ME 408 – STRING VIBRATIONS – INTERACTIVE WEBSITE
1) WAVE EQUATION FOR THIN STRING VIBRATION FROM NEWTON’S 2ND LAW
f(x,t)
v(x,t)
x
T
T
x=0
x=L

x+dx
fdx
x
T
T
vx+dx
vx
x
dx v
x + dx
Symbols (with units) and Assumptions:










Time, t (seconds).
Location on string, 0 < x < L (meters)
Transverse string displacement, v (meters).
Mass per unit length,  (kilograms/meter)
Externally-applied transverse force per unit length, f (newtons/meter)
String tension, T (newtons).
T is the only elastic attribute.
T is constant during vibrations (very small oscillations in v).
Neglect dissipative (damping) effects, including damping due to motion in air.
For small oscillations we have:
sin() = tan() = 

  2 v
v
v
… therefore:

x dx  x  dx
v x dx  v x  dx

x x 2
x
x
x
Application of Newton’s 2nd law (force = mass x acceleration) to an infinitesimal section of the string
(shown above).
fdx  T x dx  T x  dxv
Substitute in small oscillation approximations to get an equation in terms of v(x,t)
2v
2v 1

c
 f ( x, t )
s
t 2
x 2 
where cs =
T 
is the wave speed (meters/second) (1)
2) CASE OF UNFORCED FREE VIBRATION
2v
2v
… known as the wave equation. This same equation form with different

c
s
t 2
x 2
parameters in cs and v can be used to describe wave / vibration behavior in numerous scenarios including
torsional and longitudinal vibrations of a rod, and sound propagation through fluids, like air.
Unforced case:
Assume a solution of the form: v(x, t) = g(x)h(t). Substitution into the wave equation yields the
following:
g 
h
 c s2
 constant   where
h
g



and
t 2



x 2
(2)
Since h is only a function of t and g is only a function of x, it must be true that the expressions are
constant since the equality is valid for all x and t. This leads to the following ordinary linear differential
equations.
h t   h t   0 and g  x   g x  0
c 2s
(3,4)
At the boundaries of the string we know that v(x = 0, t) = v(x = L, t) = 0 for all t (fixed boundary
conditions). To satisfy these conditions and satisfy the 2nd order differential equation in g we must have
that:
 nx 
g x  sin
 n = 1, 2, …, 
 L 
(5)
with corresponding values for  of n = n 2  2 c 2s L2 . The natural frequencies of the system are given by
n = nc s L radians/second or fn = n/2 = ncs/2L cycles/second or Hertz (Hz). The general solution for
free vibrations will be:

 nx 
v x, t    sin
 C cos n t   D n sin n t 
 L  n
n 1


(6)
where Cn and Dn are determined by the initial conditions v(x, t = 0) and v  x, t  0 according to the
following formulas.
 nx 
C n  2  v x, t  0 sin
 dx
 L 
0
L
Dn 
2
n
L
 nx 
 dx
L 
 v  x, t  0 sin
0
n = 1, 2, …, 
(7)
n = 1, 2, …, 
(8)
If we account for light damping (energy dissipation) in the string over time due to air resistance and
internal friction in the string, the following expression may be used for v(x, t) where n is known as the
modal damping factor.

 nx    n n t
v x, t    sin
C n cos d t   D n sin d t 
e
 L 
n 1


where  d   n 1   2n
(9)
The above expression assumes “light” damping with n less than 1. Thus we see that, with respect to the
position x, the motion of the string v(x,t) will consist of a summation of sinusoidal functions satisfying
the end conditions at all times with each sine shape oscillating at a unique natural frequency n (or d
when considering damping). Which sine functions are present and to what degree depends on the initial
condition.
Example of Free String Vibrations – The Plucked Guitar String.
The case of a plucked string can be treated as a free vibration problem. The string is displaced from its
rest position to an initial displacement condition and then suddenly let go. Take, for example, the
plucking of a guitar string. Either a pick or finger is used to pluck the string. As a first approximation,
we assume that the string has fixed ends, one at the bridge and the other end held fixed at a certain length
by a finger on the fret board. See illustration below. The initial velocity is zero. The initial displacement
will be in the form of two straight sections as shown here for the case when the string is plucked exactly
at its center.
Approximate finger or pick as a point
force that is released at t = 0
v(x,t=0)
x
T
T
x=0
x=L
Thus, the coefficients, Cn and Dn, are found by solving the following equations with A = v(x = L/2, t = 0).
2
Cn 
L
 8 AL n 2 2
L
2A
n

x
2
A
n

x




 

