engr3300.003 advanced engineering mathematics
TRANSCRIPT
ENGR3300.003
Advanced Engineering Mathematics
Fall 2021
Massimo (Max) V. Fischetti [email protected]: Joy Roy, [email protected]
Outline of the class:1. Vector calculus
⢠Vector algebra⢠Scalar and vector fields⢠Differential calculus: Gradient, divergence, curl⢠Integral calculus: Line integrals, surface integrals, volume integrals⢠Two important theorems: Divergence, Stokes
2. Fourier analysis: Series, transforms, generalization to âorthogonal functionsâ3. Partial differential equations (PDEs)
⢠Wave equation⢠Diffusion (heat) equation⢠Laplace and Poisson equations⢠Continuity equation
4. Complex analysis
Weâll revisit later⊠Itâs better to deal with it when we shall really need it
Partial Differential Equations (Chapter 12)1. Basic ideas and examples2. The wave equation (Sec. 12.2)
⢠One spatial dimension: A violin string⢠Separation of variables, the solution as a Fourier series (Sec. 12.3)⢠Example: A violin string plucked in the middle
⢠One-dimensional waves in free space (an infinitely-long string)⢠Two spatial dimensions (Sec. 12.8)
⢠Vibration of a square membrane (Sec. 12.9)⢠Laplacian in polar coordinates (Sec. 12.10)⢠Vibrations of a circular membrane (a drum)
⢠Bessel equation and Bessel functions⢠The Fourier-Bessel series
⢠Three spatial dimensions (Sec. 12.11)⢠Laplacian in spherical and cylindrical coordinates
Examples of partial differential equations (PDEs)ðð2ð¢ð¢ððð¡ð¡2
= ðð2ðð2ð¢ð¢ððð¥ð¥2
1D wave equation
ððð¢ð¢ððð¡ð¡
= ð ð ðð2ð¢ð¢ððð¥ð¥2
1D diffusion equation
ðð2ð¢ð¢ððð¥ð¥2
+ðð2ð¢ð¢ðððŠðŠ2
= 0 2D Laplace equation
ðð2ð¢ð¢ððð¥ð¥2
+ðð2ð¢ð¢ðððŠðŠ2
= ðð ð¥ð¥, ðŠðŠ 2D Poisson equation
ðð2ð¢ð¢ððð¡ð¡2
= ðð2ðð2ð¢ð¢ððð¥ð¥2
+ðð2ð¢ð¢ðððŠðŠ2
2D wave equation
ððð¢ð¢ððð¡ð¡
= ð ð ðð2ð¢ð¢ððð¥ð¥2
+ðð2ð¢ð¢ðððŠðŠ2
2D diffusion equation
ðð2ð¢ð¢ððð¥ð¥2
+ðð2ð¢ð¢ðððŠðŠ2
+ðð2ð¢ð¢ððð§ð§2
= 0 3D Laplace equation
ðð2ð¢ð¢ððð¥ð¥2
+ðð2ð¢ð¢ðððŠðŠ2
+ðð2ð¢ð¢ððð§ð§2
= ðð ð¥ð¥, ðŠðŠ, ð§ð§ 3D Poisson equation
ðð2ð¢ð¢ððð¡ð¡2
= ðð2ðð2ð¢ð¢ððð¥ð¥2
+ðð2ð¢ð¢ðððŠðŠ2
+ðð2ð¢ð¢ððð§ð§2
3D wave equation
ððð¢ð¢ððð¡ð¡
= ð ð ðð2ð¢ð¢ððð¥ð¥2
+ðð2ð¢ð¢ðððŠðŠ2
+ðð2ð¢ð¢ððð§ð§2
3D diffusion equation
