Before we jump to analysis of the
damped systems we need to inspect and analyze the physical behavior of the un-damped
systems which will be subjected to the harmonic force. The harmonic force which
is acting on the system can be written in the following form:
$$F={{F}_{0}}\cos \omega t$$
This force acts on the body of a
mass m of the un-damped system. The equation of motion can be written in the
following form:
$$m\ddot{x}+kx={{F}_{0}}\cos \omega t$$
First we will derive the
homogenous solution of the previous differential equation. Let’s examine the left
hand side of the differential equation. So if we have
$$m\ddot{x}+kx=0$$
then the solution of that
differential equation is:
$${{x}_{h}}(t)={{C}_{1}}\cos {{\omega }_{n}}t+{{C}_{2}}\sin {{\omega }_{n}}t$$
In the assumed solution the
$${{\omega }_{n}}=\sqrt{\frac{k}{m}}$$
is the natural frequency of the system
Now, just to be sure we need to
check if the assumed homogenous solution is the solution of the differential
equation. To do that we need to derive the first and the second derivation of
the homogenous solution:
$$\begin{align}
& {{{\dot{x}}}_{h}}(t)=-{{C}_{1}}{{\omega }_{n}}\sin {{\omega }_{n}}t+{{C}_{2}}{{\omega }_{n}}\cos {{\omega }_{n}}t, \\
& {{{\ddot{x}}}_{h}}(t)=-{{C}_{1}}\omega _{n}^{2}\cos {{\omega }_{n}}t-{{C}_{2}}\omega _{n}^{2}\sin {{\omega }_{n}}t. \\
\end{align}$$
The final step is to insert the
second derivation and the homogenous solution into the differential equation.
$$\begin{align}
& -m\omega _{n}^{2}\left( {{C}_{1}}\cos {{\omega }_{n}}t+{{C}_{2}}\sin {{\omega }_{n}}t \right)+k\left( {{C}_{1}}\cos {{\omega }_{n}}t+{{C}_{2}}\sin {{\omega }_{n}}t \right)=0/:m \\
& -\omega _{n}^{2}\left( {{C}_{1}}\cos {{\omega }_{n}}t+{{C}_{2}}\sin {{\omega }_{n}}t \right)+\frac{k}{m}\left( {{C}_{1}}\cos {{\omega }_{n}}t+{{C}_{2}}\sin {{\omega }_{n}}t \right)=0 \\
& \omega _{n}^{2}=\frac{k}{m} \\
& -\omega _{n}^{2}\left( {{C}_{1}}\cos {{\omega }_{n}}t+{{C}_{2}}\sin {{\omega }_{n}}t \right)+\omega _{n}^{2}\left( {{C}_{1}}\cos {{\omega }_{n}}t+{{C}_{2}}\sin {{\omega }_{n}}t \right)=0 \\
& 0=0. \\
\end{align}$$
The next step is to get the
particular solution. Because the exciting force F(t) is harmonic that means
that the particular solution is also harmonic and has the same frequency ω.
Thus we assume a solution in the form:
$${{x}_{p}}(t)=X\cos \omega t.$$
In the previous equation the X is
a constant that denotes the maximum amplitude of particular solution. To get
the expression of the particular solution we need to substitute the assumed
particular solution into main differential solution. But before that we need to
derive the first and second derivate of the particular solution.
$$\begin{align}
& {{{\dot{x}}}_{p}}(t)=-X\omega \sin \omega t, \\
& {{{\ddot{x}}}_{p}}\left( t \right)=-X{{\omega }^{2}}\cos \omega t. \\
\end{align}$$
Now we can implement the
particular solution into main differential equation. By the substitution we
get:
$$\begin{align}
& -mX{{\omega }^{2}}\cos \omega t+kX\cos \omega t={{F}_{0}}\cos \omega t, \\
& X\cos \omega t\left( k-m{{\omega }^{2}} \right)={{F}_{0}}\cos \omega t/:\cos \omega t, \\
& X\left( k-m{{\omega }^{2}} \right)={{F}_{0}}/:\left( k-m{{\omega }^{2}} \right) \\
& X=\frac{{{F}_{0}}}{k-m{{\omega }^{2}}}=\frac{{{\delta }_{st}}}{1-{{\left( \frac{\omega }{{{\omega }_{n}}} \right)}^{2}}} \\
\end{align}$$
In previous equation the
$${{\delta }_{st}}=\frac{{{F}_{0}}}{k}$$
denotes the deflection of the
mass under a force Fo and is sometimes called static deflection because fo is a
constant static force. The total solution of the differential equation becomes:
$${{x}_{h}}(t)={{C}_{1}}\cos {{\omega }_{n}}t+{{C}_{2}}\sin {{\omega }_{n}}t+\frac{{{F}_{0}}}{k-m{{\omega }^{2}}}\cos \omega t$$
The previous equation isn’t
finished because the unknowns C1 and C2. To determine the values of these two
unknowns we need to apply the boundary conditions.
