Small oscillations

The equation of a swinging pendulum contains a sine of the angle and cannot be solved in elementary functions, which is a problem the whole of physics solves the same way.

This lesson needs the potential energy of the earlier lessons, the fact that force is minus the slope of that potential, the rotational law $τ = I α$ , and, from calculus, the Taylor expansion of a function about a point.

Why every stable equilibrium is a spring

Take any body moving in one dimension under a conservative force with potential energy $U (x)$ , and suppose it has a stable equilibrium at $x_{0}$ , meaning a minimum of $U$ . Expand about that point:

U (x) = U (x_{0}) + U^{'} (x_{0}) (x - x_{0}) + \frac{1}{2} U^{''} (x_{0}) (x - x_{0})^{2} + \dots

The first term is a constant and can be dropped, since only differences in $U$ matter. The second vanishes, because $U^{'} (x_{0}) = 0$ is what equilibrium means. So the leading behaviour near any equilibrium is the quadratic term, and the force is

F = - \frac{d U}{d x} = - U^{''} (x_{0}) (x - x_{0})

which is Hooke's law with $k = U^{''} (x_{0})$ . The conclusion is general and worth stating in full: any system displaced slightly from a stable equilibrium experiences a restoring force proportional to the displacement, whatever the underlying physics. The stiffness is the curvature of the potential at the minimum, and it is positive precisely when the equilibrium is stable, which is the analytic version of the picture of a ball in a valley.

This is why the same equation describes a mass on a spring, a pendulum, a floating hydrometer bobbing in water, a molecule vibrating, an atom in a crystal lattice, and an electrical circuit. None of them is a spring. All of them are near a minimum.

Solving the equation

With $x$ measured from equilibrium, the second law gives

m \frac{d^{2} x}{d t^{2}} = - k x \Rightarrow \frac{d^{2} x}{d t^{2}} = - ω^{2} x, ω = \sqrt{\frac{k}{m}}

The equation asks for a function whose second derivative is itself, negated and scaled, and sine and cosine are the only elementary candidates. The general solution, with the two constants that any second order equation requires, is

x (t) = A \cos (ω t + φ)

with $A$ the amplitude and $φ$ the phase, both fixed by the initial position and velocity. This is simple harmonic motion. Differentiating gives $v = - A ω \sin (ω t + φ)$ and $a = - A ω^{2} \cos (ω t + φ)$ , so the maximum speed is $A ω$ and the maximum acceleration is $A ω^{2}$ .

The period is $T = 2 π / ω = 2 π \sqrt{m / k}$ , and the single most important feature is what is missing from it. The period does not depend on the amplitude. A spring pulled twice as far takes exactly as long to return, because the extra distance is exactly compensated by the extra force. Systems with this property are called isochronous, and it is the reason oscillators can keep time at all: a clock whose escapement delivers a slightly variable push would otherwise run at a variable rate.

Energy in the motion sloshes between the two forms. With $x = A \cos ω t$ , the potential energy is $\frac{1}{2} k A^{2} \cos^{2} ω t$ and the kinetic energy is $\frac{1}{2} m A^{2} ω^{2} \sin^{2} ω t = \frac{1}{2} k A^{2} \sin^{2} ω t$ , and since the squares of sine and cosine sum to one, the total is $\frac{1}{2} k A^{2}$ , constant, as conservation requires. Each form averages half the total over a cycle.

Example. A 0.50 kg mass on a spring of stiffness 200 N m⁻¹ is pulled 8.0 cm from equilibrium and released. Find the angular frequency, period, frequency, maximum speed, maximum acceleration and total energy.

$ω = \sqrt{200 / 0.50} = 20.0$ rad s⁻¹, so $T = 2 π / 20 = 0.314$ s and $f = 1 / T = 3.18$ Hz. With $A = 0.080$ m, the maximum speed is $A ω = 1.60$ m s⁻¹ at the equilibrium point, and the maximum acceleration is $A ω^{2} = 32.0$ m s⁻² at the extremes, over three times $g$ . The total energy is $\frac{1}{2} k A^{2} = \frac{1}{2} (200) (0.0064) = 0.640$ J, which also equals $\frac{1}{2} m v_{\max}^{2} = \frac{1}{2} (0.50) (2.56)$ , as it must.

