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At the risk of being repetitious, we also give several relations which are
derivable from the foregoing equations, but which are of significance in
themselves in Fourier transform analysis.
e i
x
(x) dx = 1
(1.155)
e i
x
'(x) dx = - i
(1.156)
e -i
x d
= 2
(x)
(1.157)
1.8 Point Force and Dipole Wave Excitation
We have discussed the physical meaning and analytical modeling of point
forces and dipoles in the context of the wave equation; we have also given
Laplace transform solutions, and introduced and '. We now combine these
ideas to solve for the transverse oscillations of a string excited by a point load at
x = x* with a varying acceleration f(t). That is, we consider
2 u/ t 2 - c 2
2 u/ x 2 = f(t) (x-x*)
(1.158)
1.8.1 Example 1-15. Finite string excited by a time-varying
concentrated point force.
We solve Equation 1.158 subject to the initial flat elevation u(x,0) = 0 and
the static condition u(x,0)/ t = 0 in the interval 0 < x < L. The string is rigidly
pinned at its end points with u(0,t) = 0 and u(L,t) = 0. We multiply each side of
Equation 1.158 by e -st and time integrate the result from t = 0 to , allowing the
solution process to proceed exactly as before. Integration by parts shows that
e -st 2 u/ t 2 dt = s 2 e -st u(x,t) dt - su(x,0) - u(x,0)/ t, but we do encounter a
new f(t) (x-x*) term. It is clear that e -st f(t) (x-x*) dt equals F(s) (x-x*),
where the function of x moves across the integral, and F(s) represents the
transform of f(t) with respect to time. These steps reduce Equation 1.158 to the
ordinary differential equation s 2 U( x ,s ) - c 2 U" ( x ,s ) = F( s ) (x-x*), where
intermediate results were simplified using initial conditions. This new equation
contains (x-x*), responsible for a slope discontinuity in the transverse string.
In elementary courses, this necessitates a break-up of the problem into two parts,
and then, the matching of two solutions assuming continuity of displacement,
plus jump conditions based on momentum considerations.
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