IB DP Physics · Theme C · Wave behavior

Simple harmonic motion.

C.1 · Oscillations, the mass–spring system, the pendulum and energy
Standard level and higher level

SL + HL · 3 hours Additional HL · 4 hours

The journey

Four parts.

Part 01

What makes motion simple harmonic

\(a = -\omega^{2} x\), period and frequency

Part 02

Two oscillators

The mass on a spring and the simple pendulum

Part 03

Energy

Kinetic and potential energy through a cycle

Part 04 · HL only

The equations of the motion

Displacement, velocity, phase angle and energy

C.1 SIMPLE HARMONIC MOTION · SL + HL

Part 01

What makes motion simple harmonic.

One condition on the force, and everything that follows from it

C.1 SIMPLE HARMONIC MOTION · SL + HL

The vocabulary

Describing an oscillation.

Equilibrium position: where the resultant force is zero
Displacement \(x\): distance from equilibrium, with a direction
Amplitude \(x_{0}\): the greatest displacement
Period \(T\): the time for one complete oscillation

A MASS ON A SPRING

C.1 SIMPLE HARMONIC MOTION · SL + HL

One complete oscillation

Follow the block from any point until it is next at the same point and moving the same way. From the right-hand end that means: to the center, to the left-hand end, back to the center, back to the right-hand end. The block passes through the center twice in each oscillation, once in each direction.

How far it travels

In one oscillation the block covers four amplitudes: out, back, out the other way and back. A common slip is to call the full width of the motion, end to end, the amplitude. That width is two amplitudes.

To measure a period, time ten or twenty oscillations and divide. Count “zero” as you start the clock, not “one”, and time from the center, where the block is moving fastest and the moment it passes is sharpest.

Three ways to say how fast

Period, frequency and angular frequency.

\(T = \dfrac{1}{f} = \dfrac{2 \pi}{\omega}\)

\(f\): oscillations per second, in hertz
\(\omega = 2 \pi f\): the angular frequency, in \(\mathrm{rad} \,\mathrm{s}^{-1}\)

A mass makes 20 oscillations in 16 s \(T = \dfrac{16}{20} = \class{res}{0.80 \,\mathrm{s}}\) \(f = \dfrac{1}{T} = \dfrac{1}{0.80} = \class{res}{1.25 \,\mathrm{Hz}}\) \(\omega = \dfrac{2 \pi}{T} = \dfrac{2 \pi}{0.80} = \class{res}{7.9 \,\mathrm{rad} \,\mathrm{s}^{-1}}\)

C.1 SIMPLE HARMONIC MOTION · SL + HL

Why a third quantity

The equations of simple harmonic motion are full of sines and cosines, and those work in angles. One complete oscillation corresponds to one complete turn, \(2 \pi\) radians. \(\omega\) is the number of radians of that cycle covered each second, so it is \(2 \pi\) times the number of cycles each second. Using \(\omega\) keeps factors of \(2 \pi\) out of every other equation. The slide after next shows where the angle comes from.

The same symbol as in A.2

There \(\omega\) was the angular velocity of a body moving in a circle. That is no coincidence, as Part 1 goes on to show.

Using \(f\) where \(\omega\) is needed. The numerical value of \(\omega\) in rad s⁻¹ is about six times the numerical value of \(f\) in Hz, and it is \(\omega\) that appears in \(a = -\omega^{2} x\).

What the force must do

The condition for simple harmonic motion.

Condition 1

The force is proportional to the displacement

Twice as far from equilibrium: twice the force

Condition 2

The force is directed towards equilibrium

Always opposite to the displacement: a restoring force

\(F = -kx\)

Since \(F = ma\), the acceleration is also proportional to the displacement and opposite to it.

C.1 SIMPLE HARMONIC MOTION · SL + HL

Both conditions are needed

A ball bouncing on a hard floor moves back and forth, but the force on it is its constant weight, not a force that grows with displacement: its motion is periodic, not simple harmonic. A force proportional to displacement but pointing away from equilibrium would not give an oscillation at all: the body would run away faster and faster.

Why so many things do it

Push almost any stable system a little way from equilibrium and the force pulling it back is, to a good approximation, proportional to how far it was pushed. Atoms in a solid, a ruler clamped to a bench, the air in a bottle, a building in an earthquake and the carbon dioxide molecules of B.2 all oscillate with simple harmonic motion for small displacements.

