Download The Vibrating String - Art of Problem Solving

The Vibrating String
Kent Merryfield
Assumptions:
1. A thin, elastic string is stretched between two fixed
supports. It is held under a tension (that is, a force that acts
lengthwise along the string) of size T.
2. The string is of uniform composition, with a constant mass
per unit length of size .
In what follows, let “horizontal” mean the same direction as the
“home” or “rest” position of the string, and let “vertical” be a
direction perpendicular to that. The motions of the string are
small in the following senses:
3. When the string moves, the horizontal displacement of any
particle of the string is insignificantly small compared to its
vertical displacement.
4. The angle  that the string makes with the horizontal is so
small that for all practical purposes we may assume that
sin  = tan  = .
5. The change in the length of the string is so small that for all
practical purposes the tension T remains unchanged.
Under these assumptions, define u(x, t) to be the “vertical”
displacement of the string as a function of x, which represents
the position along the length of the string and of t, which is time.
∂u
At any point x along the string, the string has a slope ∂x .
∂u
The angle  with the horizontal is ∂x = tan     sin .
The horizontal component of the tension is T cos   T.
∂u
The vertical component of the tension is T sin   T ∂x .
Consider a small piece of the string of length dx. The horizontal
component of tension is essentially the same at each end.
∂u
At the left end, the slope is ∂x and the vertical component of
∂u
tension is about –T ∂x .
∂u ∂2u
At the right end, the slope is ∂x + 2 dx .
∂x
∂u ∂2u 
Thus the vertical component of tension about T ∂x + 2 dx  .
∂x





∂2u
The net vertical force on this piece of string is T 2 dx.
∂x
Newton’s law of motion is F = ma.
∂2 u
The mass is m =  dx. The (vertical) acceleration is 2 .
∂t
∂2u
∂2u
T 2 dx =  2 dx
∂x
∂t
Divide by  dx to get the wave equation:
∂2u
T ∂2u
∂2u
∂2u
2
=
or
= c
 ∂x 2
∂t 2
∂t 2
∂x 2
T
. If we measure  in kg/m and T in N = kgm/s2,

then c 2 is measured in m2/s2, so c is in m/s and is a velocity.
where c 2 =
If ƒ and g are completely arbitrary functions, the function
u(x, t) = ƒ(x – ct) + g(x – ct) is a solution of the wave equation at least if you don’t think about the ends of the string.
In our musical instruments, the string is stretched over two fixed
supports that are a distance L apart, leaving a string of length L.
We can use this to say our equation has boundary values:
u(0, t) = 0 for all t, and u(L, t) = 0 for all t.
Separation of Variables
∂2 u
∂2u
2
Our problem is: 2 = c
; u(0, t) = u(L, t) = 0.
∂t
∂x 2
This is a linear homogeneous partial differential equation.
Because it is linear and homogeneous, if you have a set of
functions that are solutions, then any linear combination of those
functions is a solution. We try to find some such basis set of
solutions.
Assume there is a solution of the form u(x, t) = H(x)V(t).
H(x)
V(t)
Then H(x)V(t) = c 2H(x)V(t), or H(x) = c 2 V(t) = .
H(x) = H(x); H(0) = H(L) = 0 is an eigenvalue problem.
We can’t have  ≥ 0 as we would only get H = 0. So assume that
 = –k 2. We then get H(x) = C sin kx, and sin kL = 0.
This forces kL = nπ for some natural number n, so
nπ
n 2 π2
k = L and  = – 2 .
L
n 2 π2 c 2
Then H(t) = –
H(t). This has solutions
L2
 nπct 
nπct 
Hn(t) = An cos L  + Bn sin  L  .







 nπct 
nπct 
The function Hn(t) = An cos L  + Bn sin  L  is a simple



nc
T
harmonic motion of frequency 2L . Since c 2 = , the frequency

n
T
(in time) of this simple harmonic motion is  = 2L
.





The most general solution to the partial differential equation is
u(x, t) =
∞

n=1
nπx 
sin L  Hn(t) , where Hn(t) is as above.





The most important term of this sum is the n = 1 term, in which
the full string vibrates at the fundamental frequency. The terms
with n ≥ 2 create overtones.
1
The fundamental frequency is  = 2L
T
.

This depends on , L, and T.
If , L, and  are known, then T = 4L 22.
For the violin or viola A string, we may assume  = 440 Hz. The
other strings are tuned above or below that in “fifths”. A “pure”
fifth is a frequency ratio of 3:2 or 1.5:1, but an “equally
tempered” fifth is a frequency ratio of 27/12:1 or 1.4983:1.
Calculations of string tensions:
Violin Strings
String
 (Hz)
E
659
A
440
D
294
G
196
L = 32.4 cm
T (lbs)
 (g/cm)
16
0.0066
12
0.0112
9
0.0273
10
c (mph)
956
638
426
284
Viola Strings
String
 (Hz)
A
440
D
294
G
196
C
131
L = 37.1 cm
 (g/cm)
?
0.0104
0.0225
0.0534
c (mph)
?
487
325
217
T (lbs)
?
11
11
11