Question 4.6: The damped single degree of freedom mass–spring system shown...

From Eq. 4.98, the amplitude of the oscillation of the mass is given
by
X_{p} = Y_{0}β_{b} = Y_{0} \frac{\sqrt{1 + (2rξ)^{2}}}{\sqrt{(1 − r^2)^2 + (2rξ)^2}}
Since in this example, ωf = ω, that is, r = 1, the above equation reduces to

X_{p} = Y_{0} \frac{\sqrt{(1 + 2ξ)^2}}{2ξ}
From which the damping factor ξ is

ξ = \frac{1}{2} \frac{1}{\sqrt{\left(\frac{X_p}{Y_0}\right)^{2}− 1}} = \frac{1}{2} \frac{1}{\sqrt{\left(\frac{0.01}{0.005}\right)^{2}− 1}} = 0.28868

The critical damping coefficient C_c of the system is
C_c = 2mω = 2 \sqrt{km} = 2 \sqrt{(2500)(25)} = 500N · s/m
The damping coefficient c is
c = ξC_c = 0.28868(500) = 144.34N · s/m

From Eq. 4.101, the amplitude of the force transmitted is
(F_1)_{max} = Y_{0}kr^{2}β_{b} = Y_{0}k \frac{r^{2}\sqrt{1 + (2rξ)^{2}}}{\sqrt{(1 − r^2)^2 + (2rξ)^2}}
At resonance r = 1, therefore,

(F_1)_{max} = Y_{0}k \frac{\sqrt{1 + (2rξ)^{2}}}{2ξ} = (0.005)(2500) \frac{\sqrt{1 + (2 × 0.28868)^2}}{2(0.28868)}\\ = 24.9997N
The amplitude of the displacement of the mass relative to the support can be obtained using Eq. 4.105 as
Z = \frac{Y_0r^2}{\sqrt{(1 − r^2)^2 + (2rξ)^2}}

At resonance
Z = \frac{Y_0}{2ξ} = \frac{0.005}{2(0.28868)} = 8.66 × 10^{−3} m
Observe that Z ≠ X_{p} − Y_{0} because of the phase shift angles ψ and ψ_{b}.