To determine the damping ratio for a viscously damped spring-mass system, we need to compare the given governing equation with the standard form of a single-degree-of-freedom (SDOF) damped system.

Damping Ratio for Viscously Damped System

The given governing equation for the viscously damped spring-mass system is:

\[m\frac{{{d^2}x}}{{d{t^2}}} + c\frac{{dx}}{{dt}} + kx = F\]

Where:

• \(m\) is the mass of the system, representing inertia.
• \(c\) is the viscous damping coefficient, representing energy dissipation.
• \(k\) is the spring stiffness, representing the restoring force.
• \(x\) is the displacement from the equilibrium position.
• \(F\) is the external force acting on the system.

Standard Form of Damped System

The general standard form for a damped single-degree-of-freedom system is expressed as:

\[\frac{{{d^2}x}}{{d{t^2}}} + 2\zeta{\omega_n}\frac{{dx}}{{dt}} + {\omega_n^2}x = \frac{F}{m}\]

This standard form helps us identify key parameters of the system. In this equation:

• \(\zeta\) (zeta) is the damping ratio, a dimensionless parameter that describes how oscillations in a system decay after a disturbance.
• \(\omega_n\) is the undamped natural frequency of the system, which is the frequency at which the system would oscillate if there were no damping and no external force.

Derivation of Damping Ratio

To find the damping ratio (\(\zeta\)), we first need to transform the given equation into the standard form. We do this by dividing the entire given equation by the mass \(m\):

\[\frac{1}{m}\left( {m\frac{{{d^2}x}}{{d{t^2}}} + c\frac{{dx}}{{dt}} + kx} \right) = \frac{F}{m}\]

This simplifies the governing equation to:

\[\frac{{{d^2}x}}{{d{t^2}}} + \frac{c}{m}\frac{{dx}}{{dt}} + \frac{k}{m}x = \frac{F}{m}\]

Now, we can compare the coefficients of this transformed equation with the standard form of the damped system:

1. Comparing the coefficient of \(x\):

From the standard form, the coefficient of \(x\) is \(\omega_n^2\). From our transformed equation, it is \(\frac{k}{m}\).

Therefore, by equating these terms, we get:

\[\omega_n^2 = \frac{k}{m}\]

Taking the square root of both sides, the undamped natural frequency is:

\[\omega_n = \sqrt{\frac{k}{m}}\]

2. Comparing the coefficient of \(\frac{{dx}}{{dt}}\):

From the standard form, the coefficient of \(\frac{{dx}}{{dt}}\) is \(2\zeta\omega_n\). From our transformed equation, it is \(\frac{c}{m}\).

Therefore, by equating these terms, we get:

\[2\zeta\omega_n = \frac{c}{m}\]

Now, we can solve for the damping ratio \(\zeta\):

\[\zeta = \frac{c}{2m\omega_n}\]

Substitute the expression for \(\omega_n = \sqrt{\frac{k}{m}}\) we found earlier into this equation:

\[\zeta = \frac{c}{2m\left(\sqrt{\frac{k}{m}}\right)}\]

To simplify this expression, we can rewrite \(m\) in the denominator as \(\sqrt{m} \cdot \sqrt{m}\):

\[\zeta = \frac{c}{2\sqrt{m}\sqrt{m}\frac{\sqrt{k}}{\sqrt{m}}}\]

One \(\sqrt{m}\) in the numerator and denominator cancels out, leaving:

\[\zeta = \frac{c}{2\sqrt{m}\sqrt{k}}\]

Finally, combining the terms under a single square root sign:

\[\zeta = \frac{c}{2\sqrt{km}}\]

This derived expression for the damping ratio perfectly matches one of the provided options.

Conclusion on Damping Ratio

The damping ratio (\(\zeta\)) for the viscously damped spring-mass system, governed by the relationship \(m\frac{{{d^2}x}}{{d{t^2}}} + c\frac{{dx}}{{dt}} + kx = F\), is given by the formula \(\frac{c}{{2\sqrt {km} }}\). This formula is fundamental in understanding the damping characteristics of mechanical systems.