Now you. A 2.0 kg mass on a spring of stiffness 50 N m⁻¹ oscillates with an amplitude of 12 cm. Find the period and the maximum speed.

Answer

$ω = \sqrt{50 / 2.0} = 5.0$ rad s⁻¹, so $T = 2 π / 5.0 = 1.26$ s. The maximum speed is $A ω = 0.12 \times 5.0 = 0.60$ m s⁻¹.

The simple pendulum

A bob of mass $m$ on a light string of length $L$ , displaced by an angle $θ$ , has a restoring torque about the pivot of $- m g L \sin θ$ and a moment of inertia $m L^{2}$ , so $τ = I α$ gives

m L^{2} \frac{d^{2} θ}{d t^{2}} = - m g L \sin θ \Rightarrow \frac{d^{2} θ}{d t^{2}} = - \frac{g}{L} \sin θ

The mass has already cancelled, which is why pendulum timekeeping does not depend on the bob. The remaining obstacle is the sine, and the previous section says what to do about it: for small $θ$ in radians, $\sin θ \approx θ$ , and

\frac{d^{2} θ}{d t^{2}} = - \frac{g}{L} θ \Rightarrow T = 2 π \sqrt{\frac{L}{g}}

The period depends only on the length and on $g$ , and not on the amplitude or the mass. A pendulum 1.000 m long has a period of $2 π \sqrt{1 / 9.81} = 2.006$ s. A pendulum that beats seconds, taking one second per swing and so two seconds per full period, needs $L = g / π^{2} = 0.994$ m, which is why longcase clocks are the height they are.

Turned around, the formula measures $g$ : time a hundred swings, divide, and solve for $g = 4 π^{2} L / T^{2}$ . This was the standard method of gravimetry from Huygens, who built the first pendulum clock in 1656 and worked out the theory in 1673, until well into the twentieth century, and it is how the variation of $g$ with latitude was first mapped.

Example. How long is a pendulum with a period of 1.50 s, and what would its period be on the Moon, where $g = 1.62$ m s⁻²?

Rearranging, $L = g T^{2} / 4 π^{2} = (9.81) (2.25) / 39.48 = 0.559$ m. On the Moon the same pendulum has $T = 2 π \sqrt{0.559 / 1.62} = 3.69$ s, longer by the square root of the ratio of the two values of $g$ , which is $\sqrt{9.81 / 1.62} = 2.46$ .

Now you. Find the period of a 2.50 m pendulum on Earth, and the length needed for a period of 1.00 s.

Answer

$T = 2 π \sqrt{2.50 / 9.81} = 3.17$ s. For $T = 1.00$ s, $L = (9.81) (1.00) / 39.48 = 0.248$ m.

The physical pendulum

A real pendulum is not a point on a string. Any rigid body pivoted about a point other than its centre of mass swings, and the same derivation applies with the body's own moment of inertia. If $d$ is the distance from the pivot to the centre of mass and $I$ is the moment of inertia about the pivot,

I \frac{d^{2} θ}{d t^{2}} = - m g d \sin θ \Rightarrow T = 2 π \sqrt{\frac{I}{m g d}}

for small angles. Comparing with the simple pendulum identifies the equivalent length $L_{eq} = I / m d$ : the length of a simple pendulum that would keep the same time.

Example. A uniform rod of length 1.00 m swings about a pivot at one end. What is its period, and what simple pendulum matches it?

About the end, $I = \frac{1}{3} m L^{2}$ , and the centre of mass is at $d = L / 2$ . So $L_{eq} = (\frac{1}{3} m L^{2}) / (m L / 2) = \frac{2}{3} L = 0.667$ m, and $T = 2 π \sqrt{0.667 / 9.81} = 1.64$ s. A rod swings noticeably faster than a bob on a string of the same length, which would take 2.01 s, because much of the rod's mass is close to the pivot where it has little effect on the restoring torque but still less on the inertia.

