REVIEW: Oscillations, Waves, Electric Force, Field, and Flux
Oscillations:
• Hooke’s Law:
F x
kx
(linear restoring force)
• Simple Harmonic Motion: d 2 x o Displacement obeys differential equation dt 2
= − ω 2 x o For mass and spring,
ω 2 = k m
• Solution of differential equation is: x
( )
= A cos
( ω t + φ ) o Important: argument
( ω t + φ )
must be in radians o Constants A ,
ω
, and
φ
must be found from details of motion

Relations between Position, Velocity, and Acceleration in SHM x x
( )
≤
=
A
A cos
(
ω t + φ
) v x
= dx dt
= − A ω sin
(slope of x vs t ), v x
(
ω t + φ
)
≤ A ω a x
= d 2 x dt 2
= − A ω 2 cos
(
ω t + φ
)
(slope of v x
vs t ), a x
≤ A ω 2

• Period: time in which phase increases by 2 π o
T =
2 π
ω
and f =
1
T
=
ω
2 π
• Energy for mass m on spring k so that ω 2 = k / m o x = A cos
( ω t + φ )
and v = − A ω sin
( ω t + φ ) o Kinetic Energy:
K =
1
2 mv 2 =
1
2 kA 2 sin 2
( ω t + φ ) o Potential Energy:
U =
1
2 kx 2 =
1
2 kA 2 cos 2
(
ω t + φ
) o Mechanical Energy:
E = K + U =
1
2 kA 2
( sin 2
( ω t + φ )
+ cos 2
( ω t + φ ) )
=
1
2 kA 2

• Learned how to get
( ω , φ , and A
) from description of specific motion o Angular frequency
( )
depends on properties of oscillator
• i.e.
ω = k / m for spring/mass system o Get Amplitude and phase constant from x and v at specific time (i.e. initial displacement and velocity)
•
A = x i
2 + v i
2
ω 2
• tan φ = − v i
ω x i

• Simple Pendulum d 2 θ o for small
θ
, dt 2
= − g
θ
L
ω 2 = g o Looks like SHM with
L
• Note, does not depend on m !
o Period is
T =
2 π
ω
= 2 π
L g
• Physical Pendulum d 2 θ o for small θ , dt 2
= − mgd
I
θ o Looks like SHM with
ω 2 = mgd
I

• Underdamped Oscillation: o x = Ae − bt / 2 m cos
( ω t + φ ) o
ω = ω
0
2 −


 b
2 m 

 2
• Resonance: o Periodic applied force o Damping force r
F
= − b v
= F
0 sin o Resulting oscillation is at driving frequency x = A
( ) ( ω t + φ )
ω t o Frequency dependent amplitude is
A
( )
=
(
ω 2 − ω
F
0
0
2
)
2
/ m
+
( b / 2 m
)
2

WAVES:
• Wavefunction: o o y y
( )
( )
=
= f f
(
( x x
−
+ vt vt
)
)
for wave going to right with speed v
for wave going to left with speed v
Sinusoidal wave: two ways to represent. o y vs t at fixed x
• SHM with T = 1 / f o y vs x at fixed t )
• separation between adjacent identical points on wave is wavelength ≡
λ
• wave travels distance λ in one period (time T ) so v =
λ
T
or v = f λ

EQUIVALENT ways to write wave function for traveling sinusoidal wave
• y x , = A sin


2 π
λ
( x − vt
) 

• y x , = A sin



2 π


λ T t






• y = A sin
( kx − ω t
) where k =
2
λ
π
and
Propagation speed of a wave on a string
• for string of tension T and mass/unit length
µ
ω = 2 π f = 2 π / T
wave propagates along string with speed v =
T
µ

Can relate mass/unit length to wire density and cross-sectional area
• for wire of cross-sectional area A and length L , volume is V = AL o if density is
ρ
, mass of segment is m = ρ V = ρ A L .
o So mass/unit length is
µ = m
L
= ρ A o If radius is r then A = π r 2
and
µ = ρ π r 2
Result: can write speed of wave along string as: v =
T
ρ π r 2
Energy transmitted by a wave on a string is
P =
1
2
µ v ω 2 A 2

SOUND WAVES:

INTERFERENCE:
Superposition of two waves differing only in phase:
• resultant wave function is y
R
( )
= y
1
( )
+ y
2
( )
= A
[ sin
( kx −
Spatial Interference:
• path difference:
∆ r = r
2
− r
1
• phase diff. at obs. is
∆ φ = kr
2
− kr
1
=
2 π
λ
∆ r
• CONSTRUCTIVE Interference for: o o
∆ φ =
∆ r = n
0
λ
, 2 π , 4 π , L = 2 n π
where n =
• DESTRUCTIVE Interference for: o o
∆
∆
φ r
=
=
π
( n
, 3
+
π
1
2
)
, 5
λ
π , L =
(
2 n + 1
)
π
where
0 , n
1 ,
=
2 ,
0 , etc.
1 ,
ω t
)
+ sin
( kx
2 , etc.
− ω t + φ ) ]

STANDING WAVES:
Superposition of two waves going in opposite directions:
•
• y y
2
1
= A sin
= A sin
(
( kx kx
−
+
ω
ω t t
)
)
goes in + x direction
goes in x direction
Resultant wave
• y
R
= y
1
+ y
2
= 2 A sin k x cos ω t o Each part of string oscillates with SHM and amplitude
• Nodes where sin k x = 0
2 A sin k x

String fixed at x = 0 and x = L has a series of allowed NORMAL MODES o all have same wave speed v = T / µ o frequencies set by allowed wavelengths: f=v / λ
L =
λ
1
2
⇒ λ
1
= 2 L ⇒ f
1
=
λ v
1
=
2 v
L
(fundamental)
L = λ
2
⇒ λ
2
= L ⇒ f
2
= v
λ
2
=
2 v
2 L
(2 nd harmonic)
L =
3 λ
3
2
⇒ λ
3
=
2 L
3
⇒ f
3
= v
λ
3
=
3 v
2 L
(3 rd harmonic)
• Normal mode frequencies for string fixed at both ends are: o f n
=
λ v n
= nv
2 L
where n = 1, 2, 3, etc.

