The Study of Vibrations - Examples



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. 

Figure 1.1 – a) Model of a system, b) Free body diagrams for external and effective forces

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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