Sinusoidal waves (velocity)
y ( x, t )  A cos( kx  t )

v
y ( x, t )  A cos( kx  t )

v
x
y x, t   const  kx  t  const  v x 
  k
t
λ
T
y
y
x
t
  2 / T
k  2 / 
v   /T   / k
y
vy 
 A sin kx  t 
t
Wave equation
y ( x, t )  A cos( kx  t )
y
 A sin kx  t 
t
2 y
a y  2   A 2 coskx  t 
t
vy 
2 y 2 2 y
 2
2
t
k x 2
y
  Ak sin kx  t 
x
2 y
2


Ak
coskx  t 
2
x
d 2 y  x, t  2  2 y  x, t 
v
0
2
2
dt
x

v
k
y ( x, t )  y ( x  vt,0)
y  x, t 
y  x, t 
 v
t
x
Example: Gaussian pulse

y( x, t )  A exp  ( x  vt) 2 / s 2

Velocity and Acceleration in a Sinusoidal Traveling Wave
y(x,t)  A cos(kx   t)
Wave Moving Towards Right
v
y(x) (m)
A
t=0s
y
y(x,t)
 A sin(kx   t)
t
This vy (x,t) is required in order that
vy (x,t) 
the wave can move towards the right.
Velocity amplitude = A  Avk
y
v (m/s)
A
A
a
y
ay (x,t) 
A
2A
t
 A 2 cos(kx   t)
Acceleration amplitude = A 2  A(vk)2
2A
0
vy (x,t)
  2 y(x,t)
y
a (m/s2)
y(x) (m)
A
A
5
10
x (m)
15
20
Question
A transverse sinusoidal wave on a stretched
string has the wave function
y(x,t)  (0.10 m) cos[ (3.0 rad/m) x  (6.0 rad/s) t ].
What is the velocity v of the wave?
1. v  (18 m/s) î
2. v  (0.5 m/s) î
3. v  (2.0 m/s) î
4. v  (2.0 m/s) î
Answers:
1.
2.
3.
4.
1
2
3
4
Question
A sinusoidal transverse traveling
wave on a stretched string has:
amplitude
A = 0.10 m
wave number k = 6.0 rad/m
wave speed v = 100 m/s.
The maximum speed of each particle
in the string as time passes is ___
m/s.
1. 60
2. 80
3. 100
4. 120
Question
The speed of the pulse is ___ m/s.
(Assume g = 10 m/s2)
1. 50
2. 100
3. 150
4. 200
A pulse is moving along a
stretched string as shown. The
string goes over a massless
frictionless pulley and is
connected to a hanging block.
Energy transferred by string waves
u k  dK dx  12 μv y2  12 μy x,t  t   12 μv 2 y x,t  x 
2
uU  dU dx
2
v 2  F
u k  uU
 y x, t 
2  y  x, t  
u  u k  uU  v 
 F


 x 
 x 
2
2
u is maximum where the
slope is maximum
u is zero where the slope
is zero
For sinusoidal waves:
u  μ 2 A 2 sin kx  t 
2
u ev 
1
2
μ 2 A 2
E  u ev   vA 2
Pave

1
2
E 1

 2 v 2 A 2 
T
F 2 2
2 2
F  A   A
2v
T
1
1
2
sin kx  t  

T0
2
Question
A stretched string has a linear mass
density  = 0.010 kg/m. A
sinusoidal traveling wave moving on
the string has the wave function
y = (0.010 m)cos[(1.0 rad/m)x
 (100 rad/s)t]
Recall: v = /k.
The average power transmitted by the
wave is __ W.
1.
2.
3.
4.
0.5
1.0
1.5
2.0
Example: Gaussian Pulse
Energy Density in Gaussian Pulse
y(x, t = 0) = A exp[ (x/s) 2]
FA2/s 2
0
0
y(x, t = 0)
u(x, t = 0)
A
-4
-3
-2
-1
0
x/s
1
2
3
4
Here, s is a measure of the width of the pulse: the larger s is, the larger
the width. The energy density u(x) (dotted curve) is largest where
the slope magnitude | dy / dx | is largest and zero where the slope is zero.
Details of Example: Gaussian Pulse
A Gaussian-shaped pulse is defined as
Energy Density in Gaussian Pulse
y(x,t  0)  y(x)  A exp[(x / s)2 ] .
y(x, t = 0) = A exp[ (x/s) 2]
Energy density
FA2/s 2
0
0
y(x, t = 0)
u(x, t = 0)
Here, s is a measure of the width of the
pulse: the larger s is, the larger the width.
In the figure, y(x) is plotted vs x / s.
A
2
 dy(x) 
u(x,t  0)  F 
.

 dx 
dy(x)
 2x 
 A   2  exp[(x / s)2 ], so
 s 
dx
2
2
 A  x
u(x,t  0)  4F     exp[2(x / s)2 ] .
 s   s
u(x,t  0) is plotted versus x / s as the
dashed curve (right-hand scale).
-4
-3
-2
-1
0
x/s
1
2
3
4
Total energy in the pulse
The total mechanical energy in the pulse
E is obtained by adding up the energies
in all the parts of the pulse:

E
 u(x,t  0) dx 


2
 A
(Fs)   .
 s
2
Example: Sinusoidal Traveling Wave
Energy Density in Sinusoidal Wave
Sinusoidal wave traveling towards
2
the right:
2
u(x, t=0) = F(k A) sin (kx)
2
y(x,t)  A cos(kx   t) .
In the figure, y(x,t  0) is plotted
versus kx.
y(x, t = 0)
A
0
A
Energy density
2
 y(x,t) 
Given: u(x,t)  F 
.

 x 
y(x,t)
 Ak sin(kx   t), so
x
u(x,t)  F(Ak)2 sin 2 (kx   t) .
u(x,t  0) is plotted versus kx as the
dotted curve (right-hand scale).
0
u(x, t=0)
F(kA)
y(x, t = 0) = A cos(kx)
5
10
kx (rad)
15
20
The energy density wave is rigidly
"attached" to the wave and moves along
with the wave as the wave moves.
Example: Sinusoidal Traveling Wave, cont’d.
Rewrite the energy density. We had
Energy Density in Sinusoidal Wave
u(x,t)  F(Ak)2 sin 2 (kx   t) .
2
  2 . This gives
u(x,t)   2 A 2 sin 2 (kx   t) .
Energy in one wavelength E
0
A
0
u(x, t=0)
v
2
A
y(x, t = 0)
Fk  F
2
2
F(kA)
Using   vk, with v  F /  :
2
2
u(x, t=0) = F(k A) sin (kx)
y(x, t = 0) = A cos(kx)
5
From the figure, the average of
10
kx (rad)
15
20
u over one wavelength is
1 2 2 1
Fk A   2 A 2 . Then
2
2
using   vT , we get
uave 
E  uave   uave vT 
1
v 2 A 2T
2
Average power Pave Transmitted by the Wave
The power transmitted past a point is the energy
transmitted per unit time past a given point:
Pave 
E 1
 v 2 A 2 .
T
2