In compact notation:
â2ð¢ð¢ = 0 3D Laplace equation
â2ð¢ð¢ = ðð ð¥ð¥,ðŠðŠ, ð§ð§ 3D Poisson equation
ðð2ð¢ð¢ððð¡ð¡2
= ðð2â2ð¢ð¢ 3D wave equation
ððð¢ð¢ððð¡ð¡
= ð ð â2ð¢ð¢ 3D diffusion/heat equation
ððððððð¡ð¡
= â â ï¿œ ðð 3D continuity equation
Partial differential equations
⢠Equation in an unknown u(r) with r in a domain D⢠u(r) must satisfy boundary conditions (Laplace, Poisson) and/or
initial (wave, continuity, diffusion) on a âsurfaceâ in the domainFor example: For a circular membrane (that is, a drum), described by its vertical displacement u from equilibrium at position (r,Ï) at time t, u(r,Ï,t), one must specify
u(R,Ï,t) = 0 (the membrane does not move on the rim of the drum at r=R)
u(r,Ï,t=0) = f(r,Ï) (initial position of the membrane)âu(r,Ï,t=0)/ât = g(r,Ï) (initial velocity of the membrane)
initial condition
boundary conditions
Partial differential equations
⢠For a homogeneous linear problemL[u(r)] = 0
if u1(r) and u1(r) are two solutions, that is: L[u1(r)] = 0
and L[u2(r)] = 0,
then:au1(r)+b u2(r)
is also a solution, the same situation we have for ordinary differential equations
The wave equation in one spatial dimension
1. Homogeneous string (constant mass-density Ï)
2. Ignore gravity (strong tension)3. No sliding (no motion along x)
αT1
T2 β
x x+Îx0 Lx
y
Since the string does not slide, the total force along x acting on the infinitesimal element must be zero:
ðð1 cosðŒðŒ = ðð2 cosðœðœ = ðð a constant, the â²tensionâ² (1)The vertical acceleration must be equal to the total force along y:
ðð2 sinðœðœ â ðð1 sinðŒðŒ = ððâð¥ð¥ðð2ð¢ð¢ððð¡ð¡2
(2)Divide Eq. (2) by Eq. (1):
ðð2 sinðœðœðð2 cosðœðœ
âðð1 sinðŒðŒðð1 cosðŒðŒ
= tanðœðœ â tanðŒðŒ = ððâð¥ð¥ðððð2ð¢ð¢ððð¡ð¡2
(3)
But:
tanðœðœ =ððð¢ð¢ððð¥ð¥ ð¥ð¥+âð¥ð¥
and tanðŒðŒ =ððð¢ð¢ððð¥ð¥ ð¥ð¥
(4)
Inserting into Eq. (3):
1âð¥ð¥
ððð¢ð¢ððð¥ð¥ ð¥ð¥+âð¥ð¥
âððð¢ð¢ððð¥ð¥ ð¥ð¥
=ðððððð2ð¢ð¢ððð¡ð¡2
(5)
or:
ðð2ð¢ð¢ððð¥ð¥2
=ðððððð2ð¢ð¢ððð¡ð¡2
or ðð2ðð2ð¢ð¢ððð¥ð¥2
=ðð2ð¢ð¢ððð¡ð¡2
with â²sound velocityâ² ðð =ðððð
u(x+Îx)u(x)
General form of the solutions of the wave equation in one spatial dimension:Any function of the form f(x±ct) is a solution of the wave equation.Indeed:
