Recitation 3
Chapters 14 and 24
Problem 14.5. Two traveling sinusoidal waves are described by the wave functions
y1 = (5.00 m) · sin[π(4.00x − 1200t)]
(1)
y2 = (5.00 m) · sin[π(4.00x − 1200t − 0.250)]
(2)
and
where x, y1 , and y2 are in meters and t is in seconds. (a) What is the amplitude of the resultant wave? (b) What is the
frequency of the resultant wave?
(a)
y = y1 + y2 = 5.00 m · {sin[π(4.00x − 1200t)] + sin[π(4.00x − 1200t − 0.250)} = 5.00 m · [sin(θ) + sin(θ − π/4)] ,
(3)
where θ ≡ π(4.00x − 1200t). So y2 trails y1 by π/4 = 45◦ . In terms of the reference circle, that looks like
π/4
The amplitude of y is then given by vector addition
A = 2 · A cos(φ/2) = 2 · 5.00 m · cos(π/8) = 9.24 m ,
(4)
where φ = π/4 is the phase difference between y1 and y2 .
(b) Both y1 and y2 rotate around the reference circle with a frequency of
f=
ω
1200π rad/s
=
= 600 Hz ,
2π
2π rad/cycle
(5)
so y must also rotate around the reference circle at 600 Hz.
Problem 14.6. Two identical sinusoidal waves with wavelengths of 3.00 m travel in the same direction at a speed of 2.00 m/s.
The second wave originates from the same point as the first, but at a later time. The amplitude of the resultant wave is the
same as that of each of the two initial waves. Determine the minimum possible time interval between the starting moments
of the two waves.
This is Problem 14.6 backwards. From the amplidude of the resultant wave, we can find the phase difference between the
two constituent waves.
A = 2 · A · cos(φ)
1
π
φ = arccos
=
2
3
2π
∆θ = 2φ =
rad .
3
(6)
(7)
(8)
The angular speed of the waves is given by
ω = 2πf = 2π
So the time interval is
∆t =
v
4π
=
rad/s
λ
3
∆θ
= 0.500 s
ω
(9)
(10)
Problem 14.8. Two loudspeakers are placed on a wall 2.00 m apart. A listener stands 3.00 m from the wall directly in front
of one of the speakers. A single oscillator is driving the speakers at a frequency of 300 Hz. (a) What is the phase difference
between the two waves when they reach the observer? (b) What is the frequency closest to 300 Hz to which the oscillator may
be adjusted so that the observer hears minimal sound?
L=2m
Sb
Sa
db =
da = 3 m
p
L2 + d2a = 3.606 m
Listener
(a) From Table 13.1 we find that the speed of sound in air at 20◦ C is v = 343 m/s. The wavelength of this sound is
v
= 1.143 m
f
(11)
θa = kda =
2πda
= 16.49 rad ,
λ
(12)
θb = kdb =
2πdb
= 19.81 rad .
λ
(13)
λ=
The phase change from speaker Sa is therefore
and from speaker Sb is
The phase difference is
∆θ = θb − θa = 3.33 rad
(14)
(b) For minimal sound, we want the phase difference to be exactly π (or some odd multiple of π). We see that it’s already
close to π with our initial frequency of 300 Hz, only a bit high. Decreasing the freqency a bit will reduce the rate of dephasing
between the two waves, reducing ∆θ, so we’re looking for a frequency slightly less than 300 Hz.
2πf (db − da )
2π(db − da )
=
λ
v
vπ
v
343 m/s
v∆θ
=
=
=
= 283 Hz .
f=
2π(db − da )
2π(db − da )
2(db − da )
2 · 0.606 m
∆θ = θb − θa =
(15)
(16)
Problem 14.9. Two sinusoidal waves in a string are defined by the functions
y1 = (2.00 cm) · sin(20.0x − 32.0t)
(17)
y2 = (2.00 cm) · sin(25.0x − 40.0t)
(18)
and
where y and x are in centimeters and t is in seconds. (a) What is the phase difference between there two waves at the point
x = 5.00 cm at t = 2.00 s? (b) What is the positive x value closest to the origin for which the two phases differ by ±π at
t = 2.00 s? (This location is where the two waves add to zero.)