2 2
0 L x sin  L dx  2L2 2 A  L x sin  L dx  8 AL n 

0

L
2
0sin  nx dx  0
n = 1, 2, …, 
Dn 

L n 0
 L 
L2
n  1,5,7,...
n  3,7,9,...
n  2,4,6,...
(10,11)
Only the odd n modes of vibration have nonzero coefficients, Cn. This, perhaps, is obvious since all of
the even n modes of vibration have a zero value at x = L/2 and, at equal distances on either side of x =
L/2, have opposite signs. In other words, for n even: sin(n[L/2-]) = – sin(n[L/2+]). There is no way
that shapes with these characteristics will combine to produce the initial deflection shape that is
symmetric about x = L/2.
Also note how the values of the coefficients Cn decrease for increasing n. Click here to see them graphed.
In fact, this will always be the case for any realistic initial deflection (without any breaks in the string).
This means that the summation in equation (6) or (9) does not need to go to infinity, but can be truncated
at a finite number N.
Audio – Visual Demonstration #1 – Modes of Vibration. First, we will “see and hear” the individual
modes of vibration for cases n = 1, …, 7, for a string whose fundamental frequency is 196 Hz (G3 note).
This corresponds to the 4th string of a guitar, the 1st string of a violin or the 2nd string of the viola. The
animated vibratory motion, of course, is in slow motion. Our eyes really cannot see something moving
back and forth 196 times or more per second. Click here.
The tones you hear begin with the fundamental at G3 (196.00 Hz) followed by its harmonics. This is the
fundamental of the 4th string of guitar, 1st string of violin, and 2nd string of the violin. For a typical steel
guitar string, take its length L to be 0.65 meters, its mass per unit length M to be 5.446 x 10^-4 kg/m and
its tension T to be 35.357 N. f1 = 196 Hz (G3), f2 = 392 Hz (G4), f3 = 3 x f1, f4 = 4 x f1, etc. Note an
important quality of strings is that the natural frequencies are integer multiples of one another. They are
octaves (or partials), which makes them valuable musically since octave intervals have the highest
consonance – they sound good together to the human ear.
Audio – Visual Demonstration #2 – Response of Guitar String Plucked at its Center. Now, referring to
equation (10), we will “see and hear” the guitar string response when it is plucked at its center, x = L/2.
Click here. Goes from N = 1 up to N = 10 modes in the summation. How many modes of vibration does it
take before you cannot hear the difference? In other words, at what value N can we truncate the
summation? How many modes of vibration does it take before you cannot see the difference?
Actually, the guitar string is plucked near one end, near the bridge. An approximate location might be at
x = L/5 as shown in the diagram.
Approximate finger or pick as a point
force that is released at t = 0
v(x,t=0)
x
T
T
x = L/5
x=0
x=L
Thus, the coefficients, Cn and Dn, are found by solving the following equations with A = v(x = L/5, t = 0).
2
Cn 
L
Dn 
L5

0
2
L n
5A
5 A   nx 
 nx 
5
x sin 
x  sin 
dx  2   A 
dx  Karin and Lars - solve it
L
4
4L   L 
 L 
L 5
L
L
 nx 
dx  0
L 
 0sin 
0
n = 1, 2, …, 
(12,13)
In this case even and odd n modes of vibration have nonzero coefficients, Cn. Click here to see them
graphed. Again, note their decreasing amplitude for increasing n.
Audio – Visual Demonstration #3 – Response of Guitar String Plucked at its x = L/5. Now, referring to
equation (12), we will “see and hear” the guitar string response when it is plucked at x = L/5. Click here.
Goes from N = 1 up to N = 10 modes in the summation. How many modes of vibration does it take before
you cannot hear the difference? In other words, at what value N can we truncate the summation? How
many modes of vibration does it take before you cannot see the difference?
Other Guitar Factors
Of course, the tones you have heard do not sound exactly like that of a guitar. There is a lot more that
goes into the sound a guitar makes than just the vibrating string. In fact, one of the end conditions, the
bridge, is really not a fixed condition on the string. It is through vibration of the bridge caused by the
strings that vibratory energy in the strings is transmitted to the guitar body and its air chamber. But, the
bridge vibration amplitude is fairly small relative to that of the string and the fixed end condition is a
good approximation for the string. Even though the guitar body and its air chamber vibrate with less
amplitude than the string they much more effectively couple to the surrounding air to produce the guitar
sound. For an electric guitar (not acoustical electric) there are electromagnetic transducers (known as
pickups) located at discrete locations along and near the string. They produce a voltage signal
proportional to the motion of the string at those discrete locations. Some of the other vibrational factors
creating the acoustical guitar sound are the subject of the lesson on experimental modal analysis.
3) CASE OF FORCED VIBRATION OF A STRING DRIVEN BY A POINT EXCITATION.
Recall that the general equation for a string that is driven by a force per unit length of f(x,t) is:
2
 2v
1
2  v
 c s 2  f ( x, t )
2

