Question 3.7: The damped mass–spring system shown in Fig. 3.12 has mass m ...

The circular frequency ω of the system is

ω = \sqrt\frac{k}{m} = \sqrt{\frac{1000}{10}} = 10 rad/s

and the critical damping factor ξ is given by

C_{c} = 2mω = 2(10)(10) = 200N · s/m

Therefore, the damping factor ξ is given by

ξ = \frac{c}{C_{c}} = \frac{10}{200} = 0.05

The damped natural frequency ω_{d} is given by

ω_{d} = ω \sqrt{1 − ξ²} = 10 \sqrt{1 − (0.05)²} = 9.9875 rad/s

Substituting ω, ξ , and ω_{d} into Eq. 3.74, the solution for the damped single degree of freedom system can be expressed as

x(t) = Xe^{−ξωt} sin(ω_{d}t + Φ)                        (3.74)
x = Xe^{−0.5t} sin(9.9875t + Φ)
where X and Φ are constants which can be determined from the initial conditions.