M. J. Moloney CL-109 Office 812 877 8302 Home 812 466 1328 Revised 11/21/01

**Contents**

Approximate material in Kinsler, Frey, Coppens
and Sanders *Fundamentals of Acousics, *4th Ed.

- Chapters 1, 2, 3, 4, 5, 6, 7, 10

Last 10 days to two weeks devoted to a project

Some possible project ideas

- musical acoustics
- transmission properties of a resonator in a transmission line
- Q of a pipe resonator varying with area
- ultrasonic zone plate (acts as a lens)
- resonant frequencies of pipes with holes in them (spreadsheet theory, build and test)
- absorption coefficient measurement
- horn theory vs. experiment (spreadsheet theory, build and test)

Animation of pulse on a string fixed at
both ends (maple code)

Animation of pulse on a string ; we are not
told the boundary conditions

Driven damped oscillator (graphs, maple
code, problem)

Animation and maple code for simple vibrating
membrane

Transverse waves on rods and bars (chart,
graphs, boundary conditions, maple code)

Elasticity and Bending of Beams

- Bending moment M of a beam in terms of derivatives of y with respect to x
- Equation of motion of a vibrating beam
- Transverse Vibrations of Bars (free-free, clamped-clamped, clamped-free)
- (Digression - critical load for buckling of long thin beam)

This animation shows a pulse traveling on a string fixed at both ends.

The contents of this maple file are :

**strpulse.mws pulse on a string via eigenfunction expansion**
**Use notation of Meeks, p. 47**

**An cos n pi ct/L + Bn sin n pi ct/L**
**An = 2/L integral 0 to L sin n pi x/L y(x,0)**
**Bn = 2/m pi c integral 0 to L v(x,0) cos n pi x/L**

**pulse centered at x=7/8 @t=0 pulse isn't really 0 at boundaries,
but close to it.**
**string from 0 to 2. 11/17/97**

**restart;with(plots):**
**c:=4; # pulse speed (which must be string
wave speed!)**
**y:=3/(2+450*(x-7/8-c*t)^2);**
**animate(y,x=0..2,t=0..1/2,frames=20,color=black);
# this pulse runs off end of the string!**

**vy:=subs(t=0,diff(y,t));**
**N:=30; # of terms in series**
**L:=2; # length of string**
**for m from 1 to N do**
**B[m]:=evalf(2/(m*Pi*c)*Int(vy*sin(m*Pi*x/L),x=0..L));**
**A[m]:=evalf(2/L* Int(subs(t=0, y)*sin(m*Pi*x/L),x=0..L));
od:**

**str:=0:**
**for m from 1 to N do**
**str:=str+(A[m]*cos(m*Pi*c*t/L)+B[m]*sin(m*Pi*c*t/L))*sin(m*Pi*x/L);
od: animate(str,x=0..2,t=0..1,frames=30,color=black);**

What kind of string is this pulse travelling on? Can you modify the program above to give this behavior?

[This string is not fixed at both ends!]

Take-home problem for Exam 1 - modify previous pulse problem to give this behavior. (1 problem of 4, other 3 in class.)

The response curve
for a **driven damped oscillator** is shown below, both for amplitude
vs frequency, and phase vs frequency. The **phase** shown is the amount
by which the **driver leads the response of the driven
oscillator.**

Click here to download the maple file which generates these graphs. After it comes up, you will want to do a 'Save File', and later open the file in Maple.

The commands in this file are as follows:

drivnosc.mws 9/22/97

Plot the phase and amplitude response of a driven, damped oscillator as a function of frequency.

It is interesting that very heavy damping (b=10, k=10, m=1) crushes the response away from zero frequency.

