As mentioned earlier we are going
to use the Free body diagram to derive the second order differential equation
which describes the motion of the system. The procedure can be summarized as
follows:
1) Select
a suitable coordinate to describe the position of the mass or rigid body in the
system. Use a linear coordinate to describe the linear motion of a point mass
or the centroid of a rigid body, and an angular coordinate to describe the
angular motion of a rigid body.
2) Determine
the static equilibrium configuration of the system and measure the displacement
of the mass or rigid body from its static equilibrium position.
3) Draw
the free body diagram of the mass or rigid body when a positive displacement
and velocity are given to it. Indicate all the active and reactive forces
acting on the mass or rigid body.
4) And
finally apply Newton’s second law of motion to the mass or rigid body shown in
free body diagram. Newton’s second law of motion can be state as follows :
Rate of change of momentum of a
mass is equal to the force acting on it.
If the mass of the system is displaced at a distance x(t) when acted upon by a resultant force in the same direction, Newton’s second law of motion gives
If a mass m is constant, this
equation reduces to:
$$\overrightarrow{F}\left( t \right)=m\frac{{{d}^{2}}\vec{x}\left( t \right)}{d{{t}^{2}}}=m\ddot{\vec{x}}$$
is the acceleration of the mass. Previous
equation can be stated in words as
Resultant force on the mass = mass * acceleration
Where
$$\ddot{\vec{x}}=\frac{{{d}^{2}}\vec{x}\left( t \right)}{d{{t}^{2}}}$$
For a rigid body undergoing
rotational motion Newton’s law gives
$$\vec{M}\left( t \right)=J\ddot{\vec{\theta }}$$
Where $$\vec{M}$$
is resultant moment
acting on the body and
$$\ddot{\vec{\theta }}={{{d}^{2}}\theta \left( t \right)}/{d{{t}^{2}}}\;$$
and θ are resulting angular displacement and angular acceleration respectively. Equation or represents the equation of motion of the vibrating system.
$$\ddot{\vec{\theta }}={{{d}^{2}}\theta \left( t \right)}/{d{{t}^{2}}}\;$$
and θ are resulting angular displacement and angular acceleration respectively. Equation or represents the equation of motion of the vibrating system.
 |
FIGURE 1 - SPRING MASS SYSTEM |
The initial conditions are:
$$\begin{align}
& x(0)={{x}_{0}}, \\
& \dot{x}(0)={{{\dot{x}}}_{0}} \\
\end{align}$$
The assumed solution of
previously derived differential equation can be written in the following form
$$x(t)=A\cos {{\omega }_{n}}t+B\sin {{\omega }_{n}}t$$
By applying initial conditions to
assumed solution we will determine the constants of integration A and B. After
we determine that we can write the assumed solution in the following form
$$x(t)={{x}_{0}}\cos {{\omega }_{n}}t+\frac{{{{\dot{x}}}_{0}}}{{{\omega }_{n}}}\sin {{\omega }_{n}}t$$
As we all know the
$${{\omega }_{n}}=\sqrt{\frac{k}{m}}$$
is the natural frequency of the
system.
An alternate assumed solution can
be written in the following form:
$$x\left( t \right)=A\sin \left( {{\omega }_{n}}t+\phi \right)$$
By expanding the previous equation using
trigonometric identity we will get:
$$\begin{align}
& \sin \left( a+b \right)=\sin a\cos b+\cos a\sin b \\
& x\left( t \right)=A\cos \phi \sin {{\omega }_{n}}t+A\sin \phi \cos {{\omega }_{n}}t \\
\end{align}$$
A and ϕ can be determined from the following
expressions:
$$\begin{align}
& A=\sqrt{x_{0}^{2}+{{\left( \frac{{{{\dot{x}}}_{0}}}{{{\omega }_{n}}} \right)}^{2}}}, \\
& \phi ={{\tan }^{-1}}\left( \frac{{{x}_{0}}{{\omega }_{n}}}{{{{\dot{x}}}_{0}}} \right) \\
\end{align}$$
The free vibration response of a one-degree-of-
freedom system is shown in next figure:
 |
FIGURE 2 - Free vibration response of an undaped one-degree- of-freedom system |
The initial conditions
determine the energy initially present in the system. Potential energy is
continually converted to kinetic energy and vice versa. Since energy is
conserved, the system eventually returns to its initial state with its original
kinetic and potential energies, completing the first cycle of motion. The subsequent
motion duplicates the previous motion. The system takes the same amount of time
to execute its second cycle as it does its first. Since no energy is dissipated
from the system, the system executes cycles of motion indefinitely.
A
motion which exactly repeats after some time is said to be period. A period is
the amount of time it takes the system to execute one cycle. The frequency is
the number of cycles the system executes in a period of time and is the
reciprocal of the period.
$$\begin{align}
& T=\frac{2\pi }{{{\omega }_{n}}}, \\
& f=\frac{{{\omega }_{n}}}{2\pi } \\
\end{align}$$
Example 1.1.
An engine of mass 500 kg is mounted
on an elastic foundation of stiffness 7*10^5N/m. Determine the natural
frequency of the system.
Solution:
The system is modeled as a
hanging mass-spring system. The natural frequency will be determined using the
following expression:
$${{\omega }_{n}}=\sqrt{\frac{k}{m}}=\sqrt{\frac{7\cdot {{10}^{5}}}{500}}=37.4rad/s=5.96Hz$$
Example 2.2
A wheel is mounted on a steel
shaft (G=83*10^9 N/m^2) of length 1.5 m and radius 0.80cm. the wheel is rotated
5 degrees and released. The period of oscillation is observed as 2.3.s
Determine the mass moment of inertia of the wheel.
 |
Figure 3 – Wheel mounted on torsional spring (shaft) |
Solution:
The oscillations of the wheel
about its equilibrium position are modeled as the torsional oscillations of a
disk on a massless shaft, as illustrated in previous figure. The differential
equation is:
$$\begin{align}
& \sum{{{M}_{0}}}=I\ddot{\theta } \\
& -\frac{JG}{L}\theta =I\ddot{\theta } \\
& I\ddot{\theta }+\frac{JG}{L}\theta =0 \\
\end{align}$$
If we divide the previous
equation by I we will get:
$$\ddot{\theta }+\frac{JG}{LI}\theta =0$$
The natural frequency of the
system can be determined from:
$${{\omega }_{n}}=\sqrt{\frac{JG}{IL}}$$
The natural frequency can be also
determined from:
$${{\omega }_{n}}=\frac{2\pi }{T}=\frac{2\pi }{2.3}=2.73rad/s$$
Thus the moment of inertia of the
wheel is calculated form:
$$I=\frac{JG}{L\omega _{n}^{2}}=\frac{\frac{\pi }{2}{{\left( 0.008 \right)}^{4}}\left( 83\cdot {{10}^{9}} \right)}{1.5\cdot 2.73}=47.8kg{{m}^{2}}$$
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