducting sphere with small aluminum pie plates
on top… Approximately what charge has to be on the sphere for the last one to fly off?
Assume the pie plates had a radius of 5 cm and mass of 1 g, while the sphere has a
radius of 15 cm.
m = .001kg | R = .15m | r = .05m | F = qE
| q =
π r
2
4 π R
2
∗ Q
E =
kQ
R
2
| mg = F = qE =(
r
2
4 R
2
)( k
Q
2
R
2
)
|
Q =
√
4 mg R
4
k ∗ r
2
= 1 ∗ 10
− 6
C
Two 10cm x 10cm conducting flat plates have a total charge of +1 μC and -1 μC C and -1 μC and -1 μC C
respectively. When they are very close to each other, separated by just an insulating
oxide layer about 100 nm thick, what is the force pulling them together?
For Plates: E =
σ
2 ϵ
0
| σ =
10
− 6
( The charge )
10
− 2
( The area )
= 10
− 4
F = qE =
q ∗ σ
2 ϵ
0
=
10
− 6
10
− 4
2
8.8∗ 10
− 12
=5.7 N
A long cylindrical insulator has a uniform charge density, p, throughout its volume. What
is the electric field as a function of r, the radial distance away from the center (r<1)?
E ∗ dA =
q
enc
ϵ
0
| E
( 2 πrl
)
π r
2
lp
ϵ
0
|
E =
pr
2 ϵ
0
Two waves are present on the same string: y 1
= cos(2x + 3t) and y 2
= cos(2x – 3t). What is
the distance between adjacent nodes on the string?
k =
2 π
λ
= 2 so λ = π
Therefore, since 1 wavelength includes 2 nodes, the distance between nodes is
λ
2
=
π
2
.
A conducting sphere has a charge +2Q on it, and is located inside
a conducting shell with net charge +Q on it. Rank the outgoing
electric flux from largest to smallest through centered spherical
surfaces: A) 0 < r < R1 B) R1 < r < R2 C) R2 < r < R3 and D) r > R
E ∗ dA =
q
enc
ϵ
0
= Flux
Flux(A) = 0; Flux(B) =
2 Q
ϵ
0
; Flux(C) = 0; Flux(D) =
3 Q
ϵ
0
; NOTE: NO FLUX IN A CHARGE
OBJECT
A charge Q and a charge -2Q are located at A and B along a straight line. Which graph
correctly shows the magnitude of EX(x) along this line?
The answer is A because only A
and C represent two different
charges and only the charge at a large positive x must be negative (like in A).
A line of uniform charge exists between x = 0.70 and x = 1.00, and has a linear charge
density of 7.0 μC and -1 μC C/m. What is the magnitude of the resulting electric field at x = 0.00?
E = k
dq
r
2
= k
λdr
r
2
=− kλ
(
1
1
−
1
)
=
9 ∗ 10
9
7 ∗ 10
− 6
(
1
A sphere has a volume charge density given by p(r) =
D
r
for r < R. What is the resulting
electric field for r < R?
E ∗ dr =
q
enc
ϵ
0
|
E
4 π r
2
=
1
ϵ
0
0
r
p
4 π r
2
dr
E r
2
=
D
ϵ
0
0
r
r ∗ dr =
D
ϵ
0
∗ r
2
2
so dividing by r
2
…
E =
D
2 ϵ
0
A point charge, Q, is located a distance, d, from the center of a circle
of radius d, along the axis of the circle. What is the electric flux to the
right through the plane bounded by this circle, due to the charge?
Flux
total
=
Q
ϵ
0
Since the shape is a ring:
2 π
0
π
4
sin ( θ ) dθ
4 π
=
−cos
(
π
4
)
−(−cos ( 0 ) )
2
=
1
2
−
2
4
=.
Q
ϵ
0
If you are 4 meters from a spherical sound source, how much further must you go to
change the sound level by -6 dB?
− 6 = 10 log
4
2
( 4 + x )
2
10
−0.
=(
4
4 + x
)
2
4 + x =
4
√
10
−0.
x =3.98 m
A sound wave is traveling to the right, and the displacement of air molecules is show in
the graph. Rank the delta-pressure at points A, B, C, and D from most pos to most neg.
Put any last minute notes below:
