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+clc
+clear
+mprintf('Mechanical vibrations by G.K.Grover\n Example 4.10.3\n')
+// given data
+m=20 //mass in kgs
+k=125600 //overall eqivalent stiffness i.e 4*31400 in N/m
+c=1568 //overall damping coefficient i.e 4*392 in N-sec/m
+n=500 //vibrating speed of machine in cpm
+//y=Ysin(w*t)
+Y=0.00005 //vibrating amplitude of machine in m
+W=2*%pi*n/60 //vibrating frequency in rad/sec
+Wn=sqrt(k/m) //natural frequency in rad/sec
+bet=(W/Wn) //speed ratio
+zeta=c/(2*sqrt(k*m)) //damping factor
+//calculations
+X=Y*sqrt((1+(2*zeta*bet)^2)/((1-bet^2)^2+(2*zeta*bet)^2)) //absolute amplitude of vibration of radio from eqn (4.4.6)
+Z=Y*((bet^2)/sqrt(((1-bet^2)^2+(2*zeta*bet)^2)))//from eqn 4.4.11
+FdynT=Z*sqrt((c*W)^2+k^2)//dynamic load total
+Fdyn=FdynT/4 //dynamic load on each isolator
+FdynTmax=m*W^2*X //max dynamic load on the isolators
+Fdynmax=FdynTmax/4 //max dynamic load on each isolator
+//output
+mprintf('a) The amplitude of vibration of radio is %f metres \n b)the dynamic load on each isolator due to vibration is %3.3f N',X,Fdyn)