Page 304 - Thomson, William Tyrrell-Theory of Vibration with Applications-Taylor _ Francis (2010)

Sec. 9.7 System with Repeated Identical Sections 291

At the ground, the amplitude of motion is zero, X q= 0. Equation (9.7-2) for

mass then becomes
mo)
2 1 = 0 (9.7-6)
~2k
Substituting Eq. (9.7-4) and Eq. (9.7-5) into the previous equation, we obtain
( A cos 2^ -h 5 sin 2j8) - 2 cos p{ A cos -f 5 sin j8) = 0
^(cos2)8 - 2cos^ p) + 5(sin2j3 - 2sin cos (3) = 0
Because cos 2)8 - 2cos^/3 = 1 and sin 2)8 - 2sin )8 cos )8 = 0, we have
^(1) + B{0) = 0
A = 0
and the general solution for the amplitude reduces to
= B sin pn (9.7-7)
At the top, the boundary equation is
= - X n- i)
Because the sections at the two boundaries of the system are outside the domain
of the difference equation, the choice for the value of is arbitrary. However,
we will soon see that the choice ruj^ = m /2 simplifies the boundary equation at
the top. In terms of the amplitudes, the previous equation becomes

(9.7-8)
2k
Substituting from Eqs. (9.7-4) and (9.7-7), we obtain the following equation for
evaluating the quantity )8:
sin )8( A - 1) -h cos )8 sin )8A
This equation then reduces to
sin )8 cos )8A = 0
which is satisfied by
cos )8A = 0
77 3t7 5t7 ( 2 / — 1)77
2 4 A ’* w
The natural frequencies are then available from Eq. (9.7-4) as

(o, = 2\ — s m ^
' \ m 2
[Y . ( 2/ - 1) t7
= 2\ — sm^— jT?—... (9.7-9)
Vm AN