Hi Brent,
The modeling equation as stated does not include a forcing input, so there is no transfer fucntion to speak of. If we assume a force input, then the differential equation is
m * xddot + k*x = u(t)
Taking Laplace transform of both sides yields
m*(s^2*X(s) - s*xdot(0) - x(0)) + k*X(s) = U(s)
Now we can define the transer function H(s) as
H(s) = 1/(m*s^2 + k)
With this definition, we have
X(s) = H(s)*U(s) + m*(s*xdot(0) + x(0))/(m*s^2 + k)
as the complete expression for X(s) in term of U(s) and the initial conditions.
Not sure where to go from here or what it means to "plot the natural frequency." Can you clarify?
This can all be derived in Matab using the Symbolic Math Toolbox
syms m k t real
syms u(t) x(t)
eq1 = m*diff(x(t),t,2) + k*x(t) == u(t)
Leq1 = laplace(eq1);
syms X U s
Leq1 = subs(Leq1,[laplace(x(t),t,s) laplace(u(t),t,s)],[X U])
X = solve(Leq1,X)
as derived above.
H(s) can be represented as a tf object for specific values of m and k.
