Let’s
consider a system which consists of block with mass m and spring with stiffness
k. On the block acts a force F and this force can be represented by the
following equation.
$$F={{F}_{0}}\sin {{\omega }_{0}}t$$
This
is a periodic force which has an amplitude of F and forcing frequency ω. The
free body diagram for the block when it’s displaced a distance x is shown in
next figure.
By
applying the second Newton’s law we have:
$$\begin{align}
& \sum{{{F}_{x}}=m{{a}_{x}}} \\
& {{F}_{0}}\sin {{\omega }_{0}}t-kx=m\ddot{x}, \\
& m\ddot{x}+kx={{F}_{0}}\sin {{\omega }_{0}}t \\
\end{align}$$
If
we divide the previous differential equation with m we will get:
$$\ddot{x}+\frac{k}{m}x=\frac{{{F}_{0}}}{m}\sin {{\omega }_{0}}t$$
This
equation is a nonhomogeneous-second order differential equation. The general
solution consists of a homogenous and particular solution.
We
will get homogenous solution of the differential equation by assuming solution
of the following differential equation:
$$\ddot{x}+\omega _{n}^{2}x=0$$
In
the previous differential equation the ω is the natural frequency of the system
and it can be determined by following expression:
$${{\omega }_{n}}=\sqrt{\frac{k}{m}}$$
The
assumed solution for the homogenous part of differential equation is:
$$x=A\sin {{\omega }_{n}}t+B\cos {{\omega }_{n}}t,$$
Now
we need to prove that the assumed solution is really solution of the
differential equation but before that we need to derive the first and second
derivation of the assumed solution:
$$\begin{align}
& x=A\sin {{\omega }_{n}}t+B\cos {{\omega }_{n}}t, \\
& \dot{x}=A{{\omega }_{n}}\cos {{\omega }_{n}}t-B{{\omega }_{n}}\sin {{\omega }_{n}}t \\
& \ddot{x}=-A\omega _{n}^{2}\sin {{\omega }_{n}}t-B\omega _{n}^{2}\cos {{\omega }_{n}}t=-\omega _{n}^{2}x \\
\end{align}$$
The
next step is to substitute these values of the assumed solution into differential
equation. Whit this step we will prove if the assumption is correct or wrong.
$$\begin{align}
& -A\omega _{n}^{2}\sin {{\omega }_{n}}t-B\omega _{n}^{2}\cos {{\omega }_{n}}t+\omega _{n}^{2}\left( A\sin {{\omega }_{n}}t+B\cos {{\omega }_{n}}t \right)=0 \\
& -\omega _{n}^{2}x+\omega _{n}^{2}x=0 \\
& 0=0 \\
\end{align}$$
So
as you can see we have proved that previous assumed solution is the solution of
the homogenous part of the differential equation. All that is left is to assume
the particular solution of nonhomogenous differential equation.
The
homogenous part of the solution can be written in the following form:
$${{x}_{h}}=Csin\left( {{\omega }_{n}}t+\phi \right)$$
Since
the motion is period, the particular solution can be determined by assuming a
solution in the following form:
$${{x}_{p}}=X\sin {{\omega }_{0}}t$$
where
X is constant. Taking the second time derivate and substituting into the
nonhomogenous differential equation:
$$\begin{align}
& x=X\sin {{\omega }_{0}}t \\
& \dot{x}=X{{\omega }_{0}}\cos {{\omega }_{0}}t \\
& \ddot{x}=-\omega _{0}^{2}X\sin {{\omega }_{0}}t \\
\end{align}$$
Now
we can prove that the assumed solution of the nonhomogenous differential
equation:
$$\begin{align}
& \ddot{x}+\frac{k}{m}x=\frac{{{F}_{0}}}{m}\sin {{\omega }_{0}}t \\
& -\omega _{0}^{2}X\sin {{\omega }_{0}}t+\frac{k}{m}\left( X\sin {{\omega }_{0}}t \right)=\frac{{{F}_{0}}}{m}\sin {{\omega }_{0}}t \\
& -\omega _{0}^{2}X+\frac{k}{m}X=\frac{{{F}_{0}}}{m} \\
& X\left( \frac{k}{m}-\omega _{0}^{2} \right)=\frac{{{F}_{0}}}{m} \\
& X=\frac{{{F}_{0}}/m}{(k/m)-\omega _{0}^{2}}=\frac{{{F}_{0}}/k}{1-{{\left( {{\omega }_{0}}/{{\omega }_{n}} \right)}^{2}}} \\
\end{align}$$
Substituting
expression for X into particular solution we will obtain:
$$x=\frac{{{F}_{0}}/k}{1-{{\left( {{\omega }_{0}}/{{\omega }_{n}} \right)}^{2}}}\sin {{\omega }_{0}}t$$
Now
that we obtained homogenous and particular solution of the nonhomogenous
differential equation we can write the general solution:
$$\begin{align}
& x={{x}_{h}}+{{x}_{p}}, \\
& x=Csin\left( {{\omega }_{n}}t+\phi \right)+\frac{{{F}_{0}}/k}{1-{{\left( {{\omega }_{0}}/{{\omega }_{n}} \right)}^{2}}}\sin {{\omega }_{0}}t \\
\end{align}$$
The
homogenous solution defines the FREE VIBRATIONS which depends on the natural
frequency of the system and constants C and ϕ. The particular solution
describes the forced vibration of the block caused by the applied force
$$F={{F}_{0}}\sin {{\omega }_{0}}t$$
Since
all the vibrating systems are subjected to the friction, the free vibration
will dampen out after some time. Because of free vibration dependence on time
we can call them transient and forced vibration are steady –state. The reason
why we call forced vibrations steady state is because they are the only
vibrations that remain. As long as force which causes vibrations of the system
is present in the system these vibrations will be active.
