Waves
Pulse Propagation
The Wave Equation
Lana Sheridan
De Anza College
May 17, 2017
Last time
• oscillations
• simple harmonic motion (SHM)
• spring systems
• energy in SHM
• pendula
• introducing waves
• kinds of waves
Overview
• wave speed on a string
• pulse propagation
• the wave equation
Wave Motion
Wave
a disturbance or oscillation that transfers energy through matter or
space.
The waveform moves along the medium and energy is carried with
it.
The particles in the medium do not move along with the wave.
The particles in the medium are briefly shifted from their
equilibrium positions, and then return to them.
Wave Speed on a String
484
Chapter 16 Wave Motion
How fast does a disturbance propagate on a string under tension?
As the pulse moves along the
string, new elements of the
string are displaced from their
equilibrium positions.
Figure 16.1 A hand moves the
end of a stretched string up and
which the pebble is dropped to the
motion: energy is transferred over
16.1 Propagation o
The introduction to this chapt
fer of energy through space wit
of energy transfer mechanism
and electromagnetic radiation
nism, matter transfer, the ener
through space with no wave ch
All mechanical waves requir
taining elements that can be d
which elements of the medium
wave motion is to flick one en
opposite end fixed as shown in
a pulse) is formed and travels
represents four consecutive “sn
eling pulse. The hand is the s
through which the pulse trave
from their equilibrium positio
(16.10)
tromagnetic wave that propagates into
space at the speed of light (Chapter 34)
Wave Speed on a String
Imagine traveling with the pulse at speed v to the right. Each
small section of the rope travels to the left along a circular arc
rings from your point of view.
S
l waves is that the wave speed
through which the wave travels.
o the wavelength l can be represection, we determine the speed
.
xpression for the speed of a pulse
Consider a pulse moving to the
a stationary (with respect to the
e 16.11a. Newton’s laws are valid
view this pulse from a different
h the pulse at the same speed so
as in Figure 16.11b. In this referment of the string moves to the
v
!s
a
S
v
!s
u
u
S
S
T
T
R
u
ms an approximate arc of a cirO
n Figure 16.11b. In our moving
b
oves to the left with speed v. As
ent as a We
particle
in
uniform
cirFigure 16.11
will find find how fast
a point(a) In
on the
thereferstring
ence frame of the Earth, a pulse
eleration of v 2/R, which is suprelative
to
the
wave
pulse.
moves to the right on a string with
tude is the tension in the string.
moves backwards
Wave Speed on a String
Strings
S
v
nical waves is that the wave speed
m through which the wave travels.
ve to the wavelength l can be reprehis section, we determine the speed
ing.
e expression for the speed of a pulse
T. Consider a pulse moving to the
e to a stationary (with respect to the
gure 16.11a. Newton’s laws are valid
us view this pulse from a different
with the pulse at the same speed so
me as in Figure 16.11b. In this referelement of the string moves to the
forms an approximate arc of a cirw in Figure 16.11b. In our moving
moves to the left with speed v. As
lement as a particle in uniform cirWe ofcan
usewhich
the isforce
acceleration
v 2/R,
supgnitude is
the tension
string
∆s: in the string.
ngent to the arc, as in Figure 16.11b.
each vertical component T sin u acts
adial force on the element is 2T sin u.
!s
a
S
v
!s
u
u
S
S
T
T
R
u
O
b
Figure 16.11 (a) In the referdiagram
find
the
force on a small length of
ence frame to
of the
Earth,
a pulse
moves to the right on a string with
speed v. (b) In a frame of reference moving to the right with the
net the small element of length
pulse,
Ds moves to the left with speed v.
F
= 2T sin θ ≈ 2T θ
(1)
Wave Speed on a String
Consider the centripetal force on the piece of string.
If R is the radius of curvature and m is the mass of the small piece
of string:
mv 2
Fnet =
R
Wave Speed on a String
Consider the centripetal force on the piece of string.
