1051-455-20073
Solution Set #2
1. For the following two harmonic waves:
(a) Show on a phasor diagram:
f1 [t] = 2 cos [ω 0 t]
h
πi
f2 [t] = 7 cos ω 0 t −
4
phasor is the complex vector evaluated at t = 0:
f1 [t] = 2 exp [i · 0] = 2 + 0i
h πi
³ h πi
h π i´
f2 [t] = 7 exp −i
= 7 cos −
+ i sin −
4
4
4
³ h πi
h π i´
7
= 7 cos +
− i sin
= √ (1 − i)
4
4
2
As time increases, the two phasors rotate about the origin at the rate ω 0 .
(b) Find the mathematical expression for the superposition f1 [t] + f2 [t] in the form of a cosine.
7
Re {f1 [0] + f2 [0]} = 2 + √ ∼
= 6.9497
2
7 ∼
Im {f1 [0] + f2 [0]} = 0 − √ =
−4.9497
2
sµ
¶2 µ
¶2
7
7
+ −√
2+ √
|f1 [0] + f2 [0]| =
2
2
∼
= 8.5322
(
)
∙
¸
− √72
−1 Im {f1 [0] + f2 [0]}
−1
= tan
Φ {f1 [0] + f2 [0]} = tan
Re {f1 [0] + f2 [0]}
2 + √72
∼ −0.6189 radians =
∼ −0.197π radians ∼
=
= −35.459◦
The sinusoidal expression for this oscillation is:
f1 [t] + f2 [t] ∼
= 8.5322 · cos [ω 0 t − 0.6189]
1
2. Two plane waves of the same frequency that vibrate along the z−direction are:
∙ µ
¶¸
x
f1 [x, t] = A1 cos 2π
− ν 1 t + π ; A1 = 40 mm, X1 = 30 mm, ν 1 = 20 Hz
X1
∙ µ
¶¸
y
f2 [y, t] = A2 cos 2π
− ν 2 t + π ; A2 = 20 mm, Y2 = 40 mm, ν 2 = 20 Hz
Y2
Evaluate the resultant waveform f1 [x, t] + f2 [y, t] at [x, y] = [50 mm, 20 mm]
full credit if done either way:
(a)
=
=
=
=
=
=
f1 [50 mm, t] + f2 [20 mm, t]
∙ µ
¶
¸
∙ µ
¶
¸
50 mm
20 mm
40 mm · cos 2π
− 20 Hz · t + π + 20 mm · cos 2π
− 20 Hz · t + π
30 mm
40 mm
µ
∙ µ
¶¸¶
µ
∙ µ
¶¸¶
5
1
40 mm · − cos 2π
− 20 Hz · t
+ 20 mm · − cos 2π
− 20 Hz · t
3
2
∙
¸
10π
−40 mm · cos
− 40πt − 20 mm · cos [π − 40πt]
3
∙
¸
10π
−40 mm · cos
− 40πt + 20 mm · cos [−40πt]
3
∙
¸
10π
−40 mm · cos 40πt −
+ 20 mm · cos [40πt]
3
½
∙ µ
¶¸¾
10π
Re 20 mm · exp [+i40πt] − 40 mm · exp +i 40πt −
3
Because the temporal frequencies are the same, we can evaluate the sum at t = 0 by using phasors:
f1 [50 mm, 0] + f2 [20 mm, 0]
∙ µ
¶¸
10π
= 20 mm · exp [+i40π · 0] − 40 mm · exp +i 40π · 0 −
3
∙
¸
10π
= 20 mm − 40 mm · exp −i
3
∙
¸
10π
= +40 mm
Re {f1 [50 mm, 0] + f2 [20 mm, 0]} = 20 mm − 40 mm · cos −
3
∙
¸
√
10π
Im {f1 [50 mm, 0] + f2 [20 mm, 0]} = −40 mm · sin −
= −20 3 mm ∼
= −34.641 mm
3
q
2
2
(+40 mm) + (−34.641 mm) ∼
= 52.915 mm
∙
¸
−34.641
Φ {f1 [50 mm, 0] + f2 [20 mm, 0]} = tan−1
+40
∼
= −0.713 71 radians ∼
= −0.227 18π radians ∼
= −40.892◦
|f1 [50 mm, 0] + f2 [20 mm, 0]| =
2
(b)
=
=
=
=
=
f1 [50 mm, t] + f2 [20 mm, t]
∙ µ
¶¸
∙ µ
¶¸
50 mm
20 mm
40 mm · cos 2π
− 20 Hz · t + π
+ 20 mm · cos 2π
− 20 Hz · t + π
30 mm
40 mm
¸
∙ µ
¶¸
∙ µ
¶
5
1
2
40 mm · − cos 2π
− 20 Hz · t − 20 mm · cos 2π
− 20 Hz · t + 2π
3
2
∙µ
¶
¸
£¡
¢
¤
10π