(a) This is just plugging in
θ1 = 20.0 · 5.00 − 32.0 · 2.00 = 36 rad
(19)
θ2 = 25.0 · 5.00 − 40.0 · 2.00 = 45 rad
(20)
◦
◦
∆θ = θ2 − θ1 = 9 rad = 516 = 156
(21)
(b) We’re supposed to find some x for a given t such that
∆θ = θ2 − θ1 = (25.0 · x − 80.0) − (20.0 · x − 64.0) = 5.0 · x − 16.0 = nπ
nπ + 16.0
x=
5.0
(22)
(23)
for some odd n. 16/π = 5.09, so n = −5 will give the smallest positive x for which this is true, and
x=
−5π + 16.0
= 0.0584 cm = 584 µm
5.0
(24)
Problem 14.20. In the arrangement shown in Figure P14.20, an object can be hung from a string (with a linear mass
density µ = 2.00 g/m) that passes over a light pulley. The string is connected to a vibrator (of constant frequency f ), and
the length of the string between point P and the pulley is L = 2.00 m. When the mass m of the object is either 16.0 kg or
25.0 kg, standing waves are observed, but no standing waves are observed with any mass between these values. (a) What is
the frequency of the vibrator? (Note: The greater the tension in the string, the smaller the number of nodes in the standing
wave.) (b) What is the largest object mass for which standing waves could be observed?
P
vibrator
L
µ
m
(a) For a patricular hanging mass m, the tension in the string balances the gravitational force on the mass, so
T = mg ,
(25)
so the speed of sound in the string is
s
v=
T
=
µ
r
mg
.
µ
(26)
Standing waves on strings occur when a wave completes some number of full cycles in a round trip. In mathematical terms
n · 2π = k · 2L =
2L
λ
2L
λ=
n
n=
2π
· 2L
λ
(27)
(28)
(29)
for integer n (the number of the particular vibrational mode). With the generator operating at a fixed frequency f , the
wavelength is also related to the wave speed by
v
λ= ,
(30)
f
so
r
mg
v
1
2L
=λ= = ·
(31)
n
f
f
µ
r
µ
2Lf
=n,
(32)
mg
To put it all together we notice that the two masses, m1 = 16.0 kg and m2 = 25.0 kg are said to produce consecutive modes.
From the last equation, we see that increasing mass m decreases the mode number n, so the heavier mass must be one mode
lower than the lighter, or n1 = n2 + 1. From here on out, it’s all algebra to find f .
r
µ
n2 = 2Lf
(33)
m2 g
r
r
µ
µ
n1 = 2Lf
= n2 + 1 = 2Lf
+1
(34)
m1 g
m2 g
r µ
1
1
1 = 2Lf
−√
(35)
√
g
m1
m2
r −1
µ
1
1
f = 2L
−√
= 350 Hz
(36)
√
g
m1
m2
(b) As we saw in (a), increasing the mass decreased the vibrational mode number. The largest mass that can sustain standing
waves is the one for which the vibration is in the first mode, so
r
µ
1 = n = 2Lf
(37)
mmax g
r
mmax g
= 2Lf
(38)
µ
mmax g
= (2Lf )2
(39)
µ
µ(2Lf )2
mmax =
= 400 kg
(40)
g
Problem 24.7. Figure 24.3 shows a plane electromagnetic sinosoidal wave propogating in the x direction. Suppose the
wavelength is 50.0 m and the electric field vibrates in the xy plane with an amplitude of 22.0 V/m. Calculate (a) the
frequency of the wave and (b) the magnitude and direction of B when the electric field has its maximum value in the negative
y direction. (c) Write an expression for the B with the correct unit vector, with numerical values for Bmax , k, and ω, and
with its magnidude in the form
B = Bmax cos(kx − ωt)
(41)
(a) This is just a units conversion
f=
3.00 · 108 m/s
c
=
= 6.00 · 106 cycles/s = 6.00 MHz
λ
50.0 m/cycle
(42)
(b) The magnitude of B in an electromagnetic plane wave is given by B = E/c. The direction of the wave’s motion is given
by the Poynting vector S = E × B. Using the right-hand-rule for the cross product, we see that when E is in the −ĵ direction
and S is in the î direction, B must be in the −k̂ direction. Putting this together
B0 =
−E0
k̂ = −73.3 nT · k̂
c
(43)
(c) Because it is a sinusoidal wave moving in the î direction, we know B must look something like
B = B0 cos(kx − ωt + φ) .
(44)
We already found B0 in (b), and we don’t have any phase information, so we can drop φ. That leaves
2π
= 0.126 rad/m
λ
ω = 2πf = 3.77 · 107 rad/s
k=
(45)
(46)
so
B = −73.3 nT · cos(0.126 rad/m · x − 3.77 · 107 rad/s · t)k̂ .
(47)
Problem 24.8. In SI units, the electric field in an electromagnetic wave is described by
Ey = 100 sin(1.00 · 107 x − ωt)
(48)
Find (a) the amplitude of the corresponding magnetic field oscillations, (b) the wavelength λ, and (c) the frequency f .
