MAT 275 Laboratory 6 Forced Equations And Resonance Solution

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MAT 275 Laboratory 6 Forced Equations And Resonance Solution

In this laboratory we take a deeper look at second-order nonhomogeneous equations. We will concentrate
on equations with a periodic harmonic forcing term. This will lead to a study of the phenomenon known
as resonance. The equation we consider has the form
d2y
dt2 + c
dy
dt
+ !2
0y = cos !t: (L6.1)
This equation models the movement of a mass-spring system similar to the one described in Laboratory
5. The forcing term on the right-hand side of (L6.1) models a vibration, with amplitude 1 and frequency
! (in radians per second = 1
2 rotation per second = 60
2 rotations per minute, or RPM) of the plate
holding the mass-spring system. All physical constants are assumed to be positive.
Let !1 =

!2
0
− c2=4. When c < 2!0 the general solution of (L6.1) is
y(t) = e
−1
2 ct(c1 cos(!1t) + c2 sin(!1t)) + C cos (!t − ) (L6.2)
with
C =
1 √
(!2
0
− !2)2 + c2!2
; (L6.3)
=


arctan
(
c!
!2
0
−!2
)
if !0 > !
 + arctan
(
c!
!2
0
−!2
)
if !0 < !
(L6.4)
and c1 and c2 determined by the initial conditions. The first term in (L6.2) represents the complementary
solution, that is, the general solution to the homogeneous equation (independent of !), while the second
term represents a particular solution of the full ODE.
Note that when c > 0 the first term vanishes for large t due to the decreasing exponential factor.
The solution then settles into a (forced) oscillation with amplitude C given by (L6.3). The objectives of
this laboratory are then to understand
1. the effect of the forcing term on the behavior of the solution for different values of !, in particular
on the amplitude of the solution.
2. the phenomena of resonance and beats in the absence of friction.
The Amplitude of Forced Oscillations
We assume here that !0 = 2 and c = 1 are fixed. Initial conditions are set to 0. For each value of !, the
amplitude C can be obtained numerically by taking half the difference between the highs and the lows
of the solution computed with a MATLAB ODE solver after a sufficiently large time, as follows: (note
that in the M-file below we set ! = 1:4).
1 function LAB06ex1
2 omega0 = 2; c = 1; omega = 1.4;
3 param = [omega0,c,omega];
4 t0 = 0; y0 = 0; v0 = 0; Y0 = [y0;v0]; tf = 50;
5 options = odeset(‘AbsTol’,1e-10,’RelTol’,1e-10);
6 [t,Y] = ode45(@f,[t0,tf],Y0,options,param);
7 y = Y(:,1); v = Y(:,2);
8 figure(1)
9 plot(t,y,’b-‘); ylabel(‘y’); grid on;
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2011 Stefania Tracogna, SoMSS, ASU 1
MATLAB sessions: Laboratory 6
10 t1 = 25; i = find(t>t1);
11 C = (max(Y(i,1))-min(Y(i,1)))/2;
12 disp([‘computed amplitude of forced oscillation = ‘ num2str(C)]);
13 Ctheory = 1/sqrt((omega0^2-omega^2)^2+(c*omega)^2);
14 disp([‘theoretical amplitude = ‘ num2str(Ctheory)]);
15 %—————————————————————-
16 function dYdt = f(t,Y,param)
17 y = Y(1); v = Y(2);
18 omega0 = param(1); c = param(2); omega = param(3);
19 dYdt = [ v ; cos(omega*t)-omega0^2*y-c*v ];
When executing this program we get
computed amplitude of forced oscillation = 0.40417
theoretical amplitude = 0.40417
Lines 10-14 deserve some explanation. Line 10 defines a time t1 after which we think the contribution
of the first term in (L6.2) has become negligible compared to the second term. This depends of course
on the parameter values, in particular c. With c = 1 we obtain e−1
2 ct ≃ 3:7 × 10−6 for t = 25, so this
is certainly small enough compared to the amplitude seen on Figure L6a. The index i of time values
larger than t1 is then determined. The quantity Y(i,1) refers to the values of y associated to times
larger than t1 only. The computed amplitude is simply half the difference between the max and the min
values. This value is compared to the theoretical value (L6.3).
0 10 20 30 40 50
−0.5
−0.4
−0.3
−0.2
−0.1
0
0.1
0.2
0.3
0.4
0.5
y
Figure L6a: Forced oscillation.
1. (a) What is the period of the forced oscillation? What is the numerical value (modulo 2) of the
angle defined by (L6.4)?
(b) In this question you are asked to modify the file LAB06ex1.m in order to plot the complementary
solution of (L6.1), that is, the first term in (L6.2). First define in the file the angle
(alpha) using (L6.4), then evaluate the complementary solution yc by subtracting the quantity
C cos(!t − ) from the numerical solution y. Plot the resulting quantity. Does it look
like an exponentially decreasing oscillation? Why or why not? Include the modified M-file
and the corresponding plot.
2. We now consider C as a function of !. We use again !0 = 2, c = 1 and y(0) = y′(0) = 0. The
previous problem determined C for a specific value of !. Here we consider a range of values for !
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2011 Stefania Tracogna, SoMSS, ASU 2
MATLAB sessions: Laboratory 6
and determine numerically the corresponding amplitude C. We then plot the result as a function
of !, together with the theoretical amplitude from (L6.3). You may need the following MATLAB
program.
function LAB06ex2
omega0 = 2; c = 1;
OMEGA = 1:0.02:3;
C = zeros(size(OMEGA));
Ctheory = zeros(size(OMEGA));