ðððððð(ð¥ð¥Â±ððð¡ð¡)
=ððððððð¥ð¥
ððð¥ð¥ðð(ð¥ð¥Â±ððð¡ð¡)
=ððððððð¥ð¥
(1)
But also:ðððð
ðð(ð¥ð¥Â±ððð¡ð¡)=ððððððð¡ð¡
ððð¡ð¡ðð(ð¥ð¥Â±ððð¡ð¡)
= ±1ððððððððð¥ð¥
(2)
So, the 2nd derivatives are:ðð2ðð
ðð(ð¥ð¥Â±ððð¡ð¡)2=
ðððð(ð¥ð¥Â±ððð¡ð¡)
ðððððð(ð¥ð¥Â±ððð¡ð¡)
=ðð2ððððð¥ð¥2
using (1)
ðð2ðððð(ð¥ð¥Â±ððð¡ð¡)2
=ðð
ðð(ð¥ð¥Â±ððð¡ð¡)ðððð
ðð(ð¥ð¥Â±ððð¡ð¡)=
1ðð2ðð2ððððð¡ð¡2
using (2)
So:ðð2ððððð¥ð¥2
=1ðð2ðð2ððððð¡ð¡2
General form of the solutions of the wave equation in one spatial dimension
f(x) f(x-ct)
x
f
t=0
Separation of variablesWhenever the coefficients of the PDE (e.g., c2 in our case) do not depend on both x and t, assume:
ð¢ð¢ ð¥ð¥, ð¡ð¡ = ð¹ð¹ ð¥ð¥ ðºðº(ð¡ð¡)
Now let us assume as boundary conditions for the violin string:
ð¢ð¢ 0, ð¡ð¡ = ð¢ð¢ ð¿ð¿, ð¡ð¡ = 0 for any ð¡ð¡
We will also need two initial conditions (âtwoâ, since the PDE is 2nd-order):
ð¢ð¢ ð¥ð¥, 0 = ðð ð¥ð¥ (initial position of the string)
ððð¢ð¢(ð¥ð¥, 0)ððð¡ð¡
= ðð ð¥ð¥ (initial velocity of the string)
This is not the most general solution. However, we shall see that the general solution will be obtained by a linear combination (superposition) of solutions of this form.
Separation of variables-I
ðð2ð¢ð¢(ð¥ð¥, ð¡ð¡)ððð¡ð¡2
= ðð2ðð2ð¢ð¢(ð¥ð¥, ð¡ð¡)ððð¥ð¥2
with b. c.â² s: ð¢ð¢ 0, ð¡ð¡ = ð¢ð¢ ð¿ð¿, ð¡ð¡ = 0
Set: ð¢ð¢ ð¥ð¥, ð¡ð¡ = ð¹ð¹ ð¥ð¥ ðºðº ð¡ð¡Insert into the PDE:
ð¹ð¹(ð¥ð¥)ðð2ðºðº ð¡ð¡ððð¡ð¡2
= ðð2ðºðº(ð¡ð¡)ðð2ð¹ð¹(ð¥ð¥)ððð¥ð¥2
Divide by F(x)G(t):1
ðºðº(ð¡ð¡)ðð2ðºðº ð¡ð¡ððð¡ð¡2
= ðð21
ð¹ð¹(ð¥ð¥)ðð2ð¹ð¹(ð¥ð¥)ððð¥ð¥2
This can be true only if both sides are equal to the same constant K:1
ðºðº(ð¡ð¡)ðð2ðºðº ð¡ð¡ððð¡ð¡2
= ðð21
ð¹ð¹(ð¥ð¥)ðð2ð¹ð¹(ð¥ð¥)ððð¥ð¥2
= ðŸðŸ
Separation of variables-II
ðð2ð¢ð¢(ð¥ð¥, ð¡ð¡)ððð¡ð¡2
= ðð2ðð2ð¢ð¢(ð¥ð¥, ð¡ð¡)ððð¥ð¥2
with b. c.â² s: ð¢ð¢ 0, ð¡ð¡ = ð¢ð¢ ð¿ð¿, ð¡ð¡ = 0
Therefore:1
ðºðº(ð¡ð¡)ðð2ðºðº ð¡ð¡ððð¡ð¡2
= ðŸðŸ and1
ð¹ð¹(ð¥ð¥)ðð2ð¹ð¹(ð¥ð¥)ððð¥ð¥2
=ðŸðŸðð2
or:ðð2ðºðº ð¡ð¡ððð¡ð¡2
â ðŸðŸðºðº ð¡ð¡ = 0
and ðð2ð¹ð¹(ð¥ð¥)ððð¥ð¥2
âðŸðŸðð2ð¹ð¹ ð¥ð¥ = 0 with ð¹ð¹ 0 = ð¹ð¹ ð¿ð¿ = 0
Separation of variables-IIIConsider:
ðð2ð¹ð¹(ð¥ð¥)ððð¥ð¥2
âðŸðŸðð2ð¹ð¹ ð¥ð¥ = 0 with ð¹ð¹ 0 = ð¹ð¹ ð¿ð¿ = 0 .
