A spring-damper-mass system with mass m_t, spring constant s, and damping coeffi

Question

A spring-damper-mass system with mass m_t, spring constant s, and damping coefficient of d, is excited along the z direction by a harmonically varying force. At resonance, the amplitude of the response was measured as 0.80 mm. In addition, when excited at a frequency of a factor of 1.1 of the resonant frequency, the measured response amplitude was measured as 0.75 mm. Symbolically derive the equation of motion for the system including expressions for the natural frequency and the damped natural frequency. Determine the value of the damping ratio. Derive the solution of the system's response. Assume that the effect of damping is negligible away from resonance.

Explanation / Answer

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

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