PH 404 Acoustics Winter 1997-98

M. J. Moloney        CL-109        Office 812 877 8302         Home 812 466 1328          Revised    1/8/98

Contents

Syllabus -1st 3 weeks
Syllabus - 2nd 3 weeks

Elasticity and Bending of Beams

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

Press here to download the maple file which generated the animation above of a pulse traveling at 4 m/s along a 2-m string fixed at both ends.  (You will want to do a 'Save File', then open it later in Maple.)

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);

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.

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

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.

 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.