 Vintage Watchstraps

Straps for Vintage Fixed Wire Lug Trench Watches or Officer's Wristwatches

# Simple Harmonic Motion

Simple Harmonic Motion occurs in systems involving a mass coupled to a spring. When the mass is displaced from its rest position and released, it oscillates around the rest position at a ‘natural frequency’, which is determined by the magnitude of the mass and the stiffness of the spring. If the position of the mass is plotted against elapsed time, wave-like curves such as those in the right hand side of Figure 1 are produced. These are called sine waves because they can be drawn by the trigonometric sine function.

The oscillator at the heart of a mechanical watch comprises a mass, the balance, coupled to a spring, the balance spring. The sine wave explains how it is able to keep time despite the amplitude of the oscillation changing, which occurs as the watch is put into different positions, dial up, pendant down etc., and as the main spring runs down. The balance oscillates at its natural frequency, which does not change with changes in amplitude. This is ideal because it endows a watch with isochronism, the ability to keep the same time whatever the amplitude of the balance. The characteristics of Simple Harmonic Motion and sine waves can be used to analyse the balance and spring. To get the basic equations for Simple Harmonic Motion, start with a vector A rotating about a point O at $$\omega$$ radians per second. The height of the point of this vector above a horizontal line through O plotted against time gives the blue sine wave. Since there are $$2\pi$$ radians in a complete circle, the time $$T$$ that the vector takes to complete one revolution is:

$T = \frac{2\pi}{\omega} \quad \textrm{ and } \quad \omega = \frac{2\pi}{T}$

This is called the period and is also the time that the sine wave takes to make one complete oscillation, shown by T under the sine wave. Note that one complete oscillation always includes two peaks of amplitude, velocity and acceleration.

Frequency f is the inverse of the period, so $$f = 1/T$$ or:

$f = \frac{\omega}{2\pi} \quad \textrm{ and } \omega = 2 \pi f$

A watch escapement operating at 18,000 vibrations per hour makes 5 ticks per second. The time between each tick, 0.2 seconds, represents half an oscillation, so the period, the time taken to make one complete oscillation, is 0.4 seconds. The frequency is the inverse of the period, 2.5 oscillations per second or 2.5 Hz.

### Analysis

At time t the angle of rotation of the vector $$\phi$$ is given by $$\phi = \omega t$$. The vertical displacement x, the height of the tip of vector A above a horizontal line passing through O, is given by:

\begin{equation} x=A\sin(\phi)=A\sin(\omega t) \end{equation}

Velocity $$v$$ in the x direction at time t is given by the first differential of the equation for x with respect to time:

\begin{equation} v=\frac{dx}{dt} = \omega A \cos(\omega t) \end{equation}

Acceleration $$a$$ in the x direction at time t is given by the second differential of the equation for x with respect to time:

$a=\frac{d^2x}{dt^2} = -\omega^2A\sin(\omega t)$

Since the second term in this equation, $$A\sin(\omega t)$$, is (from the equation above) equal to x, this can be substituted to give:

\begin{equation} a=-\omega^2x \end{equation}

These equations encapsulate the fundamentals of Simple Harmonic Motion. The final equation $$a=-\omega^2x$$ defines the essential condition for Simple Harmonic Motion, which is that acceleration towards the rest position is proportional to displacement. Acceleration is in the opposite direction to displacement, shown by the mius sign.

The constant of proportionality $$\omega^2$$ is related to the frequency of oscillation by the expression $$\omega = 2 \pi f$$ and period by $$\omega = \frac{2\pi}{T}$$.

To produce an acceleration that is proportional to displacement from the rest position, the restoring force must be proportional to displacement from rest position.

These relationships can be used to examine the frequency of undamped or very lightly damped oscillating systems, such as a mass on a spring.

## Mass Suspended on Spring

To study the motion of a mass suspended on a spring, laws determined by two of the greatest British scientists are used.

Hooke's Law of springs ‘ ut tensio, sic vis’ relates force to the extension $$x$$ of a spring by a ‘spring constant’ $$k$$:

$F=kx$

Newton's Second Law of motion relates force to mass and acceleration:

$F=ma$

Equating the two forces, remembering that acceleration is in the opposite direction to the displacement, gives:

$ma=-kx \quad \textrm{and therefore} \quad a=-\frac{k}{m} x$

This is the same form as the Simple Harmonic Motion relationship between acceleration and displacement, with the constant $$\frac{k}{m}$$ taking the place of $$\omega^2.$$ Therefore:

$\frac{k}{m}=\omega^2 \quad \textrm{and} \quad \omega=\sqrt{\frac{k}{m}}$

Substituting for $$\omega$$ in the equation for period:

