Lecture 2: Introduction to wave theory (II)
Mathematical description of waves (Y&F 15.3):
 Phase velocity: the velocity (speed) at which we would have to
move to keep up with a point of constant phase on the wave.
 Right moving wave:
y ( x, t )  A sin( t  kx)  t  kx  const
 Derivative with respect to t is zero:
d
t  kx    k dx    kv  0  v    f
dt
dt
k

v

Phase velocity:
is the same as speed of wave:
k
v  f
 Transverse speed of wave:
dy d
d
vy 
  A sin( t  kx)   A cos(t  kx) t  kx 
dt dt
dt
v y  A cos(t  kx)
P1X: Optics, Waves and Lasers Lectures, 2005-06.
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 Transverse speed maximum:
v y  A cos(t  kx)  A  t  kx  0, ,2 , 
 Transverse speed minimum:

3
v y  A cos(t  kx)  0  t  kx   ,
,
2
2
y
A
/4
/2
3/4
x

y ( x, t )  A sin( t  kx)
-A
Offset by 90o
vy
Aw
v y  A cos(t  kx)
/4
/2
3/4

x
- Aw
P1X: Optics, Waves and Lasers Lectures, 2005-06.
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Example: 15-2 from Y&F (page 556)
Find the maximum transverse speed of the example shown in lecture 1.
What is the velocity at t=0, at the end of the clothes-line and at 3.0 m from
the end.
x 
 t
y ( x, t )  A sin t  kx  0.075 sin 2 

m
 0.5 6.0 
1
v y  A cos(t  kx)  v y ,max  A  0.075  4  0.94ms
At x=t=0, velocity is maximum transverse speed = +0.94 ms-1
At x=3.0 m and t=0:
v y x  3.0, t  0  A cos(0   )   A  0.94ms1
P1X: Optics, Waves and Lasers Lectures, 2005-06.
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Simple Harmonic Motion (Y&F 13.1-2, 13.4-5):
 Definition:
• Simple Harmonic Motion (SHM) is motion in which a particle is
acted on by a force proportional to its displacement from a fixed
(equilibrium) position and is in the opposite direction to the
displacement:
d 2x
k
F  kx  a  2   x   2 x
dt
m
 Examples:
• Mass vibrating on a spring.
• Simple pendulum (only
when displacement is small).
P1X: Optics, Waves and Lasers Lectures, 2005-06.
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 Simple pendulum:
• Vertical:
T cosq  W  ma y  0,
• Horizontal:
 T sin q  max
If q is small then
and
and:
sin q 
cosq  1
therefore:
W  mg
x
q
L
when
P1X: Optics, Waves and Lasers Lectures, 2005-06.
m
x
T
g
sin q   x   2 x
m
L
The same as the restoring force of a spring but with:
T
q
T  mg
ax  
L
x  L
W
mg
k
L
5
 Solution:
• What function satisfies
d 2x
a  2   2 x ?
dt
•Try x  A sin( t   ) 
dx
 A cos(t   ) 
dt
d 2x
2
2


A

sin(

t


)



x
2
dt
with  the angular frequency (rad/s):
x
A
• For the case of the spring:
T/4
T/2
3T/4
T
k

m
t
•For the case of the pendulum:
g
L

-A
P1X: Optics, Waves and Lasers Lectures, 2005-06.
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Definitions:
a) Amplitude A is maximum displacement (m).
b) Frequency f: number of oscillations per second.
(Units: 1 Hertz = 1 cycle/s = 1 s-1)

f 
2
c) Period T: time (s) between oscillations
1 2
T 
f

d) Phase constant ( ): gives position of oscillation at t=0.
x(t  0)  A sin(  )
P1X: Optics, Waves and Lasers Lectures, 2005-06.
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Example: 13-2 from Y&F
A spring is mounted horizontally. A force of 6.0 N causes a displacement
of 0.030 m. If we attach an object of 0.50 kg to the end and pull it a
distance of 0.020 m and watch it oscillate in SHM, find (a) the force
constant of the spring, (b) the angular frequency, frequency and period of
oscillation.
(a) At x = 0.030 m, F=-6.0 N
(b) m=0.50 kg, k=200 N/m:
The frequency:
The period:
k
F
 6.0 N

 200 N / m
x
0.030m
k
200kg / s 2


 20rad / s
m
0.50kg

20rad / s
f 

 3.2cycle / s  3.2 Hz
2 2rad / cycle
T
1
1

 0.31s
1
f 3.2 s
P1X: Optics, Waves and Lasers Lectures, 2005-06.
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Example: 13-8 from Y&F
Find the frequency and period of a simple pendulum that is 1.0 m long
(assume g=9.80 m/s).
The angular frequency:
The frequency: f 
The period:
T

g
9.80m / s

 3.13rad / s
L
1.0m

3.13rad / s

 0.4983cycle / s  0.4983Hz
2 2rad / cycle
1
1

 2.007s
1
f 0.4983s
P1X: Optics, Waves and Lasers Lectures, 2005-06.
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Example: Vertical SHM
Vertical oscillations from a spring hanging vertically.
1) At rest: Spring is stretched by Dl such that:
kDl
 mg
2) x above equilibrium:
Fnet  k ( Dl  x )  ( mg )  kx
3) x below equilibrium:
Fnet  k ( Dl  x )  ( mg )  kx
Fnet
Same SHM as in vertical case,
oscillations with angular frequency:
k

m
Equilibrium is at stretched
position Dl instead of x=0
P1X: Optics, Waves and Lasers Lectures, 2005-06.
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Example: 13-6 from Y&F
Shock absorbers of an old car with mass 1000 kg are worn out. When a
person weighing 100 kg climbs into the car, it sinks by 2.8 cm. When the
car is in motion and hits a bump it oscillates. What is the frequency and
period of oscillation?
F
mg
100kg  9.8ms1

 3.5  104 kgs2
The spring constant: k    
x
x
 0.028m
k
3.5  104 kgs2

 5.64rad / s
The angular frequency:  
m
1100kg
The frequency:

5.64rad / s
f 

 0.898cycle / s  0.898Hz
2 2rad / cycle
The period:
T
1
1

 1.11s
1
f 0.898s
P1X: Optics, Waves and Lasers Lectures, 2005-06.
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 Simple Harmonic Motion initiates sinusoidal waves and sets the
boundary conditions for wave motion
• For example, a string attached to a vertical spring
y (t , x  0)  A sin( t   )  y (t , x )  A sin( t  kx)
v
F  kx
mg
• A radio transmitting antenna causes electromagnetic waves by
oscillating molecules
v
P1X: Optics, Waves and Lasers Lectures, 2005-06.
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