Asked for “the conditions for simple harmonic motion”, give both: acceleration (or force) proportional to displacement, and directed towards the equilibrium position.

The two conditions in one line

The defining equation.

\(a = -\omega^{2} x\)

\(a \propto x\): the graph is a straight line through the origin
The minus sign: \(a\) is always opposite to \(x\)
The constant, \(\omega^{2}\), sets the period

ACCELERATION AGAINST DISPLACEMENT

C.1 SIMPLE HARMONIC MOTION · SL + HL

What the minus sign does

When the body is to the right of equilibrium (\(x\) positive) its acceleration is to the left (negative), whichever way it happens to be moving. If it is moving outwards, it slows. If it is moving back, it speeds up. Either way it is being pulled towards the center, and that is what keeps it oscillating instead of leaving.

Where it accelerates most

At the two ends, where \(x\) is largest, the acceleration is largest, although the body is momentarily at rest there. At the center, where it is moving fastest, the acceleration is zero. Speed and acceleration are greatest at different places.

Why the constant is written as a square

Whatever the constant is, it must be positive for the motion to be an oscillation. Writing it as \(\omega^{2}\) guarantees that, and \(\omega\) turns out to be the angular frequency: a larger constant means a stronger pull back and a quicker oscillation.

To show that a motion is simple harmonic, show that its acceleration can be written as \(-(\text{a positive constant}) \times x\). The constant is then \(\omega^{2}\).

Where \(\omega\) comes from

The shadow of circular motion.

A point moves round a circle at a steady angular speed \(\omega\)
Seen from the side, it moves back and forth: simple harmonic motion
Radius → amplitude \(x_{0}\)
One revolution → one period: \(T = \dfrac{2 \pi}{\omega}\)

A CIRCLE SEEN EDGE ON

C.1 SIMPLE HARMONIC MOTION · SL + HL

Why the shadow obeys \(a = -\omega^{2} x\)

From A.2, the point on the circle has a centripetal acceleration \(\omega^{2} r\) directed towards the center. The shadow shows only the vertical part of everything. The vertical part of the position is the displacement \(x\); the vertical part of an acceleration that points at the center and has size \(\omega^{2} r\) is \(\omega^{2} x\), directed back towards the middle. So the shadow’s acceleration is \(-\omega^{2} x\).

Watch the speed

The point on the circle never changes speed. Its shadow moves fastest through the middle, where the point is traveling straight up or down, and stops for an instant at the top and bottom, where the point is traveling sideways.

This picture is the reason for the name “angular frequency”, and at HL it is where the phase angle comes from: the angle the rotating radius has turned through.

Three graphs against time

Displacement, velocity and acceleration.

All three are sinusoidal, with the same period
Velocity: the gradient of displacement. Zero at the ends, greatest at the center
Acceleration: displacement turned upside down

RELEASED FROM THE GREATEST DISPLACEMENT

C.1 SIMPLE HARMONIC MOTION · SL + HL

Reading down the dashed lines

At the start the displacement is at its greatest, the velocity is zero and the acceleration is at its most negative: the body is at rest at one end, being pulled back hardest. A quarter of a period later it is at the center: zero displacement, the greatest speed (negative, because it is heading back) and no acceleration. After half a period it is at the other end, and everything is reversed.

Building one graph from another

Velocity is the gradient of the displacement graph, as in A.1: where displacement peaks, its gradient and so the velocity are zero. Acceleration is the gradient of the velocity graph. It is also, more simply, \(-\omega^{2}\) times the displacement, which is why it is the displacement graph flipped over.

Drawing velocity in step with displacement. They are a quarter of a cycle apart: when one is at a peak, the other is at zero.

Worked example

The greatest acceleration.

A platform oscillates with period 0.50 s and amplitude 4.0 cm \(\omega = \dfrac{2 \pi}{T} = \dfrac{2 \pi}{0.50} = 12.6 \,\mathrm{rad} \,\mathrm{s}^{-1}\) The size of \(a\) is greatest at \(x = x_{0}\) \(a = \omega^{2} x_{0} = 12.6^{2} \times 0.040\) \(a = \class{res}{6.3 \,\mathrm{m} \,\mathrm{s}^{-2}}\)

At \(x = 2.0\) cm: half the displacement, half the acceleration
At the center: zero
Halve the period and \(\omega^{2}\), and so \(a\), is four times as great

C.1 SIMPLE HARMONIC MOTION · SL + HL

A question that follows from this

A coin rests on a platform that oscillates up and down. At the top of the motion the platform is accelerating downwards with \(\omega^{2} x_{0}\). If that is more than \(g\), the platform falls away faster than the coin can follow, and the coin loses contact. Here \(6.3 < 9.8\), so the coin stays on. It would leave if the amplitude exceeded \(\frac{g}{\omega^{2}} = \frac{9.8}{12.6^{2}} = 6.2\) cm.

