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authorprashantsinalkar2017-10-10 12:27:19 +0530
committerprashantsinalkar2017-10-10 12:27:19 +0530
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+clc
+clear
+mprintf('Mechanical vibrations by G.K.Grover\n Example 4.4.2\n')
+//given data
+m=0.9//mass in kg
+K=1960//stiffness in N/m
+Y=5//amp of vibration of support in m
+N=1150//frequency in cycles per min
+//calculations
+Wn=sqrt(K/m)
+W=N*2*%pi/60//frequency of vibration of support
+bet=(W/Wn)
+//case 1
+zeta=0
+X1=Y*(sqrt(1+(2*zeta*bet)^2)/sqrt((1-bet^2)^2+(2*zeta*bet)^2))//Eqn (4.4.6)
+//case 2
+zeta =0.2
+X2=Y*(sqrt(1+(2*zeta*bet)^2)/sqrt((1-bet^2)^2+(2*zeta*bet)^2))//Eqn (4.4.6)
+//output
+mprintf('The amplitude of vibration when damping factor=0 is %4.4f mm \n If damping factor=0.2,then amplitude of vibration is %4.4f mm',X1,X2)