The boundary conditions can be satisfied only if K<0. Define k2=âK/c2. Then:
ðð2ð¹ð¹(ð¥ð¥)ððð¥ð¥2
+ ðð2ð¹ð¹ ð¥ð¥ = 0 with ð¹ð¹ 0 = ð¹ð¹ ð¿ð¿ = 0
impliesð¹ð¹ ð¥ð¥ = ðŽðŽ sin
ððððð¿ð¿ð¥ð¥
so that:
ðð2 =ððððð¿ð¿
2â ðŸðŸ = âc2
ððððð¿ð¿
2
Separation of variables-IVBack to the other equation:
ðð2ðºðº ð¡ð¡ððð¡ð¡2
+ c2ððððð¿ð¿
2ðºðº ð¡ð¡ = 0
Define ðððð = ðð ððððð¿ð¿
.
Then:
ðºðº ð¡ð¡ = ï¿œðð
ðŽðŽðððð+ððððððð¡ð¡ + ðµðµððððâððððððð¡ð¡ = ï¿œðð
ð¶ð¶ððcos(ððððð¡ð¡) + ð·ð·ððsin(ððððð¡ð¡)
Finally, the general solution is:
ð¢ð¢ ð¥ð¥, ð¡ð¡ = ð¹ð¹ ð¥ð¥ ðºðº ð¡ð¡ = ï¿œðð
ðŽðŽðððð+ððððððð¡ð¡ + ðµðµððððâððððððð¡ð¡ sinððððð¿ð¿ð¥ð¥
These two forms are equivalent, since the (in principle complex) coefficients An, Bn, Cn, and Dn, are yet to be determined
Examples of partial differential equations (PDEs)ðð2ð¢ð¢ððð¡ð¡2
= ðð2ðð2ð¢ð¢ððð¥ð¥2
1D wave equation
ððð¢ð¢ððð¡ð¡
= ð ð ðð2ð¢ð¢ððð¥ð¥2
1D diffusion equation
ðð2ð¢ð¢ððð¥ð¥2
+ðð2ð¢ð¢ðððŠðŠ2
= 0 2D Laplace equation
ðð2ð¢ð¢ððð¥ð¥2
+ðð2ð¢ð¢ðððŠðŠ2
= ðð ð¥ð¥, ðŠðŠ 2D Poisson equation
ðð2ð¢ð¢ððð¡ð¡2
= ðð2ðð2ð¢ð¢ððð¥ð¥2
+ðð2ð¢ð¢ðððŠðŠ2
2D wave equation
ððð¢ð¢ððð¡ð¡
= ð ð ðð2ð¢ð¢ððð¥ð¥2
+ðð2ð¢ð¢ðððŠðŠ2
2D diffusion equation
ðð2ð¢ð¢ððð¥ð¥2
+ðð2ð¢ð¢ðððŠðŠ2
+ðð2ð¢ð¢ððð§ð§2
= 0 3D Laplace equation
ðð2ð¢ð¢ððð¥ð¥2
+ðð2ð¢ð¢ðððŠðŠ2
+ðð2ð¢ð¢ððð§ð§2
= ðð ð¥ð¥, ðŠðŠ, ð§ð§ 3D Poisson equation
ðð2ð¢ð¢ððð¡ð¡2
= ðð2ðð2ð¢ð¢ððð¥ð¥2
+ðð2ð¢ð¢ðððŠðŠ2
+ðð2ð¢ð¢ððð§ð§2
3D wave equation
ððð¢ð¢ððð¡ð¡
= ð ð ðð2ð¢ð¢ððð¥ð¥2
+ðð2ð¢ð¢ðððŠðŠ2
+ðð2ð¢ð¢ððð§ð§2
3D diffusion equation
In compact notation:
â2ð¢ð¢ = 0 3D Laplace equation
â2ð¢ð¢ = ðð ð¥ð¥,ðŠðŠ, ð§ð§ 3D Poisson equation
ðð2ð¢ð¢ððð¡ð¡2
= ðð2â2ð¢ð¢ 3D wave equation
ððð¢ð¢ððð¡ð¡
= ð ð â2ð¢ð¢ 3D diffusion/heat equation
ððððððð¡ð¡
= â â ï¿œ ðð 3D continuity equation
We were solving this equation