(a) The amplitude is the magnitude of the oscillation, which just comes from the prefactor outside the trig function. In this
case, A = 100 V/m
(b) By comparing with the standard form of sinusoidal waves
Y = A sin(kx − ωt) ,
(49)
we see that the wavenumber k = 1.00 · 107 rad/m. Converting radians to cycles and inverting yields
λ=
2π rad/cycle
= 628 nm/cycle
k
(50)
(c) Once we know the length of a cycle, and how fast the wave is moving, we can find out how many of them occur in a
second
c
3.00 m/s
f= =
= 477 · 1012 cycles/s = 477 THz
(51)
λ
628 nm/cycle
Problem 24.9. Verify by substitution that the following equations are solutions to Equations 24.15 and 24.16 respectively:
E = Emax cos(kx − ωt)
(52)
B = Bmax cos(kx − ωt)
(53)
∂2E
∂2E
= 0 µ0 2
2
∂x
∂t
∂2B
∂2B
= 0 µ0 2
∂x2
∂t
(24.15)
(24.16)
This is just an excercise in partial derivatives.
∂E
∂x
∂2E
∂x2
∂E
∂t
∂2E
∂t2
k rad/m
ω rad/s
∂2E
∂x2
= −Emax sin(kx − ωt) · k
(54)
= −Emax k cos(kx − ωt) · k = −k 2 E
(55)
= −Emax sin(kx − ωt) · (−ω)
(56)
= Emax ω cos(kx − ωt) · (−ω) = −ω 2 E
(57)
1 √
= 0 µ0
c
k2 ∂ 2 E
∂2E
= 2 2 = 0 µ0 2
ω ∂t
∂t
=
(58)
(59)
which is what we set out to show. Note that we used Equation 24.17 in Equation 58. The situation for B is exactly the same
with the replacement E → B.
∂B
∂x
∂2B
∂x2
∂B
∂t
∂2B
∂t2
k rad/m
ω rad/s
∂2B
∂x2
= −Bmax sin(kx − ωt) · k
(60)
= −Bmax k cos(kx − ωt) · k = −k 2 B
(61)
= −Bmax sin(kx − ωt) · (−ω)
(62)
= Bmax ω cos(kx − ωt) · (−ω) = −ω 2 B
(63)
1 √
= 0 µ0
c
k2 ∂ 2 B
∂2B
= 2 2 = 0 µ0 2
ω ∂t
∂t
=
(64)
(65)
Problem 24.22. An AM radio station broadcasts isotropically (equally in all directions) with an average power of 4.00 kW.
A dipole recieving antenna 65.0 cm long is at a location 4.00 miles from the transmitter. Compute the amplitude of the emf
that is induced by this signal between the ends of the recieving antenna.
To find the signal intensity at our antenna, we note that the power broadcast from the station is spread out over a sphere of
radius R = 4.00 miles. The average intensity is then
I = Savg =
P
P
4.00 · 103 W
=
=
= 7.68 µW/m2 .
A
4πR2
4π(4.00 miles · 1.609 · 103 m/mile)2
(66)
From Equation 24.27, we see
2
Emax
2µ0 c
p
= 2µ0 cSavg = 76.1 mV/m
Savg =
Emax
(24.27)
(67)
The total voltage difference produced across our length L = 65.0 cm antenna is then
∆V = LEmax = 49.4 mV
(68)
Problem 24.25. The filament of an incandescent lamp has a 150 Ω resistance and carries a direct current of 1.00 A. The
filament is 8.00 cm long and 0.900 mm in radius. (a) Calculate thte Poynting vector at the surface of the filament, associated
with the static electric field producing the current and the curret’s static magnetic field. (b) Find the magnitude of the static
electric and magnetic fields at the surface of the filament.
(a) The hot resistor will be radiating heat, and none of the electric or magnetic fields change with time, so we expect a
constant Poynting vector of magnitude
P
I 2R
=
= 332 kW/m2 .
(69)
S=
A
2πrL
This Poynting vector will always point away from the wire (in the direction the radiation is going).
(b) The electric field is given by Ohm’s law.
V = IR
IR
1.00 A · 150 Ω
V
=
=
= 1875 V/m .
E=
L
L
8.00 · 10−2 m
(70)
(71)
The magnetic field from a long, straight wire is
I
,
2πr
(72)
B = 177 T
(73)
B=
so the magnetic field at the surface of the wire is
The electric field is along the wire, and the magnetic field is perpendicular to the current, so the Poynting vector points
directly out (perpendicular to the wire’s surface) and has a magnitude
S = EB sin(90◦ ) = EB =
which is the same expression we found in (a).
IR
I
I 2R
·
=
,
L 2πr
2πrL
(74)