Consider now the mass-spring oscillator depicted physically in Fig.D.3, and in equivalent-circuit form in Fig.D.4.
![]() |
By Newton's second law of motion, the force
applied to a mass
equals its mass times its acceleration:
By Hooke's law for ideal springs, the compression force
By Newton's third law of motion (``every action produces an equal and opposite reaction''), we have
We have thus derived a second-order differential equation governing
the motion of the mass and spring. (Note that
in
Fig.D.3 is both the position of the mass and compression
of the spring at time
.)
Taking the Laplace transform of both sides of this differential equation gives
To simplify notation, denote the initial position and velocity by
and
, respectively. Solving for
gives
denoting the modulus and angle of the pole residue
, respectively.
From §D.1, the inverse Laplace transform of
is
, where
is the Heaviside unit step function at time 0
.
Then by linearity, the solution for
the motion of the mass is
If the initial velocity is zero (
), the above formula
reduces to
and the mass simply oscillates sinusoidally at frequency
, starting from its initial position
.
If instead the initial position is
, we obtain