Understanding Engineering Mathematics

(やまだぃちぅ) #1

Show that the solution in this case is


x=Asin(ωt+φ)

whereω^2 =α/m.
In the presence of damping, but no forcing terms, the equation becomes


mx ̈+βx ̇+αx= 0

Find the general solution in this case, and consider the cases of


(i)overdamping,β^2 > 4 α
(ii) critical damping,β^2 = 4 α
(iii)underdamping,β^2 < 4 α


Iff(t) =0 in equation (15.6) then we have an additional forcing term, impelling
the particle to respond not just to the spring tension and the resistance, but to some
additional force (such as gravity for example). One of the most interesting cases occurs
when the forcing term is periodic itself, resulting in an equation of the form


mx ̈+βx ̇+αx=f 0 cosυt

Show that the general solution in this case is


x(t)=Ce−βt/^2 sin

(√
4 α−β^2
2

t+φ

)
+

f 0

(α−υ^2 )^2 +β^2 υ^2

sin(υt+δ)

where tanδ=(α−υ^2 )/(βυ). The first term represents damped oscillatory motion and
will eventually die out ifβ>0. For this reason it is called thetransient solution.The
second term, on the other hand, originating from the forcing term, persists so long as
the forcing term does, with the same frequency, but a modified amplitude and phase
(this is a characteristic oflinear systems– any sinusoidal input is output with the same
frequency but modified amplitude and phase). This second term is therefore referred to
as thesteady statesolution – it is what is left after the transients decay away. Note
that the amplitude of the steady state solution


f 0

(α−υ^2 )^2 +β^2 υ^2

depends on the relation between the dampingβ, the forcing frequencyvand the natural
frequencyα. Show that ifβ^2 < 4 α(i.e. the system is underdamped), then the amplitude
of forced motion has a maximum at


υ=υr=


α−

β^2
2

At this value ofυwe say the system is atresonanceandυr/ 2 πis called theresonant
frequency.

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