Part 1 · The elementary oscillator

1.1 The spring and the equation of motion

Source: Carboni’s notes for lesson 1, Oscillatore elementare, pp. 1–2

Objectives

Vocabulary. Constitutive law, linear stiffness, hardening and softening.
Picture. The force–displacement curve of a spring, and what its bending means.
Recipe. From the forces on a mass to the equation of motion $M \ddot{U} + K U = F (t)$ .

The course starts from the smallest mechanical system there is: one mass on one spring. Everything that comes later (a body that deforms, stresses inside it, a law linking the two) has a first, one-dimensional version here. A spring deforms (it stretches by $U$ ), it carries an internal force, and a constitutive law says how the two are related.

The setting

A mass $M$ (a number, in kg) can move along a straight line, the direction of the unit vector $e_{1}$ . A spring ties it to a fixed wall.

$U (t)$ is the displacement of the mass at time $t$ : how far it is, along $e_{1}$ , from the position where the spring is unstretched. It is a function of time, in metres.
$U > 0$ means the spring is stretched, $U < 0$ that it is compressed.
Dots are time derivatives: $\dot{U} = d U / d t$ is the velocity, $\ddot{U} = d^{2} U / d t^{2}$ the acceleration.

The constitutive law of the spring

When the spring is stretched by $U$ it pulls on the mass with the elastic force $F_{e l} = \hat{F} (U) .$

Definition 1.1.1constitutive lawnotes L1, p. 1

The function $\hat{F}$ that gives the spring’s force for each displacement $U$ is the constitutive law of the spring. It may be nonlinear. It describes the spring (its material and its shape), not the motion: the same spring obeys the same $\hat{F}$ whatever it is doing.

We rarely know $\hat{F}$ exactly, but near $U = 0$ we can expand it in a Taylor series: $F_{e l} = \hat{F} (0) + \frac{d F ^}{d U}_{0} U + \frac{1}{2 !} \frac{d ^{2} F ^}{d U ^{2}}_{0} U^{2} + \frac{1}{3 !} \frac{d ^{3} F ^}{d U ^{3}}_{0} U^{3} + O (U^{4}) .$

$\hat{F} (0) = 0$ : at the unstretched position the spring exerts no force. That is what choosing $U = 0$ there means.
The derivatives at $0$ are numbers, properties of the spring. They get names:

Definition 1.1.2linear, quadratic and cubic stiffnessnotes L1, p. 1

$K = \frac{d F ^}{d U}_{0}, K_{2} = \frac{d ^{2} F ^}{d U ^{2}}_{0}, K_{3} = \frac{d ^{3} F ^}{d U ^{3}}_{0} .$ $K$ is the linear stiffness: the slope of the force–displacement curve at the origin, $K = lim_{Δ U \to 0} Δ \hat{F} ∣_{0} /Δ U$ , in N/m. $K_{2}$ is the quadratic and $K_{3}$ the cubic stiffness.

So $F_{e l} = K U + \frac{1}{2} K_{2} U^{2} + \frac{1}{6} K_{3} U^{3} + O (U^{4}) .$

Key observation 1.1.3the linear elastic lawnotes L1, p. 1

For small displacements the powers $U^{2}, U^{3}, \dots$ are negligible next to $U$ , and only the first term is left: $F_{e l} = K U .$ This is the linear elastic constitutive law (Hooke’s law for a spring). It is an approximation, valid for small $U$ .

Hardening and softening

The notes draw $F_{e l}$ against $U$ . The linear law is a straight line of slope $K$ through the origin: at $U = \overset{ˉ}{U}$ the force is $K \overset{ˉ}{U}$ . Two nonlinear springs share that tangent line at the origin and then bend away from it:

coefficients	behaviour	curve
$K > 0$ , $K_{2} = K_{3} = 0$	linear	the straight line
$K > 0$ , $K_{2} = 0$ , $K_{3} > 0$	hardening	bends away from the $U$ axis: stiffer the more it is stretched
$K > 0$ , $K_{2} = 0$ , $K_{3} < 0$	softening	bends towards the $U$ axis: softer the more it is stretched

