\[f(t)\]
.Simple harmonic oscillations in one dimension, with no damping and no external force the equation
\[m \frac{d^2x}{dt^2}+ k x= A\]
. If a resistive force proportional to speed is present, the equation becomes \[m \frac{d^2x}{dt^2}+r \frac{dx}{dt}+ k x= A\]
. If now a forcing term \[f(t)\]
is present, the equation becomes \[m \frac{d^2x}{dt^2}+r \frac{dx}{dt}+ k x= f(t)\]
.The solution of this equation is the sum of two terms.
1. The complementary solution
\[x_c\]
of the homogeneous equation \[m\frac{d^2x}{dt^2}+r \frac{dx}{dt}+ k x=0\]
.Assume a solution of the form
\[x_c=Ae^{\lambda t}\]
then \[\frac{dx}{dt}=A \lambda e^{\lambda t}, \; \frac{d^2x}{dt^2}=\lambda^2 Ae^{\lambda t}\]
so the equation becomes \[m \lambda^2 Ae^{\lambda t}+r A \lambda e^{\lambda t}+ k A e^{\lambda t}= 0\]
.Divide by
\[e^{\lambda t} \neq 0\]
.\[m \lambda^2 +r \lambda + k = 0 \rightarrow \lambda = \frac{-r \pm \sqrt{r^2-4mk}}{2m}\]
.If
\[r^2-4mk \gt 0\]
then \[\lambda_1, \; \lambda_2\]
are real and distinct, and \[x_p=Ae^{{\frac{-r + \sqrt{r^2-4mk}}{2m}}t}+Be^{{\frac{-r - \sqrt{r^2-4mk}}{2m}}t} \]
.If
\[r^2-4mk \lt 0\]
then \[\lambda_1, \; \lambda_2\]
are complex and distinct, and \[x_p=e^{- \frac{r}{2m}t}(Acos (\frac{\sqrt{4mk-r^2}}{2m}t) + Bsin (\frac{\sqrt{4mk-r^2}}{2m}t) ) \]
If \[r^2-4mk = 0\]
then \[\lambda_1, \; \lambda_2\]
are equal, and \[x_p=e^{- \frac{r}{2m}} (A+Bt ) \]
.Choose the particular solution
\[y_p\]
from this table, then \[x=x_c+x_p\]
..The constants can be found from suitable initial conditions. If there is no resistive force then put
\[r=0\]
in the above solutions.