Example 1.1.
A
spring is stretched 100 mm by a 10 kg block. If the block is displaced 50 mm
downward from its equilibrium position and given a downward velocity of 3 m/s
determine the differential equation which describes the motion. Assume that
positive displacement is downward. Also, determine the position of the block
when t=4 s.
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Solution:
From previous figure we can derive the differential equation.
$$\begin{align}
& +\downarrow \sum{{{F}_{y}}}=m{{a}_{y}} \\
& mg-k\left( y+{{y}_{st}} \right)=m\ddot{y} \\
\end{align}$$
where
$$k{{y}_{st}}=mg$$
Now we need to
substitute the previous equation into the differential equation. With this
operation we get:
$$\begin{align}
& mg-ky-mg=m\ddot{y}, \\
& \ddot{y}+\frac{k}{m}y=0, \\
\end{align}$$
Before we can
determine the natural frequency of the
system we need to determine the stiffness of the spring. We know that the force
the stretch the spring is proportion to the stiffness of the spring and the
elongation of the spring. In this case we don’t have a force which acts on
spring but we have a block which is attached to the sprig. So we will determine
the stiffness of the spring by following expression
$$\begin{align}
& F=ky,F=G=mg \\
& k=\frac{G}{y}=\frac{mg}{y}=\frac{10\cdot 9.81}{0.1}=981\text{ N/m}{{\text{m}}^{2}} \\
\end{align}$$
The
next step is to determine the natural frequency and we will do that by
following expression:
$$\omega =\sqrt{\frac{k}{m}}=\sqrt{\frac{981}{10}}=9.9045\text{ }{{\text{s}}^{-1}}$$
When
we have determined the natural frequency we can insert it into differential
equation. Whit this operation we get:
$$\begin{align}
& \ddot{y}+{{(9.9045)}^{2}}y=0, \\
& \ddot{y}+98.1y=0, \\
\end{align}$$
All
that is left is to apply the boundary conditions. At
$$t=0,y=0,05$$
By applying this
boundary conditions we get:
$$\begin{align}
& 0.05=A\sin 0+b\cos 0 \\
& b=0.05 \\
\end{align}$$
At
$$t=0,{{v}_{0}}=3m/s$$
$$\begin{align}
& {{v}_{0}}=A\omega \cos 0-0=A \\
& A=\frac{{{v}_{0}}}{\omega }=\frac{3}{9.9045}=0.30289 \\
\end{align}$$
Example
1.2.
A 3 kg block is
suspended from a spring having a stiffness of k=200N/m. If the block is pushed
50 mm upward from its equilibrium position and then released from rest,
determine the equation that describes the motion. What are the amplitude and
the frequency of the vibration? Assume that the positive displacement is
downward.
From the given data we
can determine the frequency of the vibrations.
$$\begin{align}
& \omega =\sqrt{\frac{k}{m}}=\sqrt{\frac{200}{3}}=8.165 \\
& f=\frac{\omega }{2\pi }=\frac{8.165}{2\pi }=1.299=1.3Hz \\
& x=A\sin \omega t+B\cos \omega t, \\
\end{align}$$
Boundary conditions
are:
$$\begin{align}
& t=0,x=-0.05, \\
& t=0,v=0 \\
\end{align}$$
By applying the
boundary condition to the solution of the differential equation we get:
$$\begin{align}
& 0.05=A\sin 0+B\cos 0, \\
& b=-0.05. \\
& v=A\omega \cos \omega t-B\omega \sin \omega t, \\
& 0=A\omega \cos 0-B\omega \sin 0 \\
& A=0 \\
\end{align}$$
In order to determine
the amplitude of the system we will apply the following equation.
$$C=\sqrt{{{A}^{2}}+{{B}^{2}}}=\sqrt{2.5\cdot {{10}^{-3}}}=0.05m=50mm$$
Example
1.3
A spring has a
stiffness of 800 N/m. If a 2 kg block is attached to the spring, pushed 50 mm
above its equilibrium position, and released from rest, determine the equation
that describes the block’s motion. Assume that positive displacement is
downward.
$$\begin{align}
& \omega =\sqrt{\frac{k}{m}}=\sqrt{\frac{800}{2}}=20{{s}^{-1}}, \\
& y=A\sin \omega t+B\cos \omega t, \\
\end{align}$$
For t=0, y=-0.05 m
$$\begin{align}
& -0.05=A\sin 0+B\cos 0, \\
& B=-0.05 \\
\end{align}$$
$$\begin{align}
& y=A\sin \omega t+B\cos \omega t, \\
& v=A\omega \cos \omega t-B\omega \sin \omega t, \\
\end{align}$$
For t=0, v=0
$$\begin{align}
& v=A\omega \cos \omega t-B\omega \sin \omega t, \\
& 0=A\omega \cos 0-B\omega \sin 0, \\
& A=0 \\
\end{align}$$
Thus we get:
$$y=-0.05\cos \left( 20t \right)$$
Example 1.4.
A spring is stretched
200 mm by a 15 kg block. If the bloc is displaced 100 mm downward from its
equilibrium position and given downward velocity of 0.75 m/s determine the
equation which describes the motion. What is a phase angle? Assume that
positive displacement is downward
$$\begin{align}
& F=ky\Rightarrow k=\frac{F}{y}=\frac{15\cdot 9.81}{0.2}=735.75N/m \\
& \omega =\sqrt{\frac{k}{m}}=\sqrt{\frac{735.75}{15}}=7{{s}^{-1}} \\
& y=A\sin \omega t+B\cos \omega t \\
\end{align}$$
When t=0 then y=0.1 m and
When t=0 then v=0.75
m/s
$$\begin{align}
& y=A\sin \omega t+B\cos \omega t \\
& 0.1=A\sin 0+B\cos 0 \\
& B=0.1m \\
\end{align}$$
$$\begin{align}
& v=A\omega \cos \omega t-B\omega \sin \omega t \\
& 0.75=A\omega \cos 0-B\omega \sin 0 \\
& 0.75=A\omega \\
& A=\frac{0.75}{7}=0.107 \\
\end{align}$$
Now we can insert the
values of A and B into the solution.
$$\begin{align}
& y=0.107\sin (7t)+0.100\cos (7t) \\
& \phi =arctan\left( \frac{B}{A} \right)=\arctan \left( \frac{0.100}{0.107} \right)={{43.0}^{\circ }} \\
\end{align}$$
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