Partial differential equations
⢠Equation in an unknown u(r) with r in a domain D⢠u(r) must satisfy boundary conditions (Laplace, Poisson) and/or
initial (wave, continuity, diffusion) on a âsurfaceâ in the domainFor example: For a circular membrane (that is, a drum), described by its vertical displacement u from equilibrium at position (r,Ï) at time t, u(r,Ï,t), one must specify:
u(R,Ï,t) = 0 (the membrane does not move on the rim of the drum at r=R)
u(r,Ï,t=0) = f(r,Ï) (initial position of the membrane)âu(r,Ï,t=0)/ât = g(r,Ï) (initial velocity of the membrane)
initial conditions
boundary conditions
Now we must satisfy the initial conditions:
Example A: initial position given, zero initial velocity:ð¢ð¢ ð¥ð¥, ð¡ð¡ = 0 = ðð ð¥ð¥ (initial position)ððð¢ð¢ ð¥ð¥, ð¡ð¡ = 0
ððð¡ð¡= 0 (initial velocity)
ð¢ð¢ ð¥ð¥, ð¡ð¡ = ï¿œðð
ðŽðŽððcos(ððððð¡ð¡) + ðµðµððsin(ððððð¡ð¡) sinððððð¿ð¿ð¥ð¥
At t=0
ð¢ð¢ ð¥ð¥, 0 = ðð ð¥ð¥ = âðð ðŽðŽðð sin ððððð¿ð¿ð¥ð¥
so:
ðŽðŽðð=2ð¿ð¿ï¿œ0
ð¿ð¿dð¥ð¥ ðð ð¥ð¥ sin
ððððð¿ð¿ð¥ð¥ =
1ð¿ð¿ï¿œâð¿ð¿
ð¿ð¿dð¥ð¥ ðð ð¥ð¥ sin
ððððð¿ð¿ð¥ð¥
Complex exponentials written as trig functionsOf course, the coefficients An and Bn are different from those in a previous slide, but they are unknowns, so we may use the same symbols
All Bn are zero because of the initial condition on the velocity:At t=0, âu(x,0)/ât = Σn Bn Ïn sin (nÏx/L) = 0
assuming a period 2Lwith and odd extension of f(x) for x<0
Example B: initial velocity given, zero initial displacement:ð¢ð¢ ð¥ð¥, ð¡ð¡ = 0 = 0 (initial position)
ððð¢ð¢(ð¥ð¥, ð¡ð¡ = 0)ððð¡ð¡
= ðð ð¥ð¥ initial velocitySince
ððð¢ð¢(ð¥ð¥, ð¡ð¡)ððð¡ð¡
= ï¿œðð
âðŽðŽððððððsin(ððððð¡ð¡) + ðµðµððððððcos(ððððð¡ð¡) sinððððð¿ð¿ð¥ð¥
then, at t=0:ððð¢ð¢(ð¥ð¥, 0)ððð¡ð¡
= ï¿œðð
ðµðµðððððð sinððððð¿ð¿ð¥ð¥ = ðð(ð¥ð¥)
