Wave on a string
Any way to calculate the wave speed? What is it likely to depend on?
Amplitude of the wave? Wave length? Mechanical properties of the
string?
All of those options are plausible, but it turns out the wave speed
only depends on mass of the string (rope) and its tension.
Wave on a string
Waves on a string resemble very
much harmonic oscillations of a
mass on a spring.
Tension provides the restoring
force, which wants to make the
oscillations more frequent.
Mass of the string provides the
inertia, which slows down oscillations
and wave propagation.
Speed of a wave on a string
• Speed of a wave on a stretched string under tension, F
 = Dm/DL is the mass per unit length of the string
The wave speed
is found as
v
F
The frequency of oscillations is

f 
v


cf. pendulum
1
F


1
f 
2
k
m
F  mload g  20 N
mstring mstring


L
6
m
How does v change if we increase tension by
a factor of 2? make the string twice thicker?
Frequency of a wave on a guitar string
f 
v


1
F


Wavelength,  = 2L, where L is the length
of the string.
If the string has a mass m, 
= m/L, and we have
1
F
1 F
F
f 


2L m / L 2 Lm
4Lm
Piano: thick, heavy, and long
strings – low pitch.
Tight strings – high pitch.
Compare:
a mass on a
spring.
Spring constant
(restoring force)
F
4L
k
f 
m
vs.
k
A propagating wave communicates motion and carries away
energy.
We can define wave power as amount of energy carried
away from the source (and transported through the medium)
per unit time
Wave power: how much energy is transported per unit time.
Measured in Joules/sec, J/s or Watts W.
For a harmonic wave propagating along a string
P  FkA sin (kx  t )
2
2
If we average it over time, at any position, x, we find that
 sin (kx  t )   1 / 2
2
1
2
Therefore the power averaged over time
P  FkA
2
2
We can plug in
k   / v and F  v to get
1
2 2
P   A v
2
Let’s take a closer look at the
power equation.
What kind of sense does it make?
 is the mass per unit length;
1
1 2
2 2
P   A v  v0 v
2
2
 is the angular frequency of oscillations in the wave
A is the amplitude of the oscillations
v is the wave speed
A = v0
is the amplitude of the velocity of oscillations of
the material of the string along the y-axis
1 2
v0 is the kinetic energy of oscillations per unit length
2
The power is proportional to the mass per unit length, to the
square of the amplitude of oscillations of the velocity of string
material and to the wave speed.
Wave front is a continuous line or a surface connecting
nearby wave crests.
Plane waves
Wave fronts are flat surfaces for
well directed light, radar or
sound beams, which propagate
in one direction without
spreading
Wave fronts are straight lines for
ripples on water surface at shore
line.
Wave front is a continuous line or a surface connecting
nearby wave crests.
Wave fronts are spherical
surfaces for spherical waves
originating from a point source
and propagating in 3D space.
Wave fronts are circles for
waves on water surface
originating from a point source.
Spherical and circular waves becomes less intense as they
travel further away from their source, because the same
power emitted by the source is spread over larger area
(circumference).
Wave intensity is the wave
power per unit area
P
I
A
Measured in
J
W
or 2
2
sm
m
For a plane wave the intensity
remains constant.
For a spherical wave it
decreases with the distance, r,
from the source, like
P
I
2
4r
Power P of the source is 1000 watts.
The energy spreads outward through the
air.
At a distance R from the source the area A
of the spherical front is 4R2
Intensity = number of joules per second
per square meter
= power/area = P/A
Example: 1000 Joules per second of sound energy is generated.
Spherical wave fronts spread outward from sound source at the center.
What is the intensity of sound at a distance of 10 m?
R  10 m
P  1000 W
I  P / A  P /( 4R 2 )  0.8 W/m 2
Is this a rock concert or a whisper?
At the Earth surface the radius of
curvature of the wave front is
150,000,000 km,
The intensity is the same everywhere,
and the wave is practically flat.
It is a spherical wave on the scale of
the Solar system, though.
P
I
2
4r
A light bulb is a source of a
spherical wave.
If you are 5 ft away from the bulb
and walk another 5 ft away the
light intensity decreases by a
factor of 22 = 4.
In terms of intensity of sunlight, it
is about the same as going from
Venus to mars