For the system maximum potential energy and kinetic energies are: 1 ∑n Vmax migyi  2 i 1 1 ∑n Tmax mi y&i2  2 i 1 where, mi- masses of the system, yi at mass points. –displacements Considering the system vibrates with SHM, y&i yi  ω2 From above equations ω2 ∑n Tmax mi yi2  2 i 1 According to Rayleigh’s method, Vmax Tmax  substitute Eqn. (4) and (5) in (6) n ∑migyi  ω2 i n 1 ∑mi yi2 i 1 The deflections at point 1, 2 and 3 can be found by. y1    α11m1g α12m2g α13m3g    α21m1g α22m2g α23m3g    α31m1g α32m2g α33m3g Examples: 1 Estimate the approximate fundamental natural frequency of the system shown in Fig.8 using Rayleigh’s method. Take: m=1kg and K=1000 N/m. 2K 2m x1 K 2m x2 K m x3 Fig.8 Linear vibratory system Obtain influence coefficients,
For the system maximum potential energy and kinetic energies are: Vmax  1 ∑n migyi 2 i 1 Tmax  1 ∑n mi y&i2 2 i 1 where, mi- masses of the system, yi –displacements at mass points. Considering the system vibrates with SHM, y&i  ω2 yi From above equations Tmax  ω2 ∑n mi yi2 2 i 1 According to Rayleigh’s method, Vmax  Tmax substitute Eqn. (4) and (5) in (6) n ∑migyi ω2  i 1 n ∑mi yi2 i 1 The deflections at point 1, 2 and 3 can be found by. y1  α11m1g  α12m2g  α13m3g  α21m1g  α22m2g  α23m3g  α31m1g  α32m2g  α33m3g Examples: 1 Estimate the approximate fundamental natural frequency of the system shown in Fig.8 using Rayleigh’s method. Take: m=1kg and K=1000 N/m. 2K 2m x1 K 2m x2 K m x3 Fig.8 Linear vibratory system Obtain influence coefficients,