Loading the demo…

Figure 1.1.1 The constitutive law

F_{e l} = K U + \frac{1}{6} K_{3} U^{3}

. Drag the point along the curve, or change

K_{3}

: the dashed line is the linear law with the same

K

. Near the origin all three curves agree, which is why the linear law is good for small

U

Remarkbeyond the lectures

Why $K_{2} = 0$ in both nonlinear cases. A term $\frac{1}{2} K_{2} U^{2}$ has the same sign for $U > 0$ and $U < 0$ , so it would make the spring respond differently to being stretched and being compressed. A spring that behaves the same both ways has an odd $\hat{F}$ ( $\hat{F} (- U) = - \hat{F} (U)$ ), and only odd powers survive.

The equation of motion

Treat the mass as a rigid body and draw the forces on it. With $F_{e l} = K U$ , the spring pulls the mass back towards $U = 0$ : the force on the mass is $- K U$ along $e_{1}$ .

Recipe 1.1.4equation of motion of a mass on a springnotes L1, p. 2

Kinematics. Velocity $v = \dot{U}$ , acceleration $a = \overset{v}{˙} = \ddot{U}$ .
Newton’s second law. The sum of the forces is the rate of change of momentum: $\sum F_{i} = \frac{d ( M v )}{d t} = \frac{d M}{d t} v + M \frac{d v}{d t} = M a,$ since the mass does not change ( $d M / d t = 0$ ).
The forces. Only the spring: $\sum F_{i} = - K U$ .
Equate and rearrange: $- K U = M \ddot{U}$ , that is $M \ddot{U} + K U = 0.$

Definition 1.1.5harmonic oscillatornotes L1, p. 2

$M \ddot{U} + K U = 0$ is the unforced, undamped harmonic oscillator. With an external force $F (t)$ acting on the mass along $e_{1}$ , $M \ddot{U} + K U = F (t)$ is the forced, undamped harmonic oscillator.

Remark

D’Alembert’s reading. The term $M \ddot{U}$ can be drawn on the free-body diagram as one more force, the inertial force $- M \ddot{U}$ , pointing against the acceleration. Then the forces on the mass balance: $- K U - M \ddot{U} = 0$ . The notes draw it this way, and later sections use it too: a body in motion is “in equilibrium” with its inertial forces included.

Warningslip in the notes

The second time the notes write this equation (p. 2) it reads $M \ddot{U} + K \dot{U} = 0$ : the dot over the second $U$ is a slip. The stiffness multiplies the displacement, $K U$ , as in the line just above it. The same line also writes the velocity as a capital $V$ in $\frac{d V}{d t} M$ ; it is $v$ .

Exercises

Exercise 1.1.1

A spring has $\hat{F} (U) = 200 U + 4 \times 1 0^{5} U^{3}$ (N, with $U$ in m). Find $K$ , $K_{2}$ and $K_{3}$ . Is it hardening or softening? At what displacement is the cubic term 10% of the linear one?

Solution

Compare with $K U + \frac{1}{2} K_{2} U^{2} + \frac{1}{6} K_{3} U^{3}$ : $K = 200$ N/m, $K_{2} = 0$ , $\frac{1}{6} K_{3} = 4 \times 1 0^{5}$ , so $K_{3} = 2.4 \times 1 0^{6}$ N/m³.
$K_{3} > 0$ : hardening.
$4 \times 1 0^{5} U^{3} = 0.1 \cdot 200 U$ gives $U^{2} = 5 \times 1 0^{- 5}$ , $U = 7.07$ mm. Below a few millimetres the linear law is good to better than 10%.

Exercise 1.1.2

A mass $M = 2$ kg on a spring $K = 50$ N/m is pushed by a constant force $F = 10$ N. Write the equation of motion. Where is the mass at rest?

Solution

$2 \ddot{U} + 50 U = 10$ .
At rest $\ddot{U} = 0$ , so $50 U = 10$ and $U = 0.2$ m: the spring stretches until its force balances $F$ .