Consider now the massspring oscillator depicted physically in Fig.D.3, and in equivalentcircuit 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 applied to a spring is equal to the spring constant times the displacement :
By Newton's third law of motion (``every action produces an equal and opposite reaction''), we have . That is, the compression force applied by the mass to the spring is equal and opposite to the accelerating force exerted in the negative direction by the spring on the mass. In other words, the forces at the massspring contactpoint sum to zero:
We have thus derived a secondorder 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
(s) &=& sx_0 + v_0s^2 + km
&Delta#Delta;= rs+j&omega#omega;_0 + rsj&omega#omega;_0,
&omega#omega;_0&Delta#Delta;=k/m, and
r&=& x_02 + j v_02&omega#omega;_0
&Delta#Delta;= R_r e^j&thetas#theta;_r, with
R_r &&Delta#Delta;=& v^2_0 + &omega#omega;_0^2 x^2_02&omega#omega;_0,
&thetas#theta;_r &Delta#Delta;= ^1(v_0&omega#omega;_0x_0)
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