When a string is fixed at both ends and plucked, standing waves called normal modes are produced. The fundamental frequency corresponds to the longest wavelength, which is twice the string length. Higher harmonics have shorter wavelengths and higher frequencies that are integer multiples of the fundamental. An example calculates the wave speed, wavelengths, and frequencies of the first four normal modes of a guitar string with given length, tension, and fundamental frequency.

1. Standing
waves
on
a
string
Key
concepts:
-­‐ a
string
that
is
plucked
with
both
ends
fixed
results
in
standing
waves
on
the
string
(aka
normal
modes)
o amplitudes
are
0
at
x=0
and
x=L
(L
=
length)
-­‐ fundamental
frequency
aka
first
harmonic:
the
lowest
frequency
corresponding
to
the
longest
wavelength
(λ = 2L)
-­‐ wavelength:
𝜆! = 2𝐿/𝑚
where
m
is
a
positive,
nonzero
integer
-­‐ fundamental
frequency:
𝑓! =
!
!
=
!
!!
𝑣 =
!
!!
!
!
,
where
v
=
wavespeed,
T
=
tension,
𝜇
=
linear
mass
density
-­‐ frequency:
fm
=
mf1
Example:
A
guitar
string
is
0.61m
long,
has
a
fundamental
frequency
of
500Hz,
and
a
tension
kept
at
80.0N
a) Find
the
wave
speed
of
the
string
(hint:
find
linear
mass
density)
b) Find
the
wavelengths
and
frequencies
for
the
2nd,
3rd,
and
4th
normal
modes
of
vibration
ANSWER
a) 𝑣 =
!
!
𝑓! =
!
!!
!
!
→ 500 =
!
!(!.!"!)
∗
!".!
!
𝜇 = 0.0013kg/m
𝑣 =
𝑇
𝜇
=
80.0
0.013
= 245m/s
b) for
the
2nd
harmonic:
wavelength:
𝜆! = 2𝐿/𝑚
à
𝜆! =
!!
!
=
!(!.!")
!
= 0.61𝑚
frequency:
fm
=
mf1
à
𝑓! = 2 500 = 1000Hz
For
the
3rd
harmonic:
wavelength:
𝜆! = 2𝐿/𝑚
à
𝜆! =
!!
!
=
!(!.!")
!
= 0.41𝑚
frequency:
fm
=
mf1
à
𝑓! = 3 500 = 1500Hz
For
the
4th
harmonic:
wavelength:
𝜆! = 2𝐿/𝑚
à
𝜆! =
!!
!
=
!(!.!")
!
= 0.31𝑚
frequency:
fm
=
mf1
à
𝑓! = 4 500 = 2000Hz