Standing waves on a string

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