Ing Ciencias

SECTION 10.3 STEADY PERIODIC SOLUTIONS AND NATURAL FREQUENCIES
In Problems 1-6 we substitute u ( x, t ) = X ( x) cos ω t in utt = a2 uxx (a2 = E/ δ)

and then cancel the factor cos ω t to obtain the ordinary differential equation a2X'' + ω2X = 0 with general solution X ( x) = A cos

ωx ωx . + B sin a a

(*)

It then remains only to apply the given endpoint conditions to determine thenatural (circular) frequencies — the values of ω for which a non-trivial solution exists. 1. Endpoint conditions: X(0) = X(L) = 0

With conditions X(0) = 0 in (*) implies that A = 0, so X ( x) = sin(ω x / a). Then X ( L) = sin(ω L / a) = 0 implies that ω L / a = nπ , an integral multiple of π. Hence the nth natural frequency is ω n = 3. Endpoint conditions: nπ a nπ = L L E . δ

X(0) = X'(L) = 0The condition X(0) = 0 gives A = 0 in (*), so we have X ( x) = sin

ωx , a

so

X ′( x) =

ω ωx cos . a a

Hence the condition X'(L) = 0 implies that ωL/a is an odd integral multiple of π/2. (2n − 1)π a (2n − 1)π E Thus the nth natural frequency is ω n = . = 2L 2L δ 5. Endpoint conditions: ux(0, t) = ku(L, t) + AEux(L, t) = 0

The condition X'(0) = 0 gives B = 0 in (*), so we have X( x) = cos Then

ωx , a

so

u ( x, t ) = cos

ωx cos ω t. a

u x ( x, t ) = − so the other endpoint condition is k cos

ω ωx sin cos ω t , a a

ωL ω ωL cos ω t − AE sin cos ω t = 0. a a a ωL ωL tan = kL. a a

Upon canceling the cos ω t factor, we find that AE

Thus β = ω L / a is a positive root of the equation AEx tan x = kL, and the nth natural frequency is given by β a β Eωn = n = n L L δ where βn is the nth positive root of this equation. 7. Endpoint conditions: u(0, t) = mutt(L, t) + AEux(L, t) + ku(L, t) = 0 The condition u(0, t) = 0 implies that X ( x) = sin

ωx , a

so

X ′( x) =

ω ωx cos . a a

When we substitute u(x, t) = X(x)cos ωt in the endpoint condition at x = L and cancel the cos ωt factor we get -mω2 X(L) + AEX'(L) + kX(L) = 0. Next wesubstitute z = ωL/a, X(L) = sin z,

ω = az/L,
X'(L) = (z/L) cos z.

a2 = E/ δ,

The result simplifies readily to the frequency equation (mEz2 - kδL2)sin z = MEz cos z. If βn is the nth positive root, then the nth natural frequency is ω n =

βna β = n L L

E . δ

In Problems 8-14 we substitute y(x, t) = X(x)cos ωt in ytt + a4 yxxxx = 0 (a4 = EI/ ρ)

and then cancel the factor cos ωt toobtain the ordinary differential equation a4 X(4) - ω2 X = 0 with general solution X ( x) = A cosh

θx θx θx θx + B sinh + C cos + D sin a a a a

(**)

where θ = ω . We then get the natural frequencies of vibration by applying the given endpoint conditions. 9. Endpoint conditions: y(0, t) = yx(0, t) = 0, y(L, t) = yxx(L, t) = 0

Just as in Problem 21 of Section 10.1, the endpoint conditionsX(0) = X'(0) = 0 and X(L) = X''(L) = 0 imply that
2 ωn β  λn = 4 =  n  a  L  4

where β n is the nth positive zero of the frequency equation tanh x = tan x. Therefore the nth natural frequency ωn is given by

β n2 β  ωn =  n  a2 = 2 L  L 
2

EI . ρ yx(L, t) = yxxx(L, t) = 0

11.

Endpoint conditions:

y(0, t) = yx(0, t) = 0,

Here we have the equation X(4) - λX = 0 withλ = ω 2 / a 4 = θ 4 / a 4 = α 4 and endpoint conditions X(0) = X'(0) = X'(L) = X(3)(L) = 0. The left-endpoint conditions readily give C = –A and D = –B in (**), so X ( x) = A cosh α x + B sinh α x − A cos α x − B sin α x. = A ( cosh α x − cos α x ) + B (sinh α x − sin α x ) .

Then the right-endpoint conditions give A (sinh α L + sin α L ) + B ( cosh α L − cos α L ) = 0,

A (sinh α L − sin αL ) + B ( cosh α L + cos α L ) = 0. The determinant of coefficients of A and B must vanish if there is to be a nontrivial solution, so

(sinh α L + sin α L )(cosh α L + cos α L ) − (sinh α L − sin α L )( cosh α L − cos α L )

= 0.

This equation simplifies to 2sinh α L cos α L + 2 cosh α L sin α L = 0, which upon division by cosh α L cos α L gives the frequency equation tanh x + tan x =...