Chapter 16 Wave Motion
Mechanical waves are waves that disturb and propagate through a medium; the ripple
in the water due to the pebble and a sound wave, for which air is the medium, are
examples of mechanical waves.
Electromagnetic waves are a special class of waves that do not require a medium in
order to propagate, light waves and radio waves are two familiar examples.
16.1 Propagation of a Disturbance
All mechanical waves require (1) some source of disturbance, (2) a medium that can
be disturbed, and (3) some physical mechanisms through which particles of the
medium can influence one another.
What is phonon?
Transverse waves
Longitudinal waves
y(x, 0) = f(x)
y(x, t) = y(x - vt, 0)
y(x, t) = f(x – vt); moving to the right
y(x, t) = f(x + vt); moving to the left
Sample Example: A pulse moving to the right
A wave pulse moving to the right along the x-axis is represented by the wave function
2 .0
y ( x, t ) =
where x and y are measured in cm and t is in sec. Let us
( x − 3.0t ) 2 + 1
plot the wave form at t = 0, t = 1, and t = 2 s.
1
16.2 The Traveling Wave Model
crest: highest displacement
trough: lowest displacement
wave length
frequency
wave speed
Each particle of the string oscillates vertically in the y
direction with simple harmonic motion.
at t = 0
y = A sin(
y = A sin[
k=
2π
λ
2π
λ
2π
λ
x) , if the wave moves to the right with a speed of v
( x − vt )] , v =
, ω=
λ
T
, y = A sin[
2π
λ
x−
2π
t]
T
2π
ω
, y = A sin(kx − ωt ) , v = = λf
T
k
kx − ωt = const , k
dx
dx ω
ω
−ω = 0,
= , v = = λf
dt
dt k
k
if a wave moving in the negative direction of x.
dx
dx
ω
ω
+ω = 0 ,
= − , v = − = −λf
dt
dt
k
k
all traveling waves must be of the form: y ( x, t ) = h(kx ± ωt )
kx + ωt = const , k
Sample Example: A traveling sinusoidal wave
A sinusoidal wave traveling in the positive x direction has an amplitude of 15 cm, a
wavelength of 40 cm, and a frequency of 8 Hz.
medium at t = 0 and x = 0 is also 15 cm,
The vertical displacement of the
(a) Find the angular wave number, period,
angular frequency, and speed of the wave.
(b) Determine the phase constant.
2π
1 1
2π
(a) k =
, T= = , ω=
= 50.3rad / s
40
f 8
T
v=
ω
k
= fλ = 8 ⋅ 40 = 320cm / s
(b) y = 15 sin(
π
20
x − 16πt + φ ) at t = x = 0, y = 15
2
sin φ = 1 , φ =
π
2
16.3 The Speed of Waves on Strings
Fr = 2T sin θ ≈ 2Tθ
m = µ∆s = 2 µRθ
Fr = m
v2
v2
, 2Tθ = 2 µRθ
R
R
T = µv 2 , v =
T
µ
Example: the speed of a pulse on a cord
A uniform cord has a mass of 0.3 kg and
a total length of 6 m.
Tension is
maintained in the cord by suspending an
object of mass 2 kg from one end.
Find
the speed of a pulse on the cord.
Assume that the tension is not affected
by the mass of the cord.
v=
T
µ
=
2 ⋅ 9 .8
= 19.8m / s
0 .3 / 6
The Wave Equation
y = A sin(kx − ωt )
v=
dy
= −ωA cos(kx − ωt ) , a = −ω 2 A sin(kx − ωt )
dt
dy
d2y
= kA cos(kx − ωt ) ,
= − k 2 A sin( kx − ωt )
2
dx
dx
d2y k2 d2y 1 d2y
=
= 2
dx 2 ω 2 dt 2
v dt 2
Example: A solution to the linear wave eq
Verify that the wave function presented in previous example is a solution to the linear
wave eq.
