# 2.2.3 Linearity for small vibration

The justification for treating small vibration by linear theory lies in Lagrangian mechanics. The equations of vibration of any system can be derived from expressions for the potential and kinetic energy in the system. For small motion, it is justifiable to make a series expansion of those energy expressions and keep only the first term that does not vanish. The result is that the kinetic energy is approximately a quadratic expression in the velocities, while the potential energy is a quadratic expression in the displacements. Applying Lagrange’s equations to these approximate energy expressions leads automatically to linear differential equations describing the motion.

Some details of this calculation will be given in section 2.2.5, but the essential idea can be illustrated using the simple harmonic oscillator discussed in section 2.2.2, by approaching that problem in terms of energy. The system is conservative (i.e. has no dissipative forces acting), so that the total energy is constant. The kinetic energy is

$$T=\frac{1}{2} m \dot{x}^2 \tag{1}$$

where $\dot{x}=dx/dt$. The potential energy stored in the spring is

$$V=\frac{1}{2} k x^2. \tag{2}$$

This may be seen by noting that the work done in a small displacement $\delta y$ from displacement $y$ is $ky\delta y$, and integration from zero to the desired displacement $x$ gives the expression (2). Now energy conservation requires

$$\dfrac{d}{dt} [T+V]=0 \tag{3}$$

which gives

$$m \ddot{x} +kx=0 \tag{4}$$

exactly as in equation (1) of section 2.2.2, derived via Newton’s laws.

The energy approach gives a natural way to generalise this simple example to cover small vibrations of a wide range of systems. Consider any system with just one degree of freedom, in other words one whose displaced position can be completely described by a single parameter $q$ (which might, for example, describe the displacement of a point or the angle of rotation of a rigid body). It will generally be the case that the kinetic energy of the system takes a similar quadratic form to expression (1):

$$T=\frac{1}{2} M \dot{q}^2 \tag{5}$$

where the constant $M$ will be a mass if $q$ describes a linear displacement, or a moment of inertia if $q$ describes a rotation.

The potential energy is less immediately obvious, but if the system is in a position of stable equilibrium when $q = 0$ then certainly the function $V(q)$ must have a minimum at $q = 0$. Provided this function is reasonably well-behaved, we can expect it to be well approximated by the dominant term in its Taylor expansion provided $q$ remains sufficiently small:

$$V(q) \approx V_0+\frac{1}{2} V_2 q^2 + \mathrm{terms~of~order~} q^3. \tag{6}$$

There is no linear term in this expansion by virtue of the equilibrium at $q = 0$, and in order for that equilibrium to be stable we must have

$$V_2 \ge 0. \tag{7}$$

The constant $V_0$ is of no dynamical consequence, and can be set to zero for simplicity. Provided $V_2$ is non-zero we are left with an approximate expression for the potential energy which takes the same form as that of the simple mass-spring oscillator in (2).

In some cases a similar approximation may be needed in the kinetic energy: the quantity $M$ may depend on the displacement $q$, but for the case of small displacements away from equilibrium the dominant term is obtained simply by setting $q = 0$ within this expression. The next term in a Taylor expansion would be of order $q\dot{q}^2$, small by comparison.

Having thus obtained approximate expressions for both kinetic and potential energy which match those of the simple spring-mass system, it follows that for any such system, provided the displacement away from equilibrium is sufficiently small, the free motion will obey a differential equation analogous to (4), and the solution will take the sinusoidal form of equation (5) of section 2.2.2, with a natural frequency

$$\Omega=\sqrt{V_2/M}. \tag{8}$$