Page 561 - Advanced_Engineering_Mathematics o'neil

15.3 Bessel Functions 541

This can be written
1 ω 2

f (z) + f (z) + f (z) = 0.
z gz
This is in the form of equation (15.8) if we can solve for a, b, and c so that
ω 2
2 2 2 2 2
−(2a − 1) = 1,2c − 2 =−1,b c = , and a − n c = 0.
g
Choose
1 2ω
a = n = 0,c = and b = √ .
2 g
The solution for f (z) is

z z
f (z) = c 1 J 0 2ω + c 2 Y 0 2ω .
g g
√

Now Y 0 2ω z/g →−∞ as z → 0 (which occurs if x → L, the lower end of the chain). This is
not realistic physically, so choose c 2 = 0. Then f (z) must have the form

z
f (z) = c 1 J 0 2ω .
g
Therefore,
z

u(z,t) = c 1 J 0 2ω cos(ωt − δ)
g
so y(x,t) is

L − x
y(x,t) = c 1 J 0 2ω cos(ωt − δ)
g
The frequencies of the normal vibrations are determined by using the fact that the upper end of
the chain does not move. This means that, for all times t, y(0,t) = 0. Assuming that c 1 = 0(or
else the solution vanishes), this requires that

L
J 0 2ω = 0.
g
We will see shortly that the zero order Bessel function of the first kind has infinitely many positive
zeros. If these zeros are labeled ω 1 , ω 2 , ··· in increasing order, then ω must satisfy

L
2ω = ω j
g
for some positive integer j. This means that ω can take on the values

1 g
ω = ω j .
2 L
These are the frequencies of the normal modes of vibration of the end of the chain, one normal
mode for each positive zero of J 0 . The periods of the oscillation are

4π L
.
g
ω j

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October 14, 2010 15:20 THM/NEIL Page-541 27410_15_ch15_p505-562