Chapter4-3
4.3 Vibrations of a Symmetric Circular Membrane
Thevibrations of a symmetric circular membrane where the displacement u(r,t) depends on r and t only can be describe by the one-dimensional wave equation in polar coordinates
= c20 < ra, t > 0(4.3-1)
The boundary condition for the membrane clamped at the circumference is
u(a,t) = 0, for t 0
The radially symmetric initial conditions are
u(r,0) = f(r)and(r,0) = g(r), for 0 ra
We assume that u(r,t) can be separated into R(r), a function of r alone, and T(t), a function of t alone.
u(r,t) = R(r) T(t)(4.3-2)
Differentiating the partial derivative, we get
= R = R
= T = T
In terms of R(r) and T(t), Eq. (4.3-1) becomes
R= c2T+ T(4.3-3)
Divide Eq. (4.3-3) by c2RT to obtain
= + = 2(4.3-4)
Since the RHS of Eq. (4.3-4) depends on r only, and the LHS depends on t only, they must equal to a negative constant2. The constant must be negative for non-trivial solution. The ODE on r is
+ = 2R
Multiplying the equation by r2 and rearranging yields
r2+ r+ (r)2R = 0
The ODE is Bessel’s equation of zero order with independent variable (r)
(r)2+ (r)+ (r)2R = 0
The solution for the r coordinate is
R(r)= C1J0(r) + C2 Y0(r)
The constant C2 can be determined from the boundary condition at r = 0 where R(r) must have a finite value since the displacement u(r,t) = R(r) T(t) is finite.
At r = 0, Y0(0) = C2 = 0
At r = a, R(a) = 0 = C1J0(a)
J0(a) must be equal to zero to avoid trivial solution
J0(a) = 0 a must be a root of the Bessel function J0 that has infinite number of roots. The first five roots n of J0 are: 1 = 2.4048, 2 = 5.5201, 3 = 8.6537, 4 = 11.7915, and 5 = 14.9309. The separation constant is then given by
n = , n = 1, 2, 3,
The solution for the r variable with the constant C1 set equal to one is
Rn(r)= J0()
The ODE in terms of t from (4.3-4) is now evaluated
=
The solution for T(t) is then
Tn = Ancos(ct) + Bnsin(ct)
The membrane displacement u(r,t) = R(r) T(t) is given by
u(r,t) = [Ancos(ct) + Bnsin(ct)] J0()
The constants An and Bn can be determined from the initial conditions
t = 0 u(r,0) = f(r) = An J0()(4.3-5)
Bessel functions are orthogonal with the following properties
(r)J0(r)rdr = 0,for m n
(r)2rdr = , where =
Multiplying the initial condition (4.3-5) by J0(r)r and integrating the expression with respect to r yields
J0(r) rdr = An(r)2rdr = An
An = J0(r) rdr
The initial condition on velocity (r,0) is obtained by taking the derivative of the displacement with respect to time
(r,t) = [cAnsin(ct) + cBncos(ct)] J0(r)
t = 0 (r,0) = g(r) = cBn J0()(4.3-6)
Multiplying the initial condition (4.3-6) by J0(r)r and integrating the expression with respect to r yields
J0(r) rdr = cBn(r)2rdr = cBn = cBn
Bn = J0(r) rdr
For a numerical example, let the initial position u(r,0) = f(r) = 0, the initial velocity (r,0) = g(r) = 100 m/s, the membrane radius a = 1 m, and the wave velocity c = 100 m/s.
u(r,t) = [Ancos(ct) + Bnsin(ct)] J0(r), where =
u(r,t) = [Ancos(100t) + Bnsin(100t)] J0(r)
An = J0(r) rdr = 0
Bn = J0(r) rdr = J0(nr) rdr
Bessel functions can be integrated terms by terms from the series to provide the following formula
J0(nr) rdr = =
Bn = =
The final solution for this membrane is
u(r,t) = sin(100t) J0(r)
The Matlab program listed in Table 4.3-1 evaluates u(r,t) at various time using the first five terms of the series. The program finds the first five roots of the Bessel function J0(x) by Newton method. The results are plotted in Figure 4.3-1 for the displacement u(r,t) as a function of r at various time.
____ Table 4.3-1Matlab program to plot u(r,t) = sin(100t) J0(r)_____
if exist('alfa') == 0
% Roots of Bessel function Jo(x)
x=2;alfa=zeros(1,5);
for i=1:5
ex=1;
while abs(ex)>.00001
ex=-besselj(0,x)/besselj(1,x);
x=x-ex;
end
alfa(i)=x;
fprintf('Root # %g = %8.5f ex = %9.6f \n',i,x,ex)
x=x+pi;
end
end
r=[0:20]/20;row=length(r);rt=r';
t=[0 .008 .016 .024 .032 .04];
col=length(t);y=zeros(row,col);
%
% Evaluate the first five terms of the series
%
for n=1:5
a=alfa(n);
y=y+besselj(0,a*rt)*sin(100*a*t)/(a*a*besselj(1,a));
end
y=-2*y;
ax='t=.000t=.008t=.016t=.024t=.032t=.040';
%
% Plot the results
%
for i=1:6
ib=1+(i-1)*6;ie=ib+5;
axi=ax(ib:ie);
axis([0 1 -1 1]);
subplot(3,2,i),plot(r,y(:,i))
xlabel(axi);ylabel('u(r,t)')
end
Root # 1 = 2.40483 ex = -0.000000
Root # 2 = 5.52008 ex = -0.000000
Root # 3 = 8.65373 ex = -0.000004
Root # 4 = 11.79153 ex = -0.000001
Root # 5 = 14.93092 ex = -0.000000
Note: ‘ex’ is the tolerance in the evaluation of the root
Figure 4.3-1 Vibrating circular membrane with radial symmetry
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