Undaped Forced Vibration

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.