Contents

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

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.

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);

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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.)

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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.`);

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Homework problem P6

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

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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!

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

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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.

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

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1/8/98