A springdampermass system with mass mt spring constant s and
Solution
solution:
1)here mass spring damper system is exicted by external harmonic force=Fosinwt
2) here equation of motion is
mz\'\'+dz\'+kz=Fosinwt
here z=Zsinwt as system oscillating with harmonic motion
solution of above equation is givenas
z=zc+zp
where complimentry part is obtain by
mz\'\'+dz\'+kz=0
zc=Zexp^(-zeta*wn*t)*sin(wdt+a1)
zp=Zsin(wt-a)
on differenting and putting value in equation of motion we get
amplitude of vibration as
Zp=Fo/k/((1-(w/wn)^2)^2+(2*zeta*(w/wn))^2)^.5
3)for this wquation when we putting value of amplitude we get for resonanace w=wn
Fo/k=1.6*zeta
4) for second value of frequency we get\\ equation as on putting value of Fo/k in second eqaution as
.1625/zea^2+.02480=0
solving we get
zeta=.3906
damping ratio=.3906
4) steady state deflection mean ratio of Fo/k=Zst=1.6*zeta=1.6*.3906=.62496 mm
5) solution is given by
z=Zc+Zp=Zexp^(-zeta*wn*t)*sin(wdt+a1)+.62496sin(wt-a1)/((1-(w/wn)^2)^2+(2*zeta*(w/wn))^2)^.5