t
x
where cs =
T 
is the wave speed (meters/second).
Mathematically, a force applied at a single point x = xs is represented using a Dirac delta function .
f(x,t) = F(t) (x-xs)
where
 0 x  x s
x  x s   
 x  x s
L
and
 x  x dx  1
s
(14)
0
Integration of a function of x multiplied by the Dirac function leads to the following useful result.
L
 gx x  x dx  gx 
s
s
(15)
0
HARMONIC EXCITATION
Utilizing the complex domain with j =
F(t) = Real Part of {F0ejt}
 1 we can express a harmonic force F(t) as the following.
(16)
where F0 may be a complex number denoting amplitude and  is the excitation frequency in
radians/second. The phrase “real part of” is implicit in the following expressions. Consequently, we have
2
2v
1
2  v

c
 x  x s F0 e jt
s
2
2

t
x
(17)
For a linear time-invariant (LTI) system such as this one, we know that if it is driven at a particular
frequency  it will only respond at that frequency under steady state conditions and once initial transient
responses at the natural frequencies n die down (assuming positive damping). We also know that its
response will be a sum of its mode shapes. Hence, it will have the following form.
v x , t  

V
n
n 1
 nx  jt
sin 
e
 L 
(18)
 mx 
If we multiply the equation of motion by sin 
 where m = 1, … , integrate over x from 0 to L, and
 L 
replace m with n we get the following expression for coefficients Vn.
Vn 
F0 sin nx s L 
2
n  2

 1  2  2 j n
 
n
n





for n = 1, … 
(19)
Thus, the response v(x, t) is the following:
 nx s   nx 
F0 sin 
 sin 

2
L   L  jt

v ( x, t ) 
e
 2

n 1  n
 1  2  2 j n

 
n 
n



(20)
POINT EXCITATION BY BOWING – BOWING OF A VIOLIN (OR VIOLA, CELLO, OR STRING BASS)
When a violinist uses their bow to play a violin, somehow a steady motion of the bow in one direction
across the string at one point results in the vibration of the string at its resonant (damped natural)
frequencies. It is not a free vibration case like that of a plucked string. Nor, is it forced harmonic
excitation, as this would result in the string vibrating at the harmonic excitation frequency.
Over a hundred years ago, Helmholtz (1877) showed that the bowed string of a violin may be
approximately described as forming two straight lines with a sharp bend at the point of intersection. This
bend point races around a curved oval-like path back and forth between the two ends of the string, making
one trip for each fundamental period of the vibration. See animation (no sound just basic animation, bow
at L/5).
To understand how this is caused by the bow being drawn straight across the string, we need to
understand something about sliding friction. Part of what we need to understand is that it is a nonlinear
phenomenon and it is this nonlinearity upon which the violinist depends. It is because of this nonlinearity
in sliding friction that the effective excitation force ends of being periodic, containing the resonant
frequencies of the string.
See figure of the relative velocity between the string and bow versus the resulting force of friction
between them.
Linearized damping
Viscous
Friction force (slope is
friction
coefficient c)
becomes
dominant
Sliding (Coulomb)
friction dominant
Relative
velocity
Stiction at
zero velocity
Violin operating
range
The friction force between to objects not moving relative to each other (static case: stiction) is higher than
when they are moving relative to each other (dynamic case: Coulomb friction) at low velocities. As
relative velocity increase, viscous effects start to become dominant and the friction force will increase as
a function of relative velocity.
When the violin bow initially starts to move it pulls the string with it and there is zero relative velocity.
Eventually, the elastic restoring force of the string will overcome the stiction force and the string will start
to return to its equilibrium position. Then there is relative velocity between the bow and string and a
reduced friction force. Due to the inertia of the string it travels through its equilibrium position.
Eventually, elastic restoring forces and the friction force act to slow the string until it becomes “stuck”
once again to the bow. The bow then brings the string with it through equilibrium and the process is
repeated. Depending on where the bow is on the string, somewhere between 0 < x < L, it will spend
different percentages of its cycle moving in unison with the bow or sliding in the opposite direction. In
fact, an approximate formula for the motion of the string as a function of location, x, and time, t, is:
v x , t  