LONGITUDINAL MODES OF A TUBE OPEN AT BOTH ENDS
• Boundary condition is a displacement antinode at each end
λ
1
= 2 L ⇒ f
1
=
λ v
1
=
2 v
L
(fundamental)
λ
2
= L ⇒ f
2
= v
λ
2
= v
L
= 2 f
1
(2 nd harmonic)
λ
3
=
2 L
3
⇒ f
3
= v
λ
3
=
3 v
2 L
= 3 f
1
(3 rd harmonic)
• For a tube open at both ends f n
= n v
2 L
where n = 1 , 2 , 3 , L o all harmonics possible

LONGITUDINAL MODES OF A TUBE WITH ONE OPEN & ONE CLOSED END
• Boundary condition is a displacement node at the closed end and a displacement antinode at the other end
λ
1
= 4 L ⇒ f
1
=
λ v
1
=
4 v
L
(fundamental)
λ
3
=
4 L
3
⇒ f
3
= v
λ
3
=
3 v
4 L
= 3 f
1
(3 rd harmonic)
λ
5
=
4 L
5
⇒ f
5
= v
λ
5
=
5 v
4 L
= 5 f
1
(5 th harmonic)
• For a tube with one open and one closed end, f n
= n v
4 L
where n = 1 , 3 , 5 , L o Only ODD harmonics possible o Fundamental frequency half that for same length tube, both ends open

BEATS:
Resultant of superposing two waves with different frequencies is y
R
= 2 A cos



2 π 
 2 f
2


 t


 cos



2 π 

1
2 f
2


 t



Beat frequency is f beat
= f
1
− f
2

NON-SINUSOIDAL WAVES → COMPLEX WAVES
• tone from a tuning fork (or recorder) is mostly fundamental → sinusoidal
• can get non-sinusoidal waves (more complex sounds) by adding harmonics
EXAMPLE: Can make a square wave by adding odd harmonics.
• y = ∑ n
1 n sin
(
2 π nf
1 t
)
• This is a Fourier series o Get spectrum
(intensity vs harmonic number) by Fourier analysis
• Character of sound depends on harmonics o Trumpet, clarinet, etc. act like tubes open both ends (all harmonics)
1.0
0.5
0.0
-0.5
-1.0
sin(2
π f
1 t)
(1/3)sin(2
π
*3*f
1 t)
(1/5)sin(2
π
*5*f
1 t)
(1/7)sin(2
π
*7*f
1 t)
sum
0 20 40 60 80 100
TIM E

ELECTRIC FORCE:
Coulomb’s law:
• force on charge 1 due to charge 2 is
F
12
= k e q
1 r q
2
2 r ˆ
12
Net force on a charge due to several other charges:
• VECTOR SUM of all forces on that charge due to other charges
• Called Principle of SUPERPOSITON
• Each charge exerts a force on charge 1
• Resultant force is r
1
= r
21
+ r
31
+ r
41
• says net force on charge 1 equals sum of force on 1 from 2, force on 1 from 3, and force on 1 from 4

ELECTRIC FIELD
• If the force on q
0
at a point is r
, then electric field at that point is
• If the electric field at a point is E , then the force on q
0
at point is
• Electric field at P due to a point charge is o Unit vector r ˆ points from q → P
P
= k e q r 2 r ˆ
• Electric field points away from positive charge
• Electric field points toward negative charge
= q
0
= q
0

Superposition:
Total E at point P due to an arrangement of point charges is the VECTOR SUM of the electric field contributions from all charges around P
• Total electric field at P is: r
T
= k e
∑ i q i r i
2 r ˆ i = r
1
+
2
+ r
3
+
4 o q i
is the charge at i o r i
is the distance from q i
→ P o r ˆ i
is the unit vector from q i o the sum is a VECTOR SUM
Did example with electric dipole
→ P

ELECTRIC FIELD LINES:
• E vector at a point in space is tangent to the EFL through that point
• “Density” of EFL is proportional to E (magnitude) in that region o Larger E → closer packing of lines
• EFL start on positive charges and end on negative charges
• Number of EFL starting/ending on charge is proportional to its magnitude
• Electric field lines do not cross

Looked at motion of a particle in a uniform electric field:

ELECTRIC FLUX
General result for Electric Flux through element of area
∆ A i
∆Φ
E i
= E i
∆ A i cos
i
= r i
⋅ ∆ r i
Total flux through a closed surface:
Φ
E
= ∫ ⋅ = over closed surface d r
∫ E n dA

GAUSS’S LAW (general statement):
Φ e
= ∫ ⋅ = closed surface d r q enclosed
ε
0
• Powerful way to calculate electric field if we can factor E n out of integral o Trick is to choose surface so that E n is uniform over all or part of surface
No charge inside:
• net number of lines leaving = 0
• all lines go through
Positive charge inside:
• non-zero net number of lines leaving
• lines start on charge inside sphere
Used Gauss’s Law to calculate electric field around a point charge