Page 96 - Essentials of applied mathematics for scientists and engineers

book Mobk070 March 22, 2007 11:7

86 ESSENTIALS OF APPLIED MATHEMATICS FOR SCIENTISTS AND ENGINEERS
and the solution when ω = ω n is

2C
B 2n−1 = C 1 cos(ω n τ) + C 2 sin(ω n τ) − τ cos(ω n τ) (5.72)
ω 2 n

The initial condition on position implies that C 1 = 0. The initial condition that the initial
velocity is zero gives

2C
ω n C 2 − = 0 (5.73)
ω n 2

The solution for B 2n−1 is
2C
B 2n−1 = [sin(ω n τ) − ω n τ cos(ω n τ)] (5.74)
ω 3
n
Superposition now gives

∞
2C

y(ξ, τ) = sin(ω n ξ)[sin(ω n τ) − ω n τ cos(ω n τ)] (5.75)
ω 3
n=1 n
An interesting feature of the solution is that there are an infinite number of natural frequencies,
a
η = [π, 3π, 5π, ..., (2n − 1)π, ...] (5.76)
L
If the system is excited at any of the frequencies, the magnitude of the oscillation will grow
(theoretically) without bound. The smaller natural frequencies will cause the growth to be
fastest.
Example 5.7 (Vibration of a circular membrane). Consider now a circular membrane (like a
drum). The partial differential equation describing the displacement y(t,r,θ) was derived in
Chapter 1.

2
2
∂ y 1 ∂ ∂y 1 ∂ y
a −2 = r + (5.77)
2
∂t 2 r ∂r ∂r r ∂θ 2
Suppose it has an initial displacement of y(0,r,θ) = f (r,θ) and the velocity y t = 0. The
displacement at r = r 1 is also zero and the displacement must be finite for all r,θ,and t.The
length scale is r 1 and the time scale is r 1 /a.r/r 1 = η and ta/r 1 = τ.

We have

2
2
∂ y 1 ∂ ∂y 1 ∂ y
= η + (5.78)
2
∂τ 2 η ∂η ∂η η ∂θ 2