$T=\frac{2\pi}{\omega} = 2\pi\sqrt{\frac{m}{k}}$

## Circular Motion

Simple Harmonic Motion can also be used to examine the oscillations of a balance and balance spring. The displacement is angular rather than vertical, denoted by $$\theta$$ instead of y. Note that $$\theta$$ is the angular displacement of the balance, which is not the same as the angle $$\phi$$ of the vector A. The other parameters of oscillating circular Simple Harmonic Motion are torque $$\tau$$, moment of inertia I and angular acceleration $$\alpha$$. The basic equation for simple harmonic motion restated in circular terms of angular acceleration and angular displacement is:

$\alpha =-\omega^2\theta$

Newton's second law restated for circular motion is that torque is equal to the moment of inertia of the balance multiplied by the angular acceleration: $$\tau=I\alpha$$. Hooke's law restated for a spiral or helical spring is that torque is proportional to the spring constant S and the angular displacement $$\theta$$, the angle that one end of the spring is turned relative to the other, $$\tau=S\theta$$. In an oscillating balance and spring assembly, inertia opposes torque, so:

$\tau = I\alpha = -S\theta$ $\therefore \alpha=-\frac{S}{I}\theta$

This is the same form as the Simple Harmonic Motion relationship between acceleration and displacement, with the constant $$\frac{S}{I}$$ taking the place of $$\omega^2.$$ Therefore:

$\omega^2 = \frac{S}{I} \quad \textrm{and} \quad \omega = \sqrt{\frac{S}{I}}$

Substituting for $$\omega$$ in the equation for period gives:

$T=2\pi\sqrt{\frac{I}{S}}$

## Balance Acceleration

The angular acceleration $$\alpha$$ of the balance is given by:

$\alpha = -\omega^2 A \sin(\omega t)$

The value of a sine function varies between 0 and 1, so acceleration is at a maximum when $$\sin( \omega t) = 1$$.

$\alpha_{max} = -\omega^2 A$

The equation for period can be used to express $$\omega\$$ in terms of the period $$T$$:

$T = \frac{2\pi}{\omega} \quad \textrm{therefore} \quad \omega = \frac{2\pi}{T}$

Substituting for $$\omega$$ into the equation for $$\alpha_{max}$$ gives:

$\alpha_{max} = -\frac{4\pi^2 A}{T^2}$

This shows that $$\alpha_{max}$$ is proportional to the amplitude, and inversely proportional to the period squared.

## Balance Velocity

This gets a bit confusing because $$\omega$$ is the angular velocity of the SHM vector, which is a constant, and $$v$$ is the angular velocity of the balance, which oscillates. The angular velocity of the balance is given by:

\begin{equation} v=\omega A \cos(\omega t) \end{equation}

The value of a cosine function varies between 0 and 1, so velocity is at a maximum when $$cos( \omega t) = 1$$.

$V_{max} = \omega A$

The equation for period can be used to express $$\omega\$$ in terms of the period $$T$$:

$T = \frac{2\pi}{\omega} \quad \textrm{therefore} \quad \omega = \frac{2\pi}{T}$

Substituting for $$\omega$$ into the equation for $$V_{max}$$ gives:

$V_{max}=\frac{2\pi A}{T }$

### Velocity Example

In watch terms, an oscillation is two vibrations of the balance, or two ticks of a lever escapement. An 18,000 vph (vibrations per hour) lever escapement movement makes 5 ticks per second. Its frequency is half that; 2.5 Hz, and its period $$T=0.4\textrm{sec}$$.

For an 18,000 vph balance with an amplitude of 270°:

$V_{max}=\frac{2\pi\times 270}{0.4} = 4,241^\circ\textrm{/sec} = 74.022 \textrm{ radians/sec} = 11.8 \textrm{ revolutions/sec}$

Of course the balance isn't actually revolving at nearly 12 revolutions per second; this is its instantaneous peak angular velocity as it passes though the central position, which occurs twice per oscillation.

The linear velocity of the rim of the balance is given by angular velocity in radians per second multiplied by the radius of the rim. A 10mm diameter balance will therefore have a peak velocity:

$V_{linmax} = 74.022 \times 5.10^{-3} = 0.37 \textrm{ m/sec}$

A Hamilton Model 21 marine chronometer has a balance 35.6mm diameter. The amplitude is 270°, which is 4.7 radians. The balance oscillates at 14,400 vph, so T = 0.5 seconds.

$V_{max} = \frac{2\pi\times 4.7}{0.5} = 59 \textrm{ radians/sec}$

## Total Energy

As the balance swings the balance spring bends along its length, coiling and uncoiling, which stores and releases energy. When the balance reaches the limits of its swing at, say, +/- 270°, the spring is at its maximum bend and the balance stops moving as its swing reverses. At the this point, all the energy of the system is stored in the spring. The spring accelerates the balance towards the neutral position. As it passes through the neutral position, the velocity of the balance is at its maximum and the spring force becomes zero as it stops accelerating the balance and starts slowing it down.