If R is the radius of curvature and m is the mass of the small piece
of string:
mv 2
Fnet =
R
Suppose the string has mass density µ (units: kg m−1 )
m = µ ∆s = µR(2θ)
Put this into our expression for centripetal force:
Fnet =
2µRθv 2
R
Wave Speed on a String
Put this into our expression for centripetal force:
Fnet = 2µθv 2
And using eq. (1), Fnet = 2T θ:
2T θ = 2µθv 2
The wave speed is:
s
v=
T
µ
For a given string under a given tension, all waves travel with the
same speed!
Question
If you stretch a rubber hose and pluck it, you can observe a pulse
traveling up and down the hose.
What happens to the speed of the pulse if you stretch the hose
more tightly?
(A) It increases.
(B) It decreases.
(C) It is constant.
(D) It changes unpredictably.
1
Serway & Jewett, page 499, objective question 2.
Question
If you stretch a rubber hose and pluck it, you can observe a pulse
traveling up and down the hose.
What happens to the speed of the pulse if you stretch the hose
more tightly?
(A) It increases.
←
(B) It decreases.
(C) It is constant.
(D) It changes unpredictably.
1
Serway & Jewett, page 499, objective question 2.
Question
If you stretch a rubber hose and pluck it, you can observe a pulse
traveling up and down the hose.
What happens to the speed if you fill the hose with water?
(A) It increases.
(B) It decreases.
(C) It is constant.
(D) It changes unpredictably.
1
Serway & Jewett, page 499, objective question 2.
Question
If you stretch a rubber hose and pluck it, you can observe a pulse
traveling up and down the hose.
What happens to the speed if you fill the hose with water?
(A) It increases.
(B) It decreases.
←
(C) It is constant.
(D) It changes unpredictably.
1
Serway & Jewett, page 499, objective question 2.
he string passes over a pulley and supExample
A uniform string has a mass of ms and a length of `. The string
passes over a pulley and supports an block of mass mb . Find the
speed of a pulse traveling along this string. (Assume the vertical
piece of the rope is very short.)
2 (Example
sion T in the
ained by the
ject. The
wave traveling
d is given by
2 m block g 5 0
T
2.00 kg
he string passes over a pulley and supExample
A uniform string has a mass of ms and a length of `. The string
passes over a pulley and supports an block of mass mb . Find the
speed of a pulse traveling along this string. (Assume the vertical
piece of the rope is very short.)
2 (Example
sion T in the
ained by the
ject. The
wave traveling
d is given by
2 m block g 5 0
T
2.00 kg
r
v=
mb g `
ms
Pulse Propagation
A wave pulse (in a plane) at a moment in time can be described in
terms of x and y coordinates, giving y (x).
We know that the pulse will move with speed v and be displaced,
say in the positive x direction, while maintaining its shape.
That means we can also give y as a function of time, y (x, t).
Consider a moving reference frame, S 0 , with the pulse at rest,
y 0 (x 0 ) = f (x 0 ), no time dependence. Galilean transformation:
x 0 = x − vt
displacements.
to earthquakes). The slower transverse
or secondary, travel through the Earth at
g the time interval between the arrivals
h, the distance from the seismograph to
rmined. This distance is the radius of an
raph. The origin of the waves is located
x 0 = x − vt
spheres from three or more monitoring
er intersect
at oneinregion
the Earth, of the string
Then
the of
rest-frame
curred.
on a long string as shown in Figure 16.5.
y (x, t) At=t !f0,(xthe0 )shape
= off the
(x
sition of the pulse at time t 5 0. At this
pulse is given by y ! f(x).
ay be, can be represented by some mathy
, 0) 5 f(x). This function describes the
S
v
string located at each value of x at time
v, the pulse has traveled to the right a
assume the shape of the pulse does not
e shape of the pulse is the same as it was
ntly, an element of the string at x at this
P
located at x 2 vt had at time t 5 0:
Pulse Propagation
ransverse position y for all positions and
the origin at O, as
a
y
S
vt
(16.1)
e transverse positions of elements of the
(x 1 vt)
x
O
(x 2 vt, 0)
(x 2 vt)
− vt)
v
P
(16.2)
wave function, depends on the two varitten y(x, t), which is read “y as a function
O
x
Pulse Propagation
The shape of the pulse is given by f (x) and can be arbitrary.