−40 mm · cos
+ 2π2 − 40πt − 20 mm · cos π + 2π2 − 40πt
3
∙
µ
¶¸
£
¡
¢¤
10π
2
−40 mm · cos 40πt −
− 20 mm · cos 40πt − π + 2π 2
+ 2π
3
½
∙ µ
µ
¶¶¸
¾
£ ¡
¡
¢¢¤
10π
2
2
Re −40 mm · exp +i 40πt −
− 20 mm · exp +i 40πt − π + 2π
+ 2π
3
Because the temporal frequencies are the same, we can evaluate the sum at t = 0 by using phasors:
∙
¸
£
¤
10π
2
Re {f1 [50 mm, 0] + f2 [20 mm, 0]} = −40 mm · cos
+ 2π − 20 mm · cos π + 2π 2
3
∙
¸
£
¤
10π
2
−40 mm · cos
+ 2π − 20 mm · cos π + 2π 2
3
∼
= −1.7237 mm ∙
¸
£
¤
10π
2
Re {f1 [50 mm, 0] + f2 [20 mm, 0]} = −40 mm · sin
+ 2π − 20 mm · sin π + 2π 2
3
∼
= +52.887 mm
q
2
2
|f1 [50 mm, 0] + f2 [20 mm, 0]| =
(−1.7237 mm) + (+52.887 mm) ∼
= 52.915 mm
½
¾
+52.887 ∼
Φ [f1 [50 mm, 0] + f2 [20 mm, 0]] = tan−1
= −1.5382 radians ∼
= −88.134◦
−1.7237
3
3. Consider the superposition of two sinusoidal traveling waves:
f1 [z, t] = A1 cos [k1 z − ω 1 t] ,
A1
= 101 mm, ν 1 = 100 Hz, v1 = 250
f2 [z, t] = A2 cos [k2 z − ω 2 t] ,
A2
= 99 mm, ν 2 = 150 Hz, v2 = 500
m
s
m
s
(a) Find an expression for the resulting wave in terms of the average wave, the modulation wave, plus
any remaining amplitude.
We have an expression for the sum of two waves with different frequences bu the same magnitude:
∙
¸
∙
¸
A+B
A−B
cos [A] + cos [B] = 2 cos
· cos
2
2
so we have:
A1 cos [k1 z − ω 1 t] + A2 cos [k2 z − ω 2 t]
= [A2 + (A1 − A2 )] cos [k1 z − ω 1 t] + A2 cos [k2 z − ω 2 t]
= A2 (cos [k1 z − ω 1 t] + cos [k2 z − ω 2 t]) + (A1 − A2 ) cos [k1 z − ω 1 t]
∙µ
¶
µ
¶ ¸
∙µ
¶
µ
¶ ¸
k1 + k2
ω1 + ω2
k1 − k2
ω1 − ω2
= 2A2 cos
z−
t · cos
z−
t
2
2
2
2
+ (A1 − A2 ) cos [k1 z − ω 1 t]
which is the sum of a traveling wave and the product of the average and modulation waves
(b) Calculate the wavelengths of the average and modulation waves.
In this case:
λ1 ν 1
=
λ2 ν 2
=
=⇒
250 ms
v1
5
=
= m
ν1
100 Hz
2
500 ms
v2
10
v2 =⇒ λ2 =
=
=
m
ν2
150 Hz
3
λ1 > λ2 and v1 < v2
v1 =⇒ λ1 =
1
kavg
=
=⇒
1
kmod
=
=⇒
1
+
2π
k1 + k2
=
= 2π λ1 λ2
λavg
2
2
à 1
!
−1
µ 2
1
3 ¶−1
20
λ1 + λ2
5 m + 10 m
λavg =
=
=
m = λavg
2
2
7
1
−
2π
k1 − k2
=
= 2π λ1 λ2
λmod
2
2
à 1
!−1 µ
¶−1
1
2
−
− 3
λmod = λ1 λ2
= 5 m 10 m
= 20 m = λmod
2
2
n.b., λmod > λavg
4
(c) Find the velocities of the average and modulation waves.
ω avg
=
ω mod
=
µ
¶
µ
¶
ω1 + ω2
ν1 + ν2
100 Hz + 150 Hz
= 2π
= 2π
= 2π · 125
2
2
2
µ
¶
µ
¶
ω1 − ω2
ν1 − ν2
100 Hz − 150 Hz
= 2π
= 2π
= −2π · 25
2
2
2
vavg
=
ω avg
20
2500 m
= λavg · ν avg =
m · 125 Hz =
= vavg
kavg
7
7 s
vmod
=
ω mod
m
= λmod · ν mod = 20 m · −25 Hz= 500
= vmod
kmod
s
so the modulation wave travels in the opposite direction
(d) Does this system exhibit normal or anomalous dispersion?
