Question 3.4: Figure 3.11(a) shows a pendulum which is supported by two sp...

As shown in Fig. 3.11(c), the force produced by the springs k_{1} and k_{2}, as the results of an angular displacement θ, are given by

F_{1} = −k_{1}a sin θ,          F_{2} = −k_{2}b sin θ

The resultant moment due to the forces F_{1} and F_{2} about point O is given by
M = F_{1}a cos θ + F_{2}b cos θ = −(k_{1}a² + k_{2}b²) sin θ cos θ

The force F_{e} produced by the equivalent spring is given by

F_{e} = −k_{e}d sin θ

and the moment about O produced by this force is
M_{e} = F_{e}d cos θ = −k_{e}d² sin θ cos θ

If the two systems in Fig. 3.11(a) and (b) are to be equivalent, the moment M produced by the original system must be equal to the moment Me produced by the equivalent system, that is, M = M_{e}, or
−(k_{1}a² + k_{2}b²) sin θ cos θ = −k_{e}d² sin θ cos θ

from which the equivalent stiffness coefficient k_{e} is defined as
k_{e} = \frac{k_{1}a² + k_{2}b²}{d²}
A moment equation is used in this example instead of a force equation to determine k_{e}, because this is the case of angular oscillations in which the system degree of freedom is the angular orientation θ.