Central difference method for SDOF systems

Let’s say that the vibrations of the system are described by the following governing equation:

$$m\frac{{{d}^{2}}x}{d{{t}^{2}}}+c\frac{dx}{dt}+kx=F\left( t \right)$$

Let the duration over which the solution of the previous equation is required be divided into n equal parts of interval h = ∆t each. To obtain a satisfactory solution, we must select a time step ∆t that is smaller than critical time step (∆t_cri)^2. Let the initial conditions be given by x(t=0) = x_0 and dx(t=0)/dt = dx_0/dt.

Replacing the derivatives by the central differences and writing differential equation at grid point I gives:

The previous formula is called the recurrence formula and this formula permits to calculate the displacement of the mass (x_i+1) if we know the previous history of displacements at t_i and t_i-1 as well as the present external force F_i. Applying the previous formula continuously we’ll get the complete time history of the behavior of the system. Certain care has to be exercised in applying previous equation when i = 0 since both x_0 and x_-1 are needed in finding x_1, and the initial conditions provide only the values of x_0 and dx_0/dt, we need to find the value of x_-1.

Substituting the known values of x_0 and dx_0/dt into differential equation which describes the motion of the system we get the following:

$$\begin{align} & m\frac{{{d}^{2}}x}{d{{t}^{2}}}+c\frac{dx}{dt}+kx=F\left( t \right) \\ & x(t=0)={{x}_{0}} \\ & \dot{x}(t=0)={{{\dot{x}}}_{0}} \\ & m{{{\ddot{x}}}_{0}}+c{{{\dot{x}}}_{0}}+k{{x}_{0}}=F(t=0) \\ & m{{{\ddot{x}}}_{0}}=F(t=0)-c{{{\dot{x}}}_{0}}-k{{x}_{0}}/:m \\ & {{{\ddot{x}}}_{0}}=\frac{1}{m}\left( F(t=0)-c{{{\dot{x}}}_{0}}-k{{x}_{0}} \right) \\ \end{align}$$

Now it’s time to apply the central difference approximation using the formulas:

$$\begin{align} & {{{\dot{x}}}_{i}}=\frac{1}{2h}\left( {{x}_{i+1}}-{{x}_{i-1}} \right) \\ & {{{\ddot{x}}}_{i}}=\frac{1}{{{h}^{2}}}\left( {{x}_{i+1}}-2{{x}_{i}}+{{x}_{i-1}} \right) \\ \end{align}$$

For i = 0 the formula can be rewritten as:

$$\begin{align} & {{{\dot{x}}}_{0}}=\frac{1}{2h}\left( {{x}_{1}}-{{x}_{-1}} \right) \\ & {{{\ddot{x}}}_{0}}=\frac{1}{{{h}^{2}}}\left( {{x}_{1}}-2{{x}_{0}}+{{x}_{-1}} \right) \\ \end{align}$$

Inserting them into differential equation we obtain value of x_-1:

$${{x}_{-1}}={{x}_{0}}-\Delta t{{\dot{x}}_{0}}+\frac{{{\left( \Delta t \right)}^{2}}}{2}{{\ddot{x}}_{0}}$$