Whatever the form of f , if the pulse moves in the +x direction:
y (x, t) = f (x − vt)
If the pulse moves in the −x direction:
y (x, t) = f (x + vt)
Wave Pulse Example 16.1
A pulse moving to the right along the x axis is represented by the
wave function
y (x, t) =
2
(x − 3.0t)2 + 1
where x and y are measured in centimeters and t is measured in
seconds.
What is the wave speed?
Find expressions for the wave function at t = 0, t = 1.0 s, and
t = 2.0 s.
1
Serway & Jewett, page 486.
This function is an unnormalized Cauchy distribution, or as physicists say
“it has a Lorentzian profile”.
2
Wave Pulse Example 16.1
A pulse moving to the right along the x axis is represented by the
wave function
y (x, t) =
2
(x − 3.0t)2 + 1
where x and y are measured in centimeters and t is measured in
seconds.
What is the wave speed? 3.0 cm/s
Find expressions for the wave function at t = 0, t = 1.0 s, and
t = 2.0 s.
1
Serway & Jewett, page 486.
This function is an unnormalized Cauchy distribution, or as physicists say
“it has a Lorentzian profile”.
2
their location, and the resultant pulse moves around the stadium. Is this wave
(a) transverse or (b) longitudinal?
Wave Pulse Example 16.1
mple 16.1
A Pulse Moving to the Right
y (cm)
e moving to the right along the x axis is represented by the wave
on
y1 x, t 2 5
2
1 x 2 3.0t 2 2 1 1
t = 0, y (x, 0) =
2
x +1
x and y are measured in centimeters and t2is measured in secFind expressions for the wave function at t 5 0, t 5 1.0 s, and
0 s.
3.0 cm/s
2.0
1.5
t!0
1.0
y (x, 0)
0.5
0
1 2 3 4 5 6 7 8
x (cm)
a
y (cm)
TION
3.0 cm/s
2.0
ptualize Figure 16.6a shows the pulse represented by this wave
on at t 5 0. Imagine this pulse moving to the right at a speed
cm/s and maintaining its shape as suggested by 2Figures 16.6b
6.6c.
t = 1, y (x, 1) =
1.5
2
(x − 3.0) + 1
orize We categorize this example as a relatively simple analysis
em in which we interpret the mathematical representation of a
(Example 16.1) Graphs
of the function y(x, t) 5
2/[(x 23.0t)2 1 1] at
(a) t 5 0, (b) t 5 1.0 s,
and (c) t 5 2.0 s.
y (x, 1.0)
0.5
0
1 2 3 4 5 6 7 8
x (cm)
b
y (cm)
2
ze The wave
of the
t =function
2, y is(x,
2)form
= y5
6.0)216.6
+1
v t). Inspection of the expression (x
for − Figure
and comparison to Equation 16.1 reveal
he wave speed is v 5 3.0 cm/s. Furtherby letting x 2 3.0t 5 0, we find that the
mum value of y is given by A 5 2.0 cm.
t ! 1.0 s
1.0
3.0 cm/s
2.0
1.5
1.0
0.5
0
c
2
t ! 2.0 s
y (x, 2.0)
1 2 3 4 5 6 7 8
x (cm)
The Wave Equation
tan uCan
to express
we find a general equation describing the displacement (y ) of
our medium as a function of position (x) and time?
