Wavesandoscillationsexampleques

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Problem 1.4 A wave traveling along a string is given by
y(x,t) = 2 sin(4π t + 10π x) (cm),
where x is the distance along the string in meters and y is the vertical displacement.
Determine: (a) the direction of wave travel, (b) the reference phase φ0 , (c) the
frequency, (d) the wavelength, and (e) the phase velocity.
Solution:
(a) We start by converting the given expression into a cosine function of the form
given by (1.17):
³
π´
y(x,t) = 2 cos 4π t + 10π x −
(cm).
2
Since the coefficients of t and x both have the same sign, the wave is traveling in the
negative x-direction.
(b) From the cosine expression, φ0 = −π /2.
(c) ω = 2π f = 4π ,
f = 4π /2π = 2 Hz.
(d) 2π /λ = 10π ,
λ = 2π /10π = 0.2 m.
(e) up = f λ = 2 × 0.2 = 0.4 (m/s).
Problem 1.8 Two waves on a string are given by the following functions:
y1 (x,t) = 4 cos(20t − 30x) (cm)
y2 (x,t) = −4 cos(20t + 30x) (cm)
where x is in centimeters. The waves are said to interfere constructively when their
superposition |ys | = |y1 + y2 | is a maximum, and they interfere destructively when
|ys | is a minimum.
(a) What are the directions of propagation of waves y1 (x,t) and y2 (x,t)?
(b) At t = (π /50) s, at what location x do the two waves interfere constructively,
and what is the corresponding value of |ys |?
(c) At t = (π /50) s, at what location x do the two waves interfere destructively,
and what is the corresponding value of |ys |?
Solution:
(a) y1 (x,t) is traveling in positive x-direction. y2 (x,t) is traveling in negative
x-direction.
(b) At t = (π /50) s, ys = y1 + y2 = 4[cos(0.4π − 30x) − cos(0.4π + 3x)]. Using
the formulas from Appendix C,
2 sin x sin y = cos(x − y) − (cos x + y),
we have
ys = 8 sin(0.4π ) sin 30x = 7.61 sin 30x.
Hence,
|ys |max = 7.61
µ
¶
π
π
2nπ
cm, where
and it occurs when sin 30x = 1, or 30x = + 2nπ , or x =
+
2
60
30
n = 0, 1, 2, . . . .
nπ
cm.
(c) |ys |min = 0 and it occurs when 30x = nπ , or x =
30
Problem 1.13 The voltage of an electromagnetic wave traveling on a transmission
line is given by υ (z,t) = 5e−α z sin(4π × 109t − 20π z) (V), where z is the distance in
meters from the generator.
(a) Find the frequency, wavelength, and phase velocity of the wave.
(b) At z = 2 m, the amplitude of the wave was measured to be 2 V. Find α .
Solution:
(a) This equation is similar to that of Eq. (1.28) with ω = 4π × 109 rad/s and
β = 20π rad/m. From Eq. (1.29a), f = ω /2π = 2 × 109 Hz = 2 GHz; from
Eq. (1.29b), λ = 2π /β = 0.1 m. From Eq. (1.30),
up = ω /β = 2 × 108 m/s.
(b) Using just the amplitude of the wave,
−α 2
2 = 5e
,
µ ¶
−1
2
α=
ln
= 0.46 Np/m.
2m
5
Problem 1.26 Find the phasors of the following time functions:
(a) υ (t) = 9 cos(ω t − π /3) (V)
(b) υ (t) = 12 sin(ω t + π /4) (V)
(c) i(x,t) = 5e−3x sin(ω t + π /6) (A)
(d) i(t) = −2 cos(ω t + 3π /4) (A)
(e) i(t) = 4 sin(ω t + π /3) + 3 cos(ω t − π /6) (A)
Solution:
(a) Ve = 9e− jπ /3 V.
(b) v(t) = 12 sin (ω t + π /4) = 12 cos (π /2 − (ω t + π /4)) = 12 cos (ω t − π /4) V,
e
V = 12e− jπ /4 V.
(c)
i(t) = 5e−3x sin (ω t + π /6) A = 5e−3x cos[π /2 − (ω t + π /6)] A
= 5e−3x cos (ω t − π /3) A,
(d)
(e)
Ie = 5e−3x e− jπ /3 A.
i(t) = −2 cos(ω t + 3π /4),
Ie = −2e j3π /4 = 2e− jπ e j3π /4 = 2e− jπ /4 A.
i(t) = 4 sin(ω t + π /3) + 3 cos(ω t − π /6)
= 4 cos[π /2 − (ω t + π /3)] + 3 cos(ω t − π /6)
= 4 cos(−ω t + π /6) + 3 cos(ω t − π /6)
= 4 cos(ω t − π /6) + 3 cos(ω t − π /6) = 7 cos(ω t − π /6),
Ie = 7e− jπ /6 A.
Problem 1.27 Find the instantaneous time sinusoidal functions corresponding to
the following phasors:
(a) Ve = −5e jπ /3 (V)
(b) Ve = j6e− jπ /4 (V)
(c) Ie = (6 + j8) (A)
(d) I˜ = −3 + j2 (A)
(e) I˜ = j (A)
(f) I˜ = 2e jπ /6 (A)
Solution:
(a)
Ve = −5e jπ /3 V = 5e j(π /3−π ) V = 5e− j2π /3 V,
v(t) = 5 cos (ω t − 2π /3) V.
(b)
Ve = j6e− jπ /4 V = 6e j(−π /4+π /2) V = 6e jπ /4 V,
v(t) = 6 cos (ω t + π /4) V.
(c)
◦
Ie = (6 + j8) A = 10e j53.1 A,
i(t) = 10 cos (ω t + 53.1◦ ) A.
