Question 4.4: A damped single degree of freedom mass–spring system has mas...

The circular frequency of the system is
ω = \sqrt{\frac{k}{m}} = \sqrt{\frac{4000}{10}} = 20 rad/s

The frequency ratio r is given by
r = \frac{ω_{f}}{ω} = \frac{40}{20} = 2
The critical damping coefficient C_{c} is defined as

C_{c} = 2mω = 2(10)(20) = 400 N · s/m
The damping factor ξ is then given by
ξ = \frac{c}{C_{c}} = \frac{40}{400} = 0.1

which is the case of an underdamped system, and the complete solution can be written in the following form
x(t) = x_{h} + x_{p} = X e^{−ξωt} sin(ω_{d}t + Φ) + X_{0}β sin(ω_{f}t − ψ)
where ω_{d} is the damped circular frequency

ω_{d} = ω \sqrt{1 − ξ²} = 20 \sqrt{1 − (0.1)²} = 19.8997 rad/s
The constants X_{0}, β, and ψ are

X_{0} = \frac{F_{0}}{k} = \frac{60}{4000} = 0.015 m

β = \frac{1}{\sqrt{(1 − r²)² + (2rξ)²}} = \frac{1}{\sqrt{(1 − (2)²)² + (2 × 2 × 0.1)²}} = 0.3304

ψ = tan^{−1} (\frac{2rξ}{1 − r²}) = tan^{−1}\frac{2(2)(0.1)}{1 − (2)²} = −0.13255 rad
The displacement can then be written as a function of time as

x(t) = Xe^{−2t} sin(19.8997t + Φ) + 0.004956 sin(40t + 0.13255)

The constants X and Φ can be determined using the initial conditions. The transmissibility β_{t} is defined by Eq. 4.58 as

β_{t} = β \sqrt{1 + (2rξ)²} = \frac{\sqrt{1 + (2rξ)²}}{\sqrt{(1 − r²)² + (2rξ)²}}       (4.58)

β_{t} = \frac{\sqrt{1 + (2rξ)²}}{\sqrt{(1 − r²)² + (2rξ)²}} = \frac{\sqrt{1 + (2 × 2 × 0.1)²}}{\sqrt{[1 − (2)²]² + (2 × 2 × 0.1)²}} = 0.35585

The amplitude of the force transmitted is given by
|F_{t}| = F_{0}β_{t} = (60)(0.35585) = 21.351 N