2. Analytical solution of the equation

When calculating the oscillator spectrum of an accelerogram [R4.05.03], we have to solve the linear differential equation of the second order:

(2.1)\[ \ ddot {q} +2\ mathrm {\ xi}\ mathrm {\ omega}\ dot {q} + {\ mathrm {\ omega}} ^ {2} q=\ mathrm {\ alpha}\ left (t\ right)\]

where	\(q(t)\)	is the relative displacement
	\(\alpha (t)\)	is the acceleration of the movement imposed at the base
	\(\omega\)	is the pulsation of the oscillator
	\(\xi\)	is the reduced damping of the oscillator

With initial conditions on \(q\) and \(\dot{q}\). The solution to this equation is written in the form:

(2.2)\[ q\ left (t\ right) = {\ int} _ {0} _ {0} ^ {t} h\ left (t-\ mathrm {\ tau}\ right)\ mathrm {.} \ mathrm {\ alpha}\ left (\ mathrm {\ tau}\ right) d\ mathrm {\ tau} +q\ left (0\ right) g\ left (t\ right) +\ dot {q}\ dot {q}\ left {q}\ left (q}\ left}\ left}\ left (t\ right)\]

where \(q(0)\) and \(\dot{q}(0)\) are the displacement and the speed at the initial instant.

We are going to give the expressions for \(h(t)\) and \(g(t)\) according to the value of the reduced depreciation \(\xi\).

If \(\xi <1\) (subcritical damping), then:

(2.3)\[ h\ left (t\ right) =\ frac {{e} ^ {} ^ {-\ mathrm {\ xi}\ mathrm {\ omega}}} {\ mathrm {\ omega}\ sqrt {1- {x}} ^ {x} ^ {2}} ^ {2}}}}\ text {2}}}}\ text {sin}}}}\ text {sin}\ left (\ mathrm {\ omega} t\ sqrt {1- {\ mathrm {\ x}}} ^ {2}}\ right)\]

And:

\[\]

: label: eq-4

gleft (tright) = {e} ^ {-mathrm {xi}mathrm {xi}mathrm {omega} tleft (mathrm {omega} tsqrt {1- {mathrm {xi}}}mathrm {xi}}}right) +frac {mathrm {xi}} {sqrt {1- {mathrm {xi}}} ^ {2}}}}text {sin}}left (mathrm {omega} tsqrt {1- {mathrm {xi}}} ^ {2}}right)right]

If \(\xi =1\) (critical damping), then:

\[\]

: label: eq-5

hleft (tright) = {text {te}}} ^ {-mathrm {omega} t}

And:

(2.4)\[ g\ left (t\ right) =\ left (1-\ mathrm {\ omega}\ right) {e} ^ {\ mathrm {\ omega} t}\]

If \(\xi >1\) (over-critical damping):

(2.5)\[ h\ left (t\ right) =\ frac {{e} ^ {-\ mathrm {\ xi}\ mathrm {\ omega} t}} {\ mathrm {\ omega}\ sqrt {{\ mathrm {{e}} ^ {\ mathrm {\ xi}}} ^ {2} -1}}\ text {\ omega}}\ sqrt {{\ mathrm {\ omega}}\ sqrt {{\ mathrm {\ mathrm {\ xi}}} ^ {2} -1}}\ text {.} \ text {sh}\ left (\ mathrm {\ omega} t\ sqrt {{\ mathrm {\ xi}} ^ {2} -1} -1}\ right)\]

And:

(2.6)\[ g\ left (t\ right) = {e} ^ {-\ mathrm {\ xi}\ mathrm {\ xi}\ mathrm {\ omega} t\ left (\ mathrm {\ omega} t\ sqrt {{\ mathrm {\ xi}}}} ^ {2} -1}\ right) +\ frac {\ mathrm {\ omega}}\ left (\ mathrm {\ omega}} t\ sqrt {\ omega}} t\ sqrt {\ mathrm {\ omega}} t\ sqrt {\ mathrm {\ xi}} {\ mathrm {\ xi}} ^ {2} -1}}}\ text {sh}\ left (\ mathrm {\ omega} t\ sqrt {{\ mathrm {\ xi}}} ^ {2} -1}}\ right)\ right]\]