λ1 > λ2 and v1 < v2 =⇒ longer wave travels slower =⇒ anomalous dispersion
5
1
4. The phase velocity of waves in some medium is proportional to ω + 2 . Find an expression for the
modulation velocity and determine whether the waves exhibit normal or anomalous dispersion.
vφ
=
=⇒
vmod
=
vmod
=
1
ω
= αω + 2 , where α is some constant
k
1
ω
2
=⇒ αk = ω + 2 =⇒ ω = (αk)
k=
1
+
αω 2
this is the dispersion relation ω [k]
à 1!
i
1
dω
ω+ 2
d h
2
2
2
=
(αk) = 2α k = 2α
= 2αω + 2
dk
dk
α
1
2αω + 2 = 2vavg =⇒ anomalous dispersion
6
5. Plot and write the equation of the superposition of the following harmonic waves:
hπ
i
E1 = sin
− ωt
18
∙
¸
5π
E2 = 3 cos
− ωt
9
hπ
i
E3 = 2 sin
− ωt
6
where the period of each is 2 s.
hπ
i
h³ π
´ πi
E1 = sin
− ωt = cos
− ωt −
18
18 ∙
µ2
¶¸
∙
¸
i
h³ π
π´
8π
4π
−
− ωt = cos ωt − −
= cos ωt +
= cos
18
2
18
9
E2
E3
∙
¸
∙
¸
5π
5π
= 3 cos
− ωt = 3 cos ωt −
9
9
hπ
i
h³ π
´ πi
= 2 sin
− ωt = 2 cos
− ωt −
2
h6 π
i
h6
πi
= 2 cos − − ωt = 2 cos ωt +
3
3
radians
since all three components have the same frequency ω = 2πν = 22π
sec = π sec , the sum has the same
frequency. Evaluate the amplitude for t = 0:
∙
¸
∙
¸
h
4π
5π
πi
E1 [t = 0] + E2 [t = 0] + E3 [t = 0] = 1 · exp +i ·
+ 3 · exp −i ·
+ 2 · exp +i ·
9
9
3
∙
¸
∙
¸
h πi
4π
5π
Re {E1 [t = 0] + E2 [t = 0] + E3 [t = 0]} = 1 · cos +
+ 3 · cos −
+ 2 · cos +
9
9
3
∙ ¸
4
= 1 − 2 · cos π ∼
= 0.6527
9
∙
¸
∙
¸
h πi
4π
5π
+ 3 · sin −
+ 2 · sin +
Im {E1 [t = 0] + E2 [t = 0] + E3 [t = 0]} = 1 · sin +
9
9
3
∙ ¸
√
4
3 − 2 · sin π ∼
=
= −0.237 56
9
7
|E1 [t = 0] + E2 [t = 0] + E3 [t = 0]| =
∼
=
Φ {E1 [t = 0] + E2 [t = 0] + E3 [t = 0]} =
sµ
∙ ¸¶2
∙ ¸¶2 µ
√
4
4
+
3 − 2 · sin π
1 − 2 · cos π
9
9
0.694 59
"√
£4 ¤#
3
−
2
·
sin
π
π
£ 9 ¤ = − radians = −20◦
tan−1
9
1 − 2 · cos 49 π
h
πi ∼
E1 + E2 + E3 ∼
= 0.694 59 · cos ωt −
= 0.695 · cos [ωt − 0.349]
9
(a) the three component sinusoids; (b) the sum using both the sum of the three functions in (a) and the
computation — the two results are identical
8
6. A laser emits a monochromatic beam of wavelength λ0 , which is reflected normally from a plane mirror
that recedes from the light source at velocity v.
(a) Determine the beat frequency between the incident and reflected light.