y=
2
d2y 1 d2y
,
=
( x − 3t ) 2 + 1 dx 2 9 dt 2
3
16.4 Reflection and Transmission
y l ( x, t ) = A cos(k l x − ωt ) , vl =
T
µl
=
ω
kl
,
yl ' ( x, t ) = B cos(−k l x − ωt )
y r ( x, t ) = C cos(k r x − ωt ) , v r =
T
µr
=
ω
kr
x = 0 , y l + y l ' = y r , ==> A + B = C
d
d
d
yl +
yl ' =
y r , ==> k l ( A − B ) = k r C
dx
dx
dx
k − kr
2 kl
B= l
A, C =
A , light string -> small µ -> small k
kl + k r
kl + k r
16.5 Rate of Energy Transfer by Sinusoidal
4
Waves on Strings
∆K =
1
1
1
1
∆mv 2 = µ∆xv 2 , dK = µv 2 dx = µω 2 A 2 sin 2 (kx − ωt )dx
2
2
2
2
at t = 0, average kinetic energy in a period of wave length
λ
1
K λ = µω 2 A 2 ∫ sin 2 (kx)dx ,
2
0
Kλ =
λ
∫ sin
λ
2
1 − cos(2kx)
λ
dx =
2
2
0
(kx)dx = ∫
0
1
µω 2 A 2 λ
4
the change is length from ∆x to l is: ∆x 1 + (
dy 2
1 dy
) − ∆x ≈ ( ) 2 ∆x ,
dx
2 dx
1 dy
∆U = T ⋅ (l − ∆x) = T ( ) 2 ∆x
2 dx
1 dy
1
dy
1 ω dy
∆U = T ( ) 2 ∆x = µv 2 ( ) 2 ∆x = µ ( ) 2 ( ) 2 ∆x ,
2 dx
2
dx
2 k
dx
dU =
1
µω 2 A 2 sin 2 (kx − ωt )dx
2
at t = 0, U λ =
Kλ =
λ
1
µω 2 A 2 ∫ sin 2 (kx)dx ,
2
0
λ
λ
1 − cos(2kx)
λ
dx =
2
2
0
2
∫ sin (kx)dx = ∫
0
1
µω 2 A 2 λ
4
E
1
1
µω 2 A 2 λ , P = λ = µω 2 A 2 v
2
T
2
Sample Example:
A string with linear mass density 5 x 10-2 kg/m is under a tension of 80 N. How
much power must be supplied to the string to generate sinusoidal waves at a
frequency of 60 Hz and an amplitude of 6 cm?
Eλ =
A = 0.06m , f = 60 , µ = 0.05kg / m , v =
P=
80
= 40
0.05
1
0.05 ⋅ 60 2 ⋅ 0.06 2 ⋅ 40 = 512W
2
5
16.6 The Linear Wave Equation – Physical
Model
∑F
y
∑F
y
∑F
µ∆x
y
= T sin θ B − T sin θ A ≈ T tan θ B − T tan θ A
 ∂y 
 ∂y 
= T  −T 
 ∂x  B
 ∂x  A
= ma y = m
∂2 y
∂2 y
∂  ∂y 
 ∂y 
 ∂y 
=
µ
∆
x
= T   − T   = T  ∆x
2
2
∂t
∂t
∂x  ∂x 
 ∂x  B
 ∂x  A
∂2 y
∂2 y
=
T
∆x the sinusoidal wave function is y ( x, t ) = A sin (kx − ωt )
∂t 2
∂x 2
∂2 y
∂2 y
2
=
−
k
A
sin
(
kx
−
ω
t
)
,
= −ω 2 A sin (kx − ωt )
∂x 2
∂t 2
µω 2 = Tk 2 -> v =
ω
k
=
T
µ
->
∂2 y T ∂2 y 1 ∂2 y
=
=
∂x 2 µ ∂t 2 v 2 ∂t 2
6