(d)
◦
Ie = −3 + j2 = 3.61 e j146.31 ,
i(t) = Re{3.61 e j146.31 e jω t } = 3.61 cos(ω t + 146.31◦ ) A.
◦
(e)
Ie = j = e jπ /2 ,
i(t) = Re{e jπ /2 e jω t } = cos(ω t + π /2) = − sin ω t A.
(f)
Ie = 2e jπ /6 ,
i(t) = Re{2e jπ /6 e jω t } = 2 cos(ω t + π /6) A.
Problem 1.9 Give expressions for y(x,t) for a sinusoidal wave traveling along a
string in the negative x-direction, given that ymax = 40 cm, λ = 30 cm, f = 10 Hz,
and
(a) y(x, 0) = 0 at x = 0,
(b) y(x, 0) = 0 at x = 3.75 cm.
Solution: For a wave traveling in the negative x-direction, we use Eq. (1.17) with
ω = 2π f = 20π (rad/s), β = 2π /λ = 2π /0.3 = 20π /3 (rad/s), A = 40 cm, and x
assigned a positive sign:
µ
¶
20π
y(x,t) = 40 cos 20π t +
(cm),
x + φ0
3
with x in meters.
(a) y(0, 0) = 0 = 40 cos φ0 . Hence, φ0 = ±π /2, and
¶
µ
20π
π
x±
y(x,t) = 40 cos 20π t +
3
2
¢
¡
½
20π
−40 sin
¡ 20π t + 3 ¢x (cm), if φ0 = π /2,
=
40 sin 20π t + 203π x (cm), if φ0 = −π /2.
(b) At x = 3.75 cm = 3.75 × 10−2 m, y = 0 = 40 cos(π /4 + φ0 ). Hence, φ0 = π /4
or 5π /4, and
¢
¡
½
40 cos ¡20π t + 203π x + π4 ¢(cm), if φ0 = π /4,
y(x,t) =
40 cos 20π t + 203π x + 54π (cm), if φ0 = 5π /4.
Problem 1.15 A laser beam traveling through fog was observed to have an intensity
of 1 (µ W/m2 ) at a distance of 2 m from the laser gun and an intensity of 0.2
(µ W/m2 ) at a distance of 3 m. Given that the intensity of an electromagnetic
wave is proportional to the square of its electric-field amplitude, find the attenuation
constant α of fog.
Solution: If the electric field is of the form
E(x,t) = E0 e−α x cos(ω t − β x),
then the intensity must have a form
I(x,t) ≈ [E0 e−α x cos(ω t − β x)]2
≈ E02 e−2α x cos2 (ω t − β x)
or
I(x,t) = I0 e−2α x cos2 (ω t − β x)
where we define I0 ≈ E02 . We observe that the magnitude of the intensity varies as
I0 e−2α x . Hence,
at x = 2 m,
at x = 3 m,
I0 e−4α = 1 × 10−6
(W/m2 ),
I0 e−6α = 0.2 × 10−6
I0 e−4α
10−6
=5
=
I0 e−6α
0.2 × 10−6
e−4α · e6α = e2α = 5
α = 0.8 (NP/m).
(W/m2 ).
Problem 1.29 The voltage source of the circuit shown in Fig. P1.29 is given by
vs (t) = 25 cos(4 × 104t − 45◦ ) (V).
Obtain an expression for iL (t), the current flowing through the inductor.
R1
i
A
iL
iR 2
+
vs(t)
L
R2
R1 = 20 Ω, R2 = 30 Ω, L = 0.4 mH
Figure P1.29:
Solution: Based on the given voltage expression, the phasor source voltage is
◦
Ves = 25e− j45
(V).
(9)
The voltage equation for the left-hand side loop is
R1 i + R2 iR2 = vs
(10)
diL
,
dt
(11)
For the right-hand loop,
R2 iR2 = L
and at node A,
i = iR2 + iL .
(12)
Next, we convert Eqs. (2)–(4) into phasor form:
R1 Ie+ R2 IeR2 = Ves
R2 IeR2 = jω LIeL
Ie = IeR2 + IeL
e we have:
Upon combining (6) and (7) to solve for IeR2 in terms of I,
IeR2 =
jω L
I.
R2 + j ω L
(13)
(14)
(15)
(16)
Substituting (8) in (5) and then solving for Ie leads to:
jR2 ω L e e
R1 Ie+
I = Vs
R2 + j ω L
µ
¶
jR
ω
L
2
Ie R1 +
= Ves
R2 + j ω L
µ
¶
R1 R2 + jR1 ω L + jR2 ω L
e
I
= Ves
R2 + j ω L
¶
µ
ω
L
R
+
j
2
Ves .
Ie =
R1 R2 + jω L(R1 + R2 )
(17)
Combining (6) and (7) to solve for IeL in terms of Ie gives
IeL =
R2
e
I.
R2 + j ω L
(18)
Combining (9) and (10) leads to
¶µ
¶
µ
R2 + j ω L
R2
e
Ves
IL =
R2 + j ω L
R1 R2 + jω L(R1 + R2 )
R2
Ves .
=
R1 R2 + + jω L(R1 + R2 )
Using (1) for Ves and replacing R1 , R2 , L and ω with their numerical values, we have
IeL =
30
j4 × 104 × 0.4 × 10−3 (20 + 30)
20 × 30 +
◦
30 × 25
=
e− j45
600 + j800
◦
7.5 − j45◦ 7.5e− j45
− j98.1◦
e
=
=
◦ = 0.75e
j53.1
6 + j8
10e
25e− j45
Finally,
iL (t) = Re[IeL e jω t ]
= 0.75 cos(4 × 104t − 98.1◦ ) (A).
(A).
◦
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