which implies:
ðµðµðð =2ððððð¿ð¿
ᅵ0
ð¿ð¿dð¥ð¥ ðð ð¥ð¥ sin
ððððð¿ð¿ð¥ð¥
All An are zero because of the initial condition on the position:At t=0, u(x,0) = Σn An sin (nÏx/L) = 0
Has the solution the form of a wave, f(x±ct)?Consider case A (given initial position):
ð¢ð¢ ð¥ð¥, ð¡ð¡ = ï¿œðð
ðŽðŽðð cos(ððððð¡ð¡) sinððððð¿ð¿ð¥ð¥
Use the trigonometric identity
cosðŒðŒ sinðœðœ =12
sin ðŒðŒ + ðœðœ + sin(ðŒðŒ â ðœðœ)so:
ð¢ð¢ ð¥ð¥, ð¡ð¡ =12ï¿œðð
ðŽðŽðð sin ððððð¡ð¡ +ððððð¿ð¿ð¥ð¥ + sin ððððð¡ð¡ â
ððððð¿ð¿ð¥ð¥
Recall that ðððð = ððððððð¿ð¿
. Thus:
ð¢ð¢ ð¥ð¥, ð¡ð¡ =12ï¿œðð
ðŽðŽðð sinððððð¿ð¿
(ð¥ð¥ + ððð¡ð¡) â sinððððð¿ð¿
(ð¥ð¥ â ððð¡ð¡)
which is indeed of the general form ð¢ð¢ ð¥ð¥, ð¡ð¡ = ðð ð¥ð¥ ± ððð¡ð¡ .Case B is similar. Good exercise to work it out.
A summary of what we have done:Given the equation:
ðð2ð¢ð¢ððð¡ð¡2
= ðð2ðð2ð¢ð¢ððð¥ð¥2
with boundary and initial conditions:1. Separate variables: u(x,t) = F(x) G(t)2. Expand F(x) into eigenfunctions sn(x) (sines in our case) that satisfy the boundary
conditions3. Determine Gn(t), obtaining a general solution that satisfies the boundary conditions:
u(x,t) = Σn [An cos (Ïnt) + Bn sin (Ïnt)] sn(x) 4. Fix the coefficients An and Bn so that the initial conditions are satisfied
Example: A violin string plucked in the middle-I
ð¢ð¢ ð¥ð¥, ð¡ð¡ = ï¿œðð
ðŽðŽðð cos(ððððð¡ð¡) + ðµðµðð sin(ððððð¡ð¡) sin ððððð¥ð¥
ðððð =ððð¿ð¿ðð
ðððð = ðððððð = ððððð¿ð¿ðð
LL/2
dx
u
0
ð¢ð¢ ð¥ð¥, 0 =2ððð¿ð¿ð¥ð¥ 0 †ð¥ð¥ â€
ð¿ð¿2
= 2ðð â2ððð¿ð¿ð¥ð¥
ð¿ð¿2†ð¥ð¥ †ð¿ð¿
Example: A violin string plucked in the middle-IIAt t=0:
ð¢ð¢ ð¥ð¥, ð¡ð¡ = 0 = ï¿œðð
ðŽðŽðð sin ððððð¥ð¥
So:
ðŽðŽðð=2ð¿ð¿ï¿œ0
ð¿ð¿dð¥ð¥ ð¢ð¢ ð¥ð¥, 0 sin
ððððð¿ð¿ð¥ð¥
andðµðµðð= 0 because the initial velocity is zero, as we saw before
