6.9 Finding the phase difference between two discs
Let us assume the responses between the two discs are as shown in Fig. 6.31 (a) and (b), respectively. Let t 1 , t 2 , t 3 , .. respectively are the time instances corresponding to 1 st , 2 nd , 3 rd , … peaks, respectively for a disc. Let T be the time period of the response.
The phase difference between two responses can be find out as,
(6.14)
Fig. 6.31: (a) Response from disc 1 and (b) Response from disc 2
6.10 Theoretical Calculations
The geometrical and material properties of the experimental setup are given in Tables 1 to 3. The masses of the two discs in the system can be calculated using the data in Tables 1 to 3.
Mass of a disc can be calculated as,
(6.15)
where is ρ the density of the material and V is the volume of the disc.
Volume of disc 1= Area * thickness
Substituting in Eq. (6.15), the mass of disc 1, m 1 = 0.572 kg.
Similarly, mass of disc 2, m 2 = 0.7865 kg.
The moment of inertia of the beam,
Substituting the values in Eq. (6.13), the influence coefficient matrix is obtained as,
Substituting these values in Eq. (6.12), we get,
Hence, the first natural frequency
and the second natural frequency is,
The mode shapes can now be obtained by substituting the values of natural frequencies into Eq. (6.8). One of the equations of motions from Eq. (6.8) is,
Where
Dividing the Eq. (6.16) throughout by y 1 gives,
(6.17)
Substituting ω 1 =15.9 Hz into Eq.(6.17) we get,
Substituting ω 2 =62.59 Hz into Eq.(6.17) we get
Therefore the two mode shapes are (1:1.023) and (1: -0.702). These mode shapes are shown in Fig. 6.32.
Fig. 6.32: Mode shapes of 2 DOF system