Discussion
The damping ratio for a viscously damped spring mass system, governed by the relationship \(m\frac{d^{2}x}{dt^{2}} + c\frac{dx}{dt} + kx = F(t)\), is given by
Answer: B
The damping ratio is \(\frac{c}{2\sqrt{km}}\).
The damping ratio is the actual damping coefficient divided by the critical damping coefficient, the value at which the system just stops oscillating:
\[ \begin{aligned} c_{c} &= 2\sqrt{km} = 2m\omega_{n} \\ \zeta &= \frac{c}{c_{c}} = \frac{c}{2\sqrt{km}} \end{aligned} \]
\(c/\sqrt{km}\) is twice the damping ratio. The square-root forms are dimensionally wrong: c/(mk) has units of s3/kg, so its square root is not a pure number.
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