Example: A violin string plucked in the middle-III
ðŽðŽðð=2ð¿ð¿ï¿œ0
ð¿ð¿dð¥ð¥ ð¢ð¢ ð¥ð¥, 0 sin
ððððð¿ð¿ð¥ð¥ with
Set:
ðŠðŠ =ððððð¿ð¿ð¥ð¥; ð¥ð¥ =
ð¿ð¿ðððð
ðŠðŠ; dð¥ð¥ =ð¿ð¿ðððð
dðŠðŠThen:
ðŽðŽðð=2ð¿ð¿ï¿œ0
ð¿ð¿dð¥ð¥ ð¢ð¢ ð¥ð¥, 0 sin
ððððð¿ð¿ð¥ð¥ =
2ð¿ð¿ï¿œ0
ð¿ð¿/2dð¥ð¥
2ððð¿ð¿
ð¥ð¥ sinððððð¿ð¿ð¥ð¥ +
2ð¿ð¿ï¿œð¿ð¿/2
ð¿ð¿dð¥ð¥ 2ðð â
2ððð¿ð¿ð¥ð¥ sin
ððððð¿ð¿ð¥ð¥ =
=2ð¿ð¿2ððð¿ð¿ï¿œ0
ðððð/2dðŠðŠ
ð¿ð¿ðððð
ð¿ð¿ðððð
ðŠðŠ sinðŠðŠ +2ð¿ð¿ï¿œðððð/2
ððððdðŠðŠ
ð¿ð¿ðððð
2ðð â2ðððððð
ðŠðŠ sinðŠðŠ =
=4ðððð2ðð2
ᅵ0
ðððð/2dðŠðŠ ðŠðŠ sinðŠðŠ +
4ðððððð
ï¿œðððð/2
ððððdðŠðŠ 1 â
1ðððð
ðŠðŠ sinðŠðŠ =
=4ðððð2ðð2
ᅵ0
ðððð/2dðŠðŠ ðŠðŠ sinðŠðŠ +
4ðððððð
ï¿œðððð/2
ððððdðŠðŠ sinðŠðŠ â
4ðððð2ðð2
ï¿œðððð/2
ððððdðŠðŠ ðŠðŠ sinðŠðŠ
ð¢ð¢ ð¥ð¥, 0 =2ððð¿ð¿ð¥ð¥ 0 †ð¥ð¥ â€
ð¿ð¿2
= 2ðð â2ððð¿ð¿ð¥ð¥
ð¿ð¿2†ð¥ð¥ †ð¿ð¿
Example: A violin string plucked in the middle-IVNow, integrating by parts:
ï¿œðððŠðŠ ðŠðŠ sinðŠðŠ = âðŠðŠ cos ðŠðŠ + ï¿œðððŠðŠ cos ðŠðŠ = âðŠðŠ cos ðŠðŠ + sinðŠðŠ
So:
ðŽðŽðð =4ðððð2ðð2
ᅵ0
ðððð/2dðŠðŠ ðŠðŠ sinðŠðŠ +
4ðððððð
ï¿œðððð/2
ððððdðŠðŠ sinðŠðŠ â
4ðððð2ðð2
ï¿œðððð/2
ððððdðŠðŠ ðŠðŠ sinðŠðŠ =
=4ðððð2ðð2
âðŠðŠ cos ðŠðŠ + sinðŠðŠ 0ðððð/2 â
4ðððð2ðð2
âðŠðŠ cos ðŠðŠ + sinðŠðŠ ðððð/2ðððð â
4ðððððð
cos ðŠðŠ ðððð/2ðððð =
=4ðððð2ðð2
âðððð2
cosðððð2
+ sinðððð2
â4ðððð2ðð2
âðððð cos ðððð + sin ðððð +ðððð2
cosðððð2
â sinðððð2
â4ðððððð
cos ðððð â cosðððð2
Example: A violin string plucked in the middle-VI
The sound velocity (i.e., the propagation speed of waves on the string):
ðð =ðððð
The fundamental frequency of the string:
ðð1 =ðð12ðð
=1
2ðððððð1 =
12ðð
ððððððð¿ð¿
To tune:1. Adjust the tension T (guitar, violin, and all other string instruments)2. Adjust the mass (thickness of the string; higher mass, lower note)3. Adjust the length (organ pies: longer pipes, lower notes)