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+clc
+clear
+mprintf('Mechanical vibrations by G.K.Grover\n Example 4.10.1\n')
+//given data
+m=1000//mass of machine in kg
+Fo=490//amp of force in N
+f=180//freq inRPM
+//calculations
+//case a)
+K=1.96*10^6//total stiffness of springs in N/m
+Wn=sqrt(K/m)
+W=2*%pi*f/60
+bet=(W/Wn)
+zeta=0
+Xst1=Fo/K//amplitude of steady state
+X1=Xst1*(1/(sqrt((1-bet^2)^2+(2*zeta*bet)^2)))//amp of vibration Eqn 4.2.15 in Sec 4.2.1
+Ftr1=Fo*sqrt(1+(2*zeta*bet)^2)/sqrt((1-bet^2)^2+(2*zeta*bet)^2)//force transmitted,Eqn 4.10.2 in Sec 4.10.1
+//case b)
+K=9.8*10^4//total stiffness of springs in N/m
+Wn=sqrt(K/m)
+W=2*%pi*f/60
+bet=(W/Wn)
+zeta=0
+Xst2=Fo/K//amplitude of steady state
+X2=Xst2*(1/(sqrt((1-bet^2)^2+(2*zeta*bet)^2)))//amp of vibration Eqn 4.2.15 in Sec 4.2.1
+Ftr2=Fo*sqrt(1+(2*zeta*bet)^2)/sqrt((1-bet^2)^2+(2*zeta*bet)^2)//force transmitted,Eqn 4.10.2 in Sec 4.10.1
+//output
+mprintf(' a)The amplitude of motion of machine is %f m and the maximum force transmitted\n to the foundation because of the unbalanced force when\n K=1.96*10^6 N/m is %4.4f N\n b)for same case as in a)if K=9.8*10^4 N/m then\n the amplitude of motion of machine is %f m\n and the maximum force transmitted to the foundation because of\n the unbalanced force %4.4f N',X1,Ftr1,X2,Ftr2)