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Model 2: Strings on the violin and guitar
The motion of vibrating structures such as the guitar or violin string commonly consists of a linear combination of motion of individual modes. We have sketched the first 3 transverse modes of the bias spring. Each mode has an associated shape and natural frequency. For instance, when we draw back a string on a guitar and release it we hear the fundamental tone of the string along with several overtones. If we know the initial shape of the string and its initial velocity it is possible to describe the subsequent total vibration after its release by summing the contribution from each mode.
2003/ 1
Step 1: Model the problem and find an equation
A musical string can be modelled as an elastic string. We will use the same equation:
2
22
2
2
x
uc
t
u
(5)
The 1 dimensional wave equation
Step 2: Derive a general solution
This has also already been done. We use equation 8:
)cos()sin()cos(sin, tcDtcCxBxAtxu (8)
Step 3: Fit the solution to the boundary conditions
The boundary conditions are the same as before for the spring:
At x=0 u(0,t) = 0
0)cos()sin()()0(,0 tcDtcCBtTXtu
Thus B= 0
L
At x=L u(L,t) = 0 0)cos()sin(sin)()(, tcDtcCLAtTLXtLu
So sin(L) = 0
L = nπ where n= 1,2,3,….
As we found before----
L
ctnb
L
ctna
L
xntxu nnn
cossinsin,
n= 1,2,3,….
(9)
u1(x,t)
u2(x,t)
u3(x,t) and the rest.
Equation 9 describes the motion in each mode:
L
ctnb
L
ctna
L
xntxu nnn
cossinsin,
n= 1,2,3,….
The resultant motion on the string is the sum of modes described by eqn. 9:
L
ctnb
L
ctna
L
xntxu nn
n
cossinsin,
1
(10)
(9)
Step 4: Fit the solution to the initial conditionsWe will model a string that is plucked at the middle.
At t=0 it will look like this:
X=0 X=L
L/2
Appropriate values for an and bn are determined by
the initial conditions: the displacement and velocity at t=0.
At t=0 u(x,0) only contains terms multiplied by bn:
L
cnb
L
cna
L
xnxu nn
n
0*cos
0*sinsin0,
1
0 1
L
xnbxu
nn
sin0,
1(11)
In a similar way we get terms with an only by writing
the derivative of u at t=0:
L
xn
L
cna
dt
xdu n
n
sin
)0,(
1
L
ctnb
L
ctna
L
xntxu nn
n
cossinsin,
1
(10)
Differentiate wrt t
L
ctn
L
cnb
L
ctn
L
cna
L
xn
dt
txdunn
n
sincossin
),(
1
t=0
(12)
Summary:We match the initial conditions (displacement and velocity) to the solution at t=0 to evaluate the an and bn constants.
L
xnbxu
nn
sin0,
1(11)
L
xn
L
cna
dt
xdu n
n
sin
)0,(
1(12)
Initial displacement
Initial velocity
We have a problem! There are an infinite number of an‘s and bn ‘s.
How do we evaluate all of them?
We can use a property of the sine function:
L L
dxL
xm
L
xn
0 2sinsin
if n=m
if n ≠ m (13)
L
dxL
xm
L
xn
0
0sinsin
The trick is to multiply both sides of equations 11 and 12 by sin (mπx) and then integrate between the limits 0 and L. All terms are equal to zero except
those where n =m. We can do this to eqn. 11:
L
xnbxu
nn
sin0,
1
dxL
xm
L
xnbdx
L
xmxu
nn
LL
sinsinsin0,
100
(14)
2
sin0,0
Lbdx
L
xmxu m
L
Rearrange to find the value of the m’th term:
dxL
xmxu
Lb
L
m
sin0,
2
0
(15)
We can also obtain the following expression for the an terms using equation 12:
L
xn
L
cna
dt
xdu n
n
sin
)0,(
1(12)
dxL
xm
dt
xdu
cma
L
m
sin)0,(2
0
(16)
Back to our example: We assume that the guitar string has been pulled back 0.5 L at its middle and then released at t=0.
X=0 X=L
L/2
0.5L
Initial conditions:
for x=0 to L/2 (17) xxu 0,
xLxu 0, for x=L/2 to L (18)
0)0,(
dt
xdufor x=0 to L (19)
Evaluate all bn ‘s. Use eqn. 15 to evaluate m’th term
dxL
xmxu
Lb
L
m
sin0,
2
0
L
L
LL
m dxL
xmxLdx
L
xmx
Ldx
L
xmxu
Lb
2/
2/
00
sinsin2
sin0,2
2sin
42
m
m
L
Evaluate all an ‘s. Use eqn. 16 to evaluate the m’th term
dxL
xm
dt
xdu
cma
L
m
sin)0,(2
0(16)
The velocity is zero at t=0 and we can see from eqn. 16 that all an = 0
too.
0
The final solution:
L
ctn
L
xnn
n
Ltxu
n
cossin2
sin4
,1
2
u(x,t)
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0 0.5 1
Distance x
Am
plit
ud
e
u(x,t)
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0 0.5 1
Distance x
Am
plit
ud
e
Plucked string at t=0 (left) and the first 3 non-zero modes that sum to produce the total motion.
Violin tuning question hint:
Consider the E and A strings:
Fundamental of A is tuned to 440 Hz
E string fundamental is around but not exactly 660Hz.
Beat frequency between fundamentals would be approximately 660-440= 220 Hz. Yet we hear beating of the order of a 1 to 10 Hz as we adjust the E string. This beat frequency reduces to zero as we improve the tuning.