**restart;with(plots):**
**assume(k,real);assume(m,real);assume(A,real);
assume(omega>0);assume(t,real);assume(b,real);assume(phi,real); eqn:=A*exp(I*(omega*t+phi))-k*x(t)-b*diff(x(t),t)
= m*diff(x(t),t,t); > s:=dsolve(eqn,x(t));**
**x:=rhs(s);**
**values:={A=1,m=1,k=10,b=1/5};**
**xs:=subs(values,x);**
**steady:=op(1,xs); **#
keep only the oscillating terms for steady state response** ****ev:=evalc(subs(t=0,phi=0,steady));**

The conventional phase response is the driver phase - the responding phase, so the amplitude response will be found as exp(-I alph)

**assume(alph,real);**
**cosal:=op(1,ev); sinal:=-op(2,ev)/I;**
**alph:=arctan(sinal/cosal);**
**plot(alph+Pi*Heaviside(-alph),omega=2..4,title=`
phase [0->Pi] vs. omega`);**

The driver is in phase with the driven system at very low frequencies, and pulls ahead until it is Pi/2 ahead at resonance, and goes on to be Pi ahead (180 degrees out of phase) at very high frequencies.

The Heaviside function forces the inverse tangent to run between 0 and Pi, rather than between -Pi/2 and +Pi/2.

**xcc:=subs(I=-I,steady); # get complex conjugate**
**xmag:=simplify(steady*xcc); # get magnitude
squared**
**xev:=simplify(evalc(xmag));**
**plot(sqrt(xev),omega=0..6,title=`Response
amplitude vs freq.`);**

**Homework problem P6**

**Turn in this problem instead
of 1.12 on Tues 12/9/97**

**a) Figure out the theoretical Q value for the
oscillator in the problem above.**

**b) Re-plot the amplitude response curve in
Maple from, say, 3 to 4 sec, and determine the Q 'experimentally' by clicking
on the plot at points where the amplitude is 0.707 of the max amplitude.
Show your calculations for Q.**

**c) Double the Q of the system (be sure to tell
me how you did this). Re-plot and 'experimentally' check that the new parameters
resulted in the correct value of Q.**

Here is an animated vibrating rectangular membrane, in its lowest mode

Click here to download the maple file which does this animation. Do a 'Save File', then open the file in Maple.

Here is the code in the maple file:

**restart;with(plots):**
**Lx:=2; Ly:=1;**
**g:=sin(Pi*x/Lx)*sin(Pi*y/Ly);**
**plot3d(g,x=0..Lx,y=0..Ly);**

**c:=100;** # wave speed on membrane
**omega:=c*sqrt((Pi/Lx)^2+(Pi/Ly)^2);**
# lowest mode angular freq
**Nframes:=10; tfac:=2*Pi/(omega*Nframes);**

**for i from 0 to Nframes-1 do**
**plt.(i):=plot3d(sin(omega*i*tfac)*g,x=0..Lx,y=0..Ly);od:**
**display([plt.(0..Nframes-1)],insequence=true);
**#
this is the animation

Transverse waves on bars and rods.

They are solutions of y'''' = k^4 y = w^2 /(kap^2 c^2),

where c = sqrt(Y/rho), and kap = sqrt(int r^2 dA/ int dA).

**frequencies: ** w = k^2 kap c.

same frequencies for free-free and clamped-clamped bars, but different
functions!

free-free bar
y''( +/- L/2)=0 y'''( +/- L/2)=0 |
clamped-clamped
y( +/- L/2) = 0 y'( +/- L/2)=0 |
clamped-free
y(0)=0=y'(0) y''(L)=0=y'''(L) |

even:
solve tan kL/2 = -tanh kL/2 kL =. 3, 7, 11, etc. Pi/2 |
even:
solve tan kL/2 = -tanh kL/2 kL =. 3, 7, 11, etc Pi/2 |
all solutions : solve
cos kL cosh kL = -1 |

odd
solve tan kL/2 = +tanh kL/2 kL = . 5, 9, 13, etc. Pi/2 |
odd
solve tan kL/2 = +tanh kL/2 kL = . 5, 9, 13, etc. Pi/2 |
kL =. 1, 3, 5, 7, etc Pi/2 |

lowest even mode | lowest even mode | lowest mode |

lowest odd mode | lowest odd mode | next-lowest mode |

**Here
is the maple file which produced the graphs shown above. **
Do a 'Save File, then open later in Maple

Some theory on beam bending. .

**Examine the neglect of Ic alpha in the derivation of equation [2]
above, and then obtain an improved equation for the bar, which enables
us to make frequency corrections at high frequencies.**

Vibration frequencies and equations - free-free, clamped-clamped, and clamped-free.

1/8/98