First try it without for v = 0; the reflected light differs from the incident light only in the direction
(assuming no losses or phase changes in the mirror):
f1 [z, t] = A0 cos [k1 z − ω 1 t]
f2 [z, t] = A0 cos [k1 z + ω 1 t]
The sum creates standing waves:
f1 [z, t] + f2 [z, t] = A0 cos [k1 z − ω 1 t] + A0 cos [k1 z + ω 1 t]
= 2A0 cos [k1 z] cos [ω 1 t]
Note that if the mirror moves with velocity v, the frequency and the wavelength of the reflected
light change due to the Doppler effect. From Eq.(4-44) n Pedrotti(s):
s
1 − vc
λaf ter
=
λbef ore
1 + vc
In the limit v << c, then Eq.(4-45) applies:
s
r³
1 − vc
λ00
v´ ³
λaf ter
v ´−1
≡
=
=
1
−
1
+
λbef ore
λ0
1 + vc
c
c
r³
r
´
v
v −1
=
1− ·
1+
c
c
r
r
v
v
∼
1 − · 1 + (−1)
=
c
c
r
r
v
v
=
1− · 1−
c
c
v
= 1−
³ cv´
0
λ0 =
1−
λ0
c
so the wavelength of the reflected beam is longer if v recedes from the source (v < 0, red shift)
and shorter if it approaches the source (v > 0, blue shift). In our case, the reflected beam exhibits
twice the Doppler shift because it has to change direction, so the corresponding wavevectors are:
k1
k10
=
=
2π
unchanged
λ0
µ
µ
¶−1
¶
2π
2π
1
2v
2v
∼
¢
=¡
= k1 ·
= k1 · 1 −
= k1 1 +
c
c
λ00
1 − 2v
1 − 2v
c
c λ0
where the identity has been used:
∞
X
1
2
αn = 1 − α + α2 − · · · ∼
=
= 1 − α if |α| << |α|
1 − α n=0
The corresponding oscillation frequencies are easy to find because the velocity of light is unchanged:
λ0 · ν 0 = λ00 · ν 00 =⇒
9
λ00
ν0
2v
= 0 =1−
λ0
ν0
c
:
ν 01
ν 1 is unchanged
µ
¶
ν1 ∼
2v
=
· ν1
= 1+
c
1 − 2v
c
So if v is positive, the wavelength decreases and the frequency increases.
The sum of the two waves is:
f1 [z, t] + f2 [z, t] = A0 cos [k1 z − ω 1 t] + A0 cos [k10 z + ω 01 t]
= 2A0 cos [kavg z − ω avg t] · cos [kmod z − ω mod t]
where:
kavg
=
kmod
=
ω avg
=
ω mod
=
ν mod
=
µ
µ
¶¶
µ
¶
k1 + k10
k1 v
2v
1
1
2v
=
k1 + k1 1 +
=
2k1 + k1
= k1 +
2
2
c
2
c
c
µ
µ
¶¶
0
k1 − k1
2v
1
k1 v
=
k1 − k1 1 +
=−
2
2
c
c
µ
¶
µ
¶
0
ω1 + ω1
2v
vω 1
1
=
ω1 + 1 +
· ω1 = ω1 +
2
2
c
c
³
´
³
´
0
ω1 − ω1
v
vω 1
1
=
ω1 − 1 + 2
· ω1 = −
2
2
c
c
vν 1
−
c
f1 [z, t] + f2 [z, t] = A0 cos [k1 z − ω 1 t] + A0 cos [k10 z + ω 01 t]
∙µ
¶
¸
∙
¸
³
k1 v
vω 1 ´
k1 v
vω 1
= 2A0 cos k1 +
z − ω1 +
t · cos −
z−−
t
c
c
c
c
∙µ
¶
¸
i
³
h v
k1 v
vω 1 ´
= 2A0 cos k1 +
z − ω1 +
t · cos − (k1 z + ω 1 t)
c
c
c
so the beat frequency in the equation is:
|ν mod | =
|ω mod |
1 vω 1
ν1
=
=
·v
2π
2π c
c
but we actually see the squared magnitude, which oscillates twice as fast (Pedrotti Eq.(5-34))
|ν mod | = 2 ·
10
ν1
·v
c
(b) Determine the beat frequency between the incident and reflected light if the light is incident on
the plane mirror at angle θ.
Here the reflected beam travels at angle 2θ, so the k−vectors are different before and after the
reflection. Again consider the case v = 0:
⎡
⎤
⎡ ⎤
0
0
2π
⎣ 0 ⎦
k1 = ⎣ 0 ⎦ =
λ
2π
1
λ
⎡
⎤
⎡ 2π
⎤
sin [θ]
λ sin [θ]
2π
⎣
⎦
⎦=
0
0
k01 = ⎣
λ
2π
cos [θ]
λ cos [θ]
⎡ sin[θ] ⎤
2
k1 + k01
2π ⎣
⎦
0
kavg =
=
2
λ
1+cos[θ]
=
kmod
k1 −
2
k01
2
⎤
− sin[θ]
2
2π ⎣
⎦
0
=
λ
1−cos[θ]
⎡
2
The sum of the two waves is:
∙
2π
f1 [x, y, z, t]+f2 [x, y, z, t] = 2 cos
λ
µ
¶
¸
∙ µ
¶¸
sin [θ]
2π
1 + cos [θ]
sin [θ]
1 − cos [θ]
x+
z − ωt ·cos
−
x+
z
2
2
λ
2
2
so the beat frequency is zero (standing waves).
If we add the movement, we get the same wavelength shifts, but now they depend on the angle:
vz
vx
= v · cos [2θ]
= v · sin [2θ]
ν mod = 2
11
ν1
· v · cos [2θ]
c