First boundary conditions:
$$\begin{align}
& x(t=0)={{x}_{0}} \\
& {{x}_{h}}(0)={{C}_{1}}\cos 0+{{C}_{2}}\sin 0+\frac{{{F}_{0}}}{k-m{{\omega }^{2}}}\cos 0, \\
& {{x}_{0}}={{C}_{1}}+\frac{{{F}_{0}}}{k-m{{\omega }^{2}}}, \\
& {{C}_{1}}={{x}_{0}}-\frac{{{F}_{0}}}{k-m{{\omega }^{2}}}, \\
\end{align}$$
Second boundary conditions:
$$\begin{align}
& \dot{x}(t=0)={{{\dot{x}}}_{0}}, \\
& {{{\dot{x}}}_{h}}(0)=-{{C}_{1}}{{\omega }_{n}}\sin 0+{{C}_{2}}{{\omega }_{n}}\cos 0, \\
& {{{\dot{x}}}_{0}}={{C}_{2}}{{\omega }_{n}}, \\
& {{C}_{2}}=\frac{{{{\dot{x}}}_{0}}}{{{\omega }_{n}}}. \\
\end{align}$$
So now that we determine the
values of C1 and C2 we can write the complete solution of the differential
equation 3.3.
$${{x}_{h}}(t)=\left( {{x}_{0}}-\frac{{{F}_{0}}}{k-m{{\omega }^{2}}} \right)\cos {{\omega }_{n}}t+\left( \frac{{{{\dot{x}}}_{0}}}{{{\omega }_{n}}} \right)\sin {{\omega }_{n}}t+\left( \frac{{{F}_{0}}}{k-m{{\omega }^{2}}} \right)\cos \omega t$$
Now let’s examine the formula for
maximum amplitude of the particular solution.
$$X=\frac{{{\delta }_{st}}}{1-{{\left( \frac{\omega }{{{\omega }_{n}}} \right)}^{2}}}$$
We can modify this formula to get
the following expression.
$$\begin{align}
& X=\frac{{{\delta }_{st}}}{1-{{\left( \frac{\omega }{{{\omega }_{n}}} \right)}^{2}}}/:{{\delta }_{st}} \\
& \frac{X}{{{\delta }_{st}}}=\frac{1}{1-{{\left( \frac{\omega }{{{\omega }_{n}}} \right)}^{2}}} \\
\end{align}$$
This formula is very important
because the ratio of X/δ represents the ration of the dynamic to the static
amplitude of motion and is called MAGNIFICATION FACTOR, AMPLIFICATION FACTOR,
or simply AMPLITUDE RATIO.
 |
Magnification factor of an undamped system |
With help of previous graph, the
response of the system can be identified to be of three types and these types
are:
CASE 1.
$$0<\frac{\omega }{{{\omega }_{n}}}<1>
The denominator in amplification factor equation
is positive and the response is given by the particular solution without a change. The
harmonic response of the system is said to be in phase with the external force
as shown below.
CASE 2.
$$\frac{\omega }{{{\omega }_{n}}}>1$$
The denominator is negative in
equation 3.10 and the steady-state solution can be expressed as:
$${{x}_{p}}(t)=-X\cos \omega t$$
Where the amplitude of motion X
is redefined to be a positive quantity as:
$$X=\frac{{{\delta }_{st}}}{{{\left( \frac{\omega }{{{\omega }_{n}}} \right)}^{2}}-1}$$
Since particular solution and the
harmonic force have the opposite signs, the response is said to be 180 degrees
out of the phase with the external force. As the ration between the harmonic
frequency and the natural frequency approaches infinity the amplitude of the
particular solution approaches to zero. So response of the system to a harmonic
force of very high frequency is close to zero.
CASE 3.
$$\frac{\omega }{{{\omega }_{n}}}=1$$
The amplitude X given in by the
equation
$$\frac{X}{{{\delta }_{st}}}=\frac{1}{1-{{\left( \frac{\omega }{{{\omega }_{n}}} \right)}^{2}}}$$
or
$$X=\frac{{{\delta }_{st}}}{{{\left( \frac{\omega }{{{\omega }_{n}}} \right)}^{2}}-1}$$
becomes infinite. This condition,
for which the forcing frequency is equal to the natural frequency of the system
is called resonance. To find the response for this condition, we use following
expression:
$$x(t)={{x}_{0}}\cos {{\omega }_{n}}t+\frac{{{{\dot{x}}}_{0}}}{{{\omega }_{n}}}\sin {{\omega }_{n}}t+{{\delta }_{st}}\left[ \frac{\cos \omega t-\cos {{\omega }_{n}}t}{1-{{\left( \frac{\omega }{{{\omega }_{n}}} \right)}^{2}}} \right]$$
Nema komentara:
Objavi komentar