From the equation for period:

$T=2\pi\sqrt{\frac{I}{S}}$

The spring term $$S$$ is the torque generated by the spring per radian of bend. It can be written as:

$S = I \left( \frac{2\pi}{T} \right)^2$

At maximum bend, the energy stored in the spring is given by $$S \theta$$. For the Hamilton balance this works out as

Kinetic energy is given in angular terms by $$\frac{1}{2}I(v)^2$$. The Hamilton balance has a mass of 5.1 gram and a diameter of 35.6mm giving a radius of 17.8mm. Not all of the mass will be in the rim, and the radius of gyration will be less that the radius of the balance, but for a first-order approximation let's take these dimensions:

$\frac{1}{2} \times 5.1\!\times\!10^{-3} \times (17.8\! \times\! 10^{-3})^2 \times 59^2 = 2.81\!\times\!10^{-3} joule$

The maximum linear velocity $$L$$ of the rim is given by its radius multiplied by the maximum angular velocity:

$L = 17.8\!\times\!10^{-3} \times 59 = 1.05 m/s \quad \textrm{which is just over 2 miles per hour.}$

So although the balance looks like it is moving pretty quickly, its small size means that its speed through the air is quite low. The peak kinetic energy of the balance is then:

$KE = \frac{1}{2} mL^2 = \frac{1}{2} 5.1\! \times\! 10^{-3} \times 1.05^2 = 2.81\!\times\!10^{-3} joule$

## Balance Amplitude

The vector angle $$\phi$$ at time t is given by:

$\phi=\frac{2\pi t}{T}$

The angular displacement $$\theta$$ of the balance is:

$\theta=A\sin(\phi) = A \sin\left( \frac{2\pi t}{T} \right)$

If $$\theta$$ is the lift angle, the amplitude A is:

$A = \frac{\theta}{\sin\left( \frac{2\pi t}{T} \right)} \quad \textrm{or} \quad A = \frac{\theta}{\sin\left( \frac{360 t}{T} \right)} \textrm{ if working in degrees}$

Since for small angles $$\sin(\theta) \approx \theta$$, this can be simplified to:

$A = \frac{\theta T}{2\pi t}$

### Amplitude Example

An 18,000 vph movement has a 52° lift angle and lift time measured on a timing machine of 0.012 seconds. The movement makes 5 ticks per second so the ticks are $$\frac{1}{5}$$ or 0.2 seconds apart. The period $$T$$ of a complete oscillation is two ticks, which is 0.4 seconds. The amplitude is then:

$A = \frac{52\times 0.4}{2\pi\times 0.012} = 275.9^\circ$

The simplified equation works with the lift angle in either degrees or radians, the result will be in the same units. The more complicated version with the sine term gives a result of 277.5° a difference of only 1.6° which is usually insignificant – the fractions after the decimal point are included only to show this. Remember to use the appropriate equation for either degrees or radians.

## Energy

As a balance swings backwards and forwards, its kinetic energy is converted into potential energy stored in the spring, and vice versa. At each end of the swing when the balance is momentarily stationary, its kinetic energy is zero and all the energy is stored in the spring. At the moment the balance passes through the mid point of its swing, it is travelling with maximum velocity and all the energy is in the kinetic energy of the balance. The total energy in the system at any moment is the sum of the kinetic energy of the balance and the potential energy of the spring.

Kinetic energy of a mass moving in a straight line is given by the equation $$E_k = \frac{1}{2}mv^2$$. The kinetic energy of a rotating object with moment of inertia $$I$$ is given by:

\begin{equation} E_k = \frac{1}{2}Iv^2 \end{equation}

The kinetic energy of the balance is greatest when it is travelling with maximum velocity. The instantaneous velocity of the balance is given by:

\begin{equation} v=\omega A \cos(\omega t) \end{equation}

The value of a cosine function varies between 0 and 1, so velocity is at a maximum when $$cos( \omega t) = 1$$.

$V_{max} = \omega A$

The vector angle $$\omega$$ is related to the period $$T$$ by the equation:

$\omega = \frac{2\pi}{T}$

Substituting these into the first equation gives:

$E_{k(max)} = \frac{1}{2}I \left ( \frac{2\pi}{T} A \right ) ^2$

This can also be written:

$E_{k(max)} = I \cdot \left ( \frac{2\pi}{T} \right ) ^2 \cdot \frac{A^2}{2}$

### Energy Example

A Hamilton Model 21 marine chronometer has a balance 35.6mm diameter with a mass of 5.1 gram. Not all of the mass will be in the rim, and the radius of gyration will be less that the radius of the balance, but for a first-order approximation let's take these dimensions. The amplitude is 270°, which is 4.7 radians. The balance oscillates at 14,400 vph, so T = 0.5 seconds.

$E_{k(max)} = 5.1\times10^{-3} \cdot (17.8 \times 10^{-3})^2 \cdot \left ( \frac{2\pi}{0.5} \right ) ^2 \cdot \frac{4.7^2}{2} = 2.82 \times 10^{-3} \textrm{joule}$

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