a
n 1
n
 nx 
sin 
 sin n t 
 L 
See animation (where you can choose the location of the bow … perhaps several discrete choices (not
interactive … animation of the string motion as the displacement of the string at the bow position is
traced out as a function of time).
Now, the above discussion does not address:
1) Why does the string complete a cycle of motion at the rate of its fundamental natural frequency even
though it is driven by an external force?
2) Why is the string profile that of two straight lines with a bend that races around?
Approximate answer to #1. Consider the force input from the bow as a function of time. It periodically
varies between that of stiction and Coulomb friction force levels. To gain some insight, one could very
crudely curve-fit the range of friction force levels that are experienced with a straight line whose slope
would represent the linearized viscous damping coefficient. See figure. This slope will be negative.
Applying modal decomposition to the string would break it up into single degree-of-freedom (SDOF)
oscillators, each having a negative linear viscous damping ratio. From SDOF theory we know that such a
system will oscillate at its natural frequency.
Approximate answer to #2. In class.
Other factors to consider.
The above discussion has significantly simplified what really happens, but captures its essence. Here are
some more factors to consider.
The limits on the bowing conditions are the limits on the conditions at which the bend can trigger the
beginning and the end of slippage between bow and string. For each position of the bow, there is a
maximum and minimum bowing force, as shown here (Rossing Figure 10.9). The closer to the bridge the
instrument is bowed, the less leeway the violinist has between minimum and maximum bowing force.
Bowing close to the bridge (sul ponticello) gives a loud, bright tone, but requires considerable bowing
force and the steady hand of an experienced player. Bowing further from the bridge (sul tasto) produces a
gentle tone with less brilliance.
While the speed of the bend around its curved path is essentially independent of the speed and position of
the bow, the amplitude of vibration can increase either by increasing the bow speed or by bowing closer
to the bridge (end of the string). Does this make sense to you? Try to explain it.
The sound radiated from the violin depends more on the alternating transverse force applied to the bridge
(which in turn transmits this force to the top plate) than on motion of the string, itself. The force applied
to the bridge is in the direction of the string. The following animation shows how the alternating
transverse component of this force varies as the string vibrates. See animation. Movie_mode_4. load
movie_bow_2; movie(M_bow_2,3). Needs work. Want top and bottom pictures animated synchronously.
In the ideal case of a completely flexible string vibrating between two fixed end supports, this force has a
sawtooth waveform with a spectrum of harmonics varying in amplitude as 1/n. This can be determined
using Fourier series – See derivation – Lars and Karin, you can work this out. Listen to sound. You can
construct this. Use G3 as fundamental … go up to 7th harmonic (n = 7). In actual practice, the wave
form of the force is modified by the string stiffness, mechanical properties of the bridge, and other factors.
The motion of the top plate, which is the source of most of the violin sound is the result of a complex
interaction between the driving force from the bridge and the various resonances of the violin body. It is
not simply proportional to the force applied to the bridge. Vibration of the violin body is considered in
the supplemental ME 408 website to Topic 7, which reviews the basics of experimental modal analysis.
Other common cases of vibration caused by “bowing” or “stick and slip” motion.
This type of “stick and slip” motion seen between the bow and string is a very common cause of many of
the vibrations we see or hear around us everyday. Examples less pleasant sounding include brake squeel,
chalk screeching on a blackboard and tool chatter in machining operations.
Another pleasant sounding musical instrument, very popular in the 19th century, that uses this principle is
the glass harp. A glass harp consists of a number of drinking glasses filled with different levels of water.
One wets their fingers and the rims of the glasses and then runs their finger along the rims causing the
glass to vibrate at its resonant frequencies. By filling the glasses with different amounts of water, one can
control (or tune) the resonant frequencies to the desired tones.