Now you. A uniform rod of length 1.50 m swings about one end. Find its period.

Answer

$L_{eq} = \frac{2}{3} (1.50) = 1.00$ m, so $T = 2 π \sqrt{1.00 / 9.81} = 2.01$ s.

Exactly how wrong the approximation is

The small angle step is where a physicist ought to be nervous, and it is easy to quantify, because the exact pendulum equation can be solved in terms of elliptic integrals. The exact period is the small angle answer multiplied by a factor that depends only on the amplitude $θ_{0}$ :

T = T_{0} (1 + \frac{1}{4} \sin^{2} \frac{θ_{0}}{2} + \frac{9}{64} \sin^{4} \frac{θ_{0}}{2} + \dots)

Evaluating it settles the question. At an amplitude of 5° the exact period is 0.048 per cent longer than $T_{0}$ ; at 10°, 0.19 per cent; at 20°, 0.77 per cent; at 30°, 1.74 per cent; at 90°, 18.0 per cent. The approximation is not merely good at small angles, it is good in a very specific and useful sense: the error grows as the square of the amplitude, so halving the swing quarters the error.

For a clock that matters a great deal. A pendulum swinging at 10° rather than infinitesimally runs slow by 0.19 per cent, which is about 165 seconds a day, and a clock that loses nearly three minutes daily is useless. Clockmakers solved it by keeping the amplitude both small, a degree or two, and constant, which is what a good escapement is for: it is not primarily a device for supplying energy but a device for supplying the same energy every swing.

The isochronism of a pendulum is therefore approximate, and Huygens knew it. His answer in 1659 was to make the bob swing on a cycloidal path rather than a circular one, using shaped cheeks at the suspension, which is exactly isochronous at every amplitude. In practice the friction the cheeks introduced cost more than the error they removed, and clockmakers went back to small circular arcs.

Damping and resonance

Real oscillators stop. Add a resistive force proportional to velocity and the equation becomes $m \ddot{x} + b \dot{x} + k x = 0$ , whose solution for light damping is an oscillation whose amplitude decays exponentially, $A e^{- b t / 2 m}$ , at a frequency slightly below the undamped one. Heavier damping kills the oscillation entirely: at critical damping the system returns to equilibrium in the shortest time without overshooting, which is what a car's shock absorbers and a door closer are tuned for.

Drive a lightly damped oscillator at a frequency near its own and the amplitude grows large, which is resonance, and the sharpness of the peak is set by how little damping there is. This is what makes a wine glass sing and a radio tune to one station.

Two famous examples deserve care, because the popular version of each is wrong. The Tacoma Narrows bridge, which destroyed itself in November 1940, is usually offered as resonance with vortex shedding, and it was not: the collapse was aeroelastic flutter, a self-excited oscillation in which the deck's own twisting motion extracted energy from a steady wind, with no external periodic driving at all. The London Millennium Bridge, closed two days after opening in June 2000, is closer to the textbook case but has its own twist: a small lateral sway made pedestrians adjust their gait in step with it, and the synchronised walking fed the sway, which is a feedback loop rather than a fixed external drive. It was fixed by hanging dampers under the deck, viscous ones to bleed energy out of the lateral sway and tuned masses to broaden the peak, which is the engineering answer to every resonance problem: if the driving cannot be removed, remove the sharpness of the peak.

What is still assumed

Every force in this course so far has been a contact force or the constant $m g$ near the ground, and $m g$ has been taken on trust for twelve lessons. Nothing has said why it is $9.81$ m s⁻², why it is the same for all bodies, or what happens far from the Earth where it is not constant at all.

That question is the last and largest part of the subject. The answer is a force that acts across empty space, falls off as the square of the distance, and is the same law for an apple, the Moon and a comet. Establishing it, and then finding what such a force does to a body over long times, is where the remaining lessons go, and the oscillation machinery just built has one more use there: a satellite in a circular orbit, seen edge on, oscillates exactly as a pendulum does.