How to establish marginal stability of a mass-spring system using Routh array?

293 Views Asked by At

After simplifications, a mass-spring system can be put in the form \begin{equation} \ddot{x} + b x = 0 \tag{1} \end{equation} where $b = {k \over m} > 0$ is a constant.

The ODE (1) can be directly solved and by looking at the solutions, we can show that the system (1) is marginally stable.

Alternatively, we can express (1) as a system of first order differential equations and use Lyapunov stability theory to show that the resulting linear system on the plane is marginally stable.

The characteristic equation of the mass-spring system (1) is $$ s^2 + b = 0 \tag{2} $$

I am interesting in using Routh array and Routh stability criterion to show that the system (1) is marginally stable.

As there is a zero coefficient in (2), we can express (2) as $$ s^2 + \epsilon s + b = 0 \tag{3} $$ with $\epsilon > 0$ and calculate Routh array.

The first column has the entries $$ \left[ \begin{array}{c} 1 \\ \epsilon \\ b \\ \end{array} \right] $$

If we take limits as $\epsilon \rightarrow 0$, we have the signs $ +, 0, +$ in the first column.

Does it mean that the system is marginally stable? Can you kindly help me on this!

Thanks a lot!

1

There are 1 best solutions below

0
On BEST ANSWER

The characteristic equation for the mass-spring equation is given by $$ s^2 + b = 0 \tag{1} $$

Though it is obvious that any second order ODE with the characteristic equation (1) is marginally stable with oscillatory solutions by just calculating the general solution of the system analytically, here the interest is how to establish the same using Routh stability criterion that involves a Routh table.

As pointed out in the comment, the calculations are easier here as we are dealing with a characteristic polynomial of second order only.

In applying the Routh stability criterion, the only difficulty is a zero row corresponding to $s^1$ row in the Routh array:

$s^2$: $1$ $ \ \ $ $b$

$s^1$: $0$ $ \ \ $ $0$

$s^0$: $b$ $ \ \ $ $0$

First of all, since the first column of the Routh array does not have all positive entries, the second order system is not asymptotically stable.

To check whether there is any pole of the system in open RHP, we calculate the auxiliary polynomial, which is given as $$ A(s) = s^2 + b $$

Note that $$ A'(s) = 2 s $$

This is used to rewrite the Routh array as follows:

$s^2$: $1$ $\ \ $ $b$

$s^1$: $2$ $ \ \ $ $0$

$s^0$: $b$ $ \ \ $ $0$

Since all the entries in the first column of the revised Routh array are positive, we conclude that there are no poles in the open RHP.

Thus, the poles are in the imaginary axis, which are given by the roots of the auxiliary polynomial $A(s)$.

Indeed, the poles are obtained by solving $$ A(s) = s^2 + b = 0 $$ viz. $s = \pm \sqrt{b} j$.

Hence, the mass-spring system is marginally stable.