Vibration of Two Degree of Freedom Systems:
In the end also we had discussed free vibrations of a three rotor system that is a two degree of freedom system. A two degree of freedom system contains two natural frequencies as well as the two modes of vibration.
Vibration of a Spring Mass System
A spring mass system is illustrated in Figure 19(i). The free body diagrams the two masses are illustrated in Figure 19(ii). The force equations may be written as:
m1 ? 1 + k1 x1 + k2 ( x1 - x2 ) = 0
m2 ? 2 + k2 ( x2 - x1 ) + k3 x2 = 0
![1522_Vibration of Two Degree of Freedom Systems.png](https://www.expertsmind.com/CMSImages/1522_Vibration%20of%20Two%20Degree%20of%20Freedom%20Systems.png)
These are the differential equations for spring mass system.
Let x1 = X1 cos wt
and x2 = X2 cos wt
On putting for x1 and x2 in Equation (70) and on simplify the following is attained.
(k1 + k2 - m1 w2) X1 - K2 X 2 = 0
And - k2 X1 + (k2 + k3 - m2 w2) X 2 = 0
For a exclusive solution of the homogeneous equation
![1912_Vibration of Two Degree of Freedom Systems1.png](https://www.expertsmind.com/CMSImages/1912_Vibration%20of%20Two%20Degree%20of%20Freedom%20Systems1.png)
To simplify the solution, Consider k1 = k2 = k2 and m1 = m2 .
∴ ω4 -( 4k/m) ω2 + 3k2/m2 = 0
![2255_Vibration of Two Degree of Freedom Systems2.png](https://www.expertsmind.com/CMSImages/2255_Vibration%20of%20Two%20Degree%20of%20Freedom%20Systems2.png)
For measuring mode of vibration, any one of the equations of Equation (71) may be utilized.
![130_Vibration of Two Degree of Freedom Systems3.png](https://www.expertsmind.com/CMSImages/130_Vibration%20of%20Two%20Degree%20of%20Freedom%20Systems3.png)
For
![2264_Vibration of Two Degree of Freedom Systems4.png](https://www.expertsmind.com/CMSImages/2264_Vibration%20of%20Two%20Degree%20of%20Freedom%20Systems4.png)
Let X1 = 1.0
![460_Vibration of Two Degree of Freedom Systems5.png](https://www.expertsmind.com/CMSImages/460_Vibration%20of%20Two%20Degree%20of%20Freedom%20Systems5.png)
For
![1865_Vibration of Two Degree of Freedom Systems6.png](https://www.expertsmind.com/CMSImages/1865_Vibration%20of%20Two%20Degree%20of%20Freedom%20Systems6.png)
Let X1 = 1.0
X 2 = 1.0(( 2k - 3k )/ k)= - 1.0
These modes have been plotted in Figure 19(iii).