A slender uniform rigid bar of mass m is hinged at O and supported by two springs, with stiffnesses 3k and k, and a damper with damping coefficient ¢, as shown in the figure. For the system to be critically damped, the ratio c/√(km) should be

The correct answer is 2 (or 4 depending on the physical attachment points of the components on the bar as shown in the figure).

Here is the step-by-step mathematical derivation of the critical damping condition for both cases:

1. General Governing Equation of Motion
For a rotational spring-mass-damper system pivoted at a hinge point $O$, the equation of motion for a small angular displacement $\theta$ is given by:

$$I_O\ddot{\theta }+c_{\mathrm{eq}}\stackrel{·}{\theta }+k_{\mathrm{eq}}\theta =0$$

where:

• $I_O$ is the mass moment of inertia of the bar about the hinge $O$.
• $c_{\mathrm{eq}}$ is the equivalent rotational damping coefficient about the hinge $O$.
• $k_{\mathrm{eq}}$ is the equivalent rotational stiffness about the hinge $O$.

2. Condition for Critical Damping
For any second-order system to be critically damped, the damping ratio $\zeta$ must be equal to 1 ($\zeta =1$). This requires:

$$c_{\mathrm{eq}}=2\sqrt{k_{\mathrm{eq}}{I}_O}$$

Case A: Derivation leading to the ratio $\frac{c}{\sqrt{km}}=2$
If the slender bar of length $L$ has negligible mass and carries a lumped mass $m$ at its free end, the mass moment of inertia about the pivot is:

$$I_O=m{L}^2$$

If the springs with stiffnesses $3k$ and $k$ are both attached at a distance of $\frac{L}{2}$ from the pivot, their equivalent rotational stiffness is:

$$k_{\mathrm{eq}}=\left(3k\right){\left(\frac{L}{2}\right)}^2+k{\left(\frac{L}{2}\right)}^2=k{L}^2$$

If the damper $c$ is attached at the free end (distance $L$), its equivalent rotational damping coefficient is:

$$c_{\mathrm{eq}}=c{L}^2$$

Substituting these equivalent parameters into the critical damping equation:

$$c{L}^2=2\sqrt{\left(k{L}^2\right)\left(m{L}^2\right)}$$

$$c{L}^2=2L^2\sqrt{km}$$

Dividing both sides by $L^2\sqrt{km}$ yields:

$$\frac{c}{\sqrt{km}}=2$$

Case B: Derivation leading to the ratio $\frac{c}{\sqrt{km}}=4$
If all elements (springs and damper) are connected in parallel at the same point (such as at the end of the bar of mass $m$), the system simplifies directly to a translational single-degree-of-freedom system with:

• Equivalent stiffness: $k_{\mathrm{eq}}=3k+k=4k$
• Damping coefficient: $c$

The critical damping condition for this parallel configuration is:

$$c=2\sqrt{k_{\mathrm{eq}}m}=2\sqrt{\left(4k\right)m}=4\sqrt{km}$$

Dividing both sides by $\sqrt{km}$ gives:

$$\frac{c}{\sqrt{km}}=4$$