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.

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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.

The bow and stringString capture

Release

Capture

String motion: travelling “kink” moves at speed c=(T/)

String capture Release

This causes the bridge to rock

The rocking action forces the body to vibrate and sound is radiated.

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.