t0 = 0; y0 = 0; v0 = 0; Y0 = [y0;v0]; tf = 50; t1 = 25;
for k = 1:length(OMEGA)
omega = OMEGA(k);
param = [omega0,c,omega];
[t,Y] = ode45(@f,[t0,tf],Y0,[],param);
i = find(t>t1);
C(k) = (max(Y(i,1))-min(Y(i,1)))/2;
Ctheory(k) = ??; % FILL-IN
end
figure(2)
plot(??); grid on; % FILL-IN
xlabel(‘\omega’); ylabel(‘C’);
%———————————————————
function dYdt = f(t,Y,param)
y = Y(1); v = Y(2);
omega0 = param(1); c = param(2); omega = param(3);
dYdt = [ v ; cos(omega*t)-omega0^2*y-c*v ];
1 1.5 2 2.5 3
0.2
0.25
0.3
0.35
0.4
0.45
0.5
0.55
w
C
Figure L6b: Amplitude as a function of !
(a) Fill in the missing parts in the M-file LAB06ex2.m and execute it. You should get a figure like
Figure L6b. Include the modified M-file in your lab report.
(b) Examine the graph obtained by running LAB06ex2.m and determine for what (approximate)
value of ! the amplitude of the forced oscillation, C, is maximal. This value of ! is called the
practical resonance frequency. Give the corresponding maximum value of C.
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2011 Stefania Tracogna, SoMSS, ASU 3
MATLAB sessions: Laboratory 6
(c) Determine analytically the value of ! for which the amplitude of the forced oscillation, C, is
maximal by differentiating the expression for C in (L6.3) as a function of !. Compare the
value you find with the value obtained in part (b).
(d) Run LAB06ex1.m with the value of ! found in part (c) (include the graph). What is the
amplitude of the forced oscillation? How does it compare with the amplitude of the forced
oscillation in problem 1.? If you run LAB06ex1.m with any other value of !, how do you
expect the amplitude of the solution to be?
(e) Are the results affected by changes in the initial conditions? Answer this question both numerically
(by modifying the initial conditions in LAB06ex2.m) and theoretically (by analyzing
the expression for C in (L6.3)). Note that the initial conditions for the DE are y0 and v0.
Resonance
We now investigate what happens to the solution (L6.2), and more specifically to the maximal amplitude
C of the forced oscillation, when we let c → 0. The value of ! corresponding to this maximal amplitude is
called pure resonance frequency. When a mechanical system is stimulated by an external force operating
at this frequency the system is said to be resonant.
3. Set c = 0 in LAB06ex2.m.
1 1.5 2 2.5 3
0
2
4
6
8
10
12
14
w
C
computed numerically
theoretical
(a) Explain what happens. What is the maximal amplitude? What is the value of ! yielding the
maximal amplitude in the forced solution? How does this value compare to !0?
(b) Run LAB06ex1.m with c = 0 and ! equal to the value found in part (a). Comment on the
behavior of the solution. Include the graph.
Beats
When c = 0 and ! ̸= !0 , the solution (L6.2) to (L6.1) reduces to
y(t) = c1 cos(!0t) + c2 sin(!0t) + C cos(!t − )
with C = 1
|!2
0
−!2| . If the initial conditions are set to zero, the solution reduces to
y(t) = C(cos(!t) − cos(!0t))
which can be rewritten as
y(t) = 2C sin
(
1
2
(!0 − !)t
)
sin
(
1
2
(!0 + !)t
)
:
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2011 Stefania Tracogna, SoMSS, ASU 4
MATLAB sessions: Laboratory 6
When ! is close to !0 we have that ! + !0 is large in comparison to |!0 − !|. Then sin
( 1
2 (!0 + !)t
)
is a very rapidly varying function, whereas sin
( 1
2 (!0 − !)t
)
is a slowly varying function. If we define
A(t) = 2C sin
( 1
2 (!0 − !)t
)
, then the solution can be written as
y(t) = A(t) sin
(
1
2
(!0 + !)t
)
and we may interpret it as a rapidly oscillating function with period T = 4
!0+! , but with a slowly varying
amplitude A(t). This is the phenomenon known as beats. Note that A(t) and −A(t) are the so-called
“envelope functions”. The period of A(t) is 4
|!0−!| , thus the length of the beats is 2
|!0−!| .
4. To see the beats phenomenon, set c = 0 and ! = 1:8 in LAB06ex1. Also extend the interval of
simulation to 100.
0 20 40 60 80 100
−3
−2
−1
0
1
2
3
y
(a) In LAB06ex1 define the “envelope” function A = 2C sin
( 1
2 (!0 − !)t
)
with C = 1
|!2
0
−!2| .
Plot A in red and −A in green, together with the solution. You should obtain Figure L6c.
Include the modified M-file.
0 20 40 60 80 100
−3
−2
−1
0
1
2
3
y
Figure L6c: Solution and envelope functions
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2011 Stefania Tracogna, SoMSS, ASU 5
MATLAB sessions: Laboratory 6
(b) What is the period of the fast oscillation (that is, the period of sin
( 1
2 (!0 + !)t
)
)? Confirm
your answer by zooming in on the graph of the solution. Include a graph to support your
answer.
(c) What is the length of the beats? Determine the length analytically using the envelope functions,
and numerically from the graph.
(d) Change the value of ! in LAB06ex1 to 1:9 (a value closer to !0) and then ! = 1:6 ( a value
farther away from !0). Include the two graphs. For each of these two values of ! find the
period of the fast oscillation and the length of the beats. How do the periods change compared
to parts (b) and (c)?
(e) If you let ! = 0:5, is the beats phenomenon still present? Why or why not?
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2011 Stefania Tracogna, SoMSS, ASU 6

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