(16.22)
Start by considering a string carrying a disturbance.
the rightSend of
he force T. This
presented by the
axis for this disinstant of time,
ngents in Equa-
S
T
!x
uA
B
uB
A
S
T
(16.23)
Figure 16.19 An element of a
string under tension T.
n sin u < tan u to express
The Wave Equation
Consider a (16.22)
small length of string ∆x.
rd from the rightSend of
senting the force T. This
an be represented by the
o the x axis for this disarticular instant of time,
or the tangents in Equa-
S
T
!x
uA
B
uB
A
S
T
(16.23)
Figure 16.19 An element of a
string under
T.
As we did for oscillations,
start tension
from Newton’s
2nd law.
Fy
= may
T sin θB − T sin θA = (µ ∆x)
For small angles
sin θ ≈ tan θ
∂2 y
∂t 2
The Wave Equation
We can write tan θ as the slope of y (x):
tan θ =
∂y
∂x
Now Newton’s second law becomes:
∂y ∂2 y
∂y −
= (µ ∆x) 2
T
∂x x=B
∂x x=A
∂t
∂y ∂x x=B −
µ ∂2 y
=
T ∂t 2
∆x
∂y ∂x x=A
The Wave Equation
We can write tan θ as the slope of y (x):
tan θ =
∂y
∂x
Now Newton’s second law becomes:
∂y ∂2 y
∂y −
= (µ ∆x) 2
T
∂x x=B
∂x x=A
∂t
∂y ∂x x=B −
µ ∂2 y
=
T ∂t 2
∆x
∂y ∂x x=A
We need to take the limit of this expression as we consider an
infinitesimal piece of string: ∆x → 0.
The Wave Equation
We can write tan θ as the slope of y (x):
tan θ =
∂y
∂x
Now Newton’s second law becomes:
∂y ∂2 y
∂y −
= (µ ∆x) 2
T
∂x x=B
∂x x=A
∂t
∂y ∂x x=B −
µ ∂2 y
=
T ∂t 2
∆x
∂y ∂x x=A
We need to take the limit of this expression as we consider an
infinitesimal piece of string: ∆x → 0.
µ ∂2 y
T ∂t 2
=
∂2 y
∂x 2
where we use the definition of the partial derivative.
The Wave Equation
µ ∂2 y
∂2 y
=
T ∂t 2
∂x 2
Remember that the speed of a wave on a string is
s
T
v=
µ
The wave equation:
1 ∂2 y
∂2 y
= 2 2
2
∂x
v ∂t
Even though we derived this for a string, it applies much more
generally!
The Wave Equation
We can model longitudinal waves like sound waves by a series of
masses connected by springs, length h.
u1
u2
u3
u is a function that gives the displacement of the mass at each
equilibrium position x, x + h, etc.
For such a case, the propagation speed is
s
KL
v=
µ
where K is the spring constant of the entire spring chain, L is the
length, and µ is the mass density.
1
Figure from Wikipedia, by Sebastian Henckel.
The Wave Equation
u1
u2
u3
u is a function that gives the displacement of the mass at each
equilibrium position x, x + h, etc.
Consider the mass, m, at equilibrium position x + h
F
= ma
∂2 u
k(u3 − u2 ) − k(u2 − u1 ) = m 2
∂t
2
m∂ u
= u3 − 2u2 + u1
k ∂t 2
1
Figure from Wikipedia, by Sebastian Henckel.
The Wave Equation
m ∂2 u
= u3 − 2u2 + u1
k ∂t 2
We can re-write m
k in terms of quantities for the entire spring
chain. Suppose there are N masses.
m=
µL
N
and k = NK and N =
L
h
u(x + 2h) − 2u(x + h) + u(x)
µ ∂2 u
=
2
KL ∂t
h2
Letting N → ∞ and h → 0, the RHS is the definition of the 2nd
derivative. Same equation!
1 ∂2 u
∂2 u
=
v 2 ∂t 2
∂x 2
Summary
• wave speed on a string
• pulse propagation
• the wave equation
• solutions to the wave equation
• sine waves
• transverse motion
Homework Serway & Jewett:
• Ch 16, onward from page 499. OQs: 5; CQs: 1, 9; Probs: 1,
3, 23, 29, 53, 59, 60