PHY 102: Waves & Quanta
Topic 2
Travelling Waves
John Cockburn (j.cockburn@... Room E15)
•What is a wave?
•Mathematical description of travelling pulses & waves
•The wave equation
•Speed of transverse waves on a string
TRANSVERSE
WAVE
LONGITUDINAL
WAVE
WATER WAVE
(Long + Trans
Combined)
•Disturbance moves (propagates) with velocity v (wave speed)
•The wave speed is not the same as the speed with which the
particles in the medium move
•TRANSVERSE WAVE: particle motion perpendicular to
direction of wave propagation
•LONGITUDINAL WAVE: particle motion parallel/antiparallel
to direction of propagation
No net motion of particles of medium from one region to
another: WAVES TRANSPORT ENERGY NOT MATTER
Mathematical description of a wave pulse
f(x+5) f(x)
f(x-10)
GCSE(?) maths:
1.0
Translation of f(x) by a distance d to
the rightf(x-d)
0.8
y
0.6
0.4
For wave pulse travelling to the right
with velocity v :
0.2
0.0
-8
-6
-4
-2
0
2
4
6
8
10
12
14
X
f(x)
f(x-vt)
1.0
f ( x, t )  f ( x  vt)
d=vt
0.8
function shown is actually:
0.6
y
f ( x, t )  e
0.4
0.2
0.0
0
X
 ( x vt ) 2
Sinusoidal waves
Periodic sinoisoidal wave produced
by excitation oscillating with SHM
(transverse or longitudinal)
Wavelength λ
Every particle in the medium
oscillates with SHM with the
same frequency and
amplitude
Sinusoidal travelling waves: particle motion
Disturbance travels with
velocity v
Travels distance λ in
one time period T
  vT

v
T
v  f
Sinusoidal travelling waves: Mathematical description
Imagine taking “snapshot” of wave at
some time t (say t=0)
Dispacement of wave given by;
 2x 
y ( x, t  0)  A cos

  
If we “turn on” wave motion to the
right with velocity v we have (see
slide 5):
 2 ( x  vt) 
y ( x, t )  A cos




Sinusoidal travelling waves: Mathematical description
 2 ( x  vt) 
y ( x, t )  A cos




We can define a new quantity called the “wave number”, k = 2/λ
y ( x, t )  A cos( kx  kvt)
2f 
v  f 

k
k
y ( x, t )  A cos( kx  t )
NB in wave motion, y is a function of both x and t
The Wave Equation
Curvature of string is a maximum
Particle acceleration (SHM) is a maximum
Curvature of string is zero
Particle acceleration (SHM) is zero
So, lets make a guess that string curvature  particle acceleration at that point……
The Wave Equation
Mathematically, the string curvature is:
And the particle acceleration is:
So we’re suggesting that:
 2 y ( x, t )
x 2
 2 y ( x, t )
t 2
 2 y ( x, t )  2 y ( x , t )

2
x
t 2
The Wave Equation
y ( x, t )  A cos( kx  t )
y ( x, t )
 kAsin( kx  t )
x
y ( x, t )
 A sin( kx  t )
t
 2 y ( x, t )
2


k
A cos( kx  t )
2
x
 2 y ( x, t )
2



A cos( kx  t )
2
t
 2 y( x, t )
x 2
1  2 y( x, t )
 2
v
t 2
Applies to ALL wave motion (not just sinusoidal waves on strings)
Wave Speed on a string
T2y
motion
y
T2
T
Small element
of string
Small element of string
(undisturbed length ∆x)
undergoes transverse motion,
driven by difference in the ycomponents of tension at each
end (x-components equal and
opposite)
T
T1
T1y
∆x
x
x+∆x
Wave Speed on a string
Net force in y-direction:
Fy  T2 y  T1 y
T2y, T1y given by:
 y 
 y 
T2 y  T   ; T1 y  T  
 x  x  x
 x  x
From Newton 2, :
2 y
2 y
Fy  m 2  x 2
dt
dt
 y 
 y 
T 
T 
2 y
 x  x  x
 x  x
 2
x
t
Wave Speed on a string
Now in the limit as ∆x0:
 y 
 y 
T 
T 
2 y
 x  x  x
 x  x
T 2
x
x
So Finally:
2 y  2 y

2
x
T t 2
Comparing with wave equation:
 2 y( x, t )
x 2
1  2 y( x, t )
 2
v
t 2
v
T