Why frequency matters so much

The acceleration depends on the square of the frequency. A loudspeaker cone moving 1 mm at 1 kHz has a greatest acceleration of \((2 \pi \times 1000)^{2} \times 0.001 \approx 4 \times 10^{4}\) m s⁻², thousands of times \(g\).

Leaving the amplitude in centimeters. 4.0 cm is 0.040 m.

Quick check

What makes motion simple harmonic.

01State the defining equation of simple harmonic motion.

\(a = -\omega^{2} x\)

02What does the minus sign mean?

\(a\) is opposite to \(x\): towards equilibrium

03An oscillator has \(f\) = 2.0 Hz. Period?

0.50 s

04Where is the speed greatest?

At the equilibrium position

05Where is the acceleration greatest?

At the greatest displacement

06How far does an oscillator travel in one period?

Four amplitudes

PRESS → TO REVEAL EACH ANSWER

Part 02

Two oscillators.

The mass on a spring and the simple pendulum, and what sets their periods

C.1 SIMPLE HARMONIC MOTION · SL + HL

First oscillator

A mass on a spring.

Hooke’s law gives the restoring force \(F = -kx\) \(a = \dfrac{F}{m} = -\dfrac{k}{m} x\) Compare with \(a = -\omega^{2} x\) \(\omega^{2} = \dfrac{k}{m}\)

\(T = 2 \pi \sqrt{\dfrac{m}{k}}\)

More mass: a longer period
A stiffer spring: a shorter period
The amplitude does not appear

C.1 SIMPLE HARMONIC MOTION · SL + HL

Why each factor acts as it does

A larger mass has more inertia: the same spring force gives it a smaller acceleration, so it takes longer to get anywhere. A stiffer spring pulls harder at every displacement, so the mass is hurried back and forth more quickly.

Why the amplitude makes no difference

Double the amplitude and the mass has twice as far to go. But the spring force at every stage is also doubled, so every acceleration and every speed is doubled. Twice the distance at twice the speed takes the same time. This property, a period independent of amplitude, is called isochronism, and it belongs to simple harmonic motion alone.

Hanging vertically

The same formula holds for a mass hanging from a spring. Gravity only shifts the equilibrium position downwards. Measured from that new equilibrium, the resultant force is still \(-kx\).

Because \(T \propto \sqrt{m}\), four times the mass is needed to double the period.

Worked example

The period of a mass on a spring.

A 0.40 kg mass on a spring of \(k = 25 \,\mathrm{N} \,\mathrm{m}^{-1}\) \(T = 2 \pi \sqrt{\dfrac{m}{k}}\) \(T = 2 \pi \sqrt{\dfrac{0.40}{25}}\) \(T = \class{res}{0.79 \,\mathrm{s}}\)

What mass would give a period of 1.0 s? \(T^{2} = \dfrac{4 \pi^{2} m}{k}\) so \(m = \dfrac{k T^{2}}{4 \pi^{2}}\) \(m = \dfrac{25 \times 1.0^{2}}{4 \pi^{2}}\) \(m = \class{res}{0.63 \,\mathrm{kg}}\)

C.1 SIMPLE HARMONIC MOTION · SL + HL

Weighing without gravity

An astronaut in orbit cannot stand on scales: nothing presses down on them. Instead the astronaut sits in a chair mounted on springs of known stiffness and is set oscillating. Rearranging the formula as in the second column gives the mass from the measured period. The method works because the period of a spring oscillator depends on inertia, not on weight.

Finding a spring constant

Hang different masses on a spring, time the oscillations, and plot \(T^{2}\) against \(m\). The graph is a straight line through the origin of gradient \(\frac{4 \pi^{2}}{k}\).

Forgetting the square root, or squaring \(2 \pi\) but not \(T\) when rearranging. Square the whole equation first, then rearrange.

Second oscillator

The simple pendulum.

\(T = 2 \pi \sqrt{\dfrac{l}{g}}\)

A small bob on a light string of length \(l\)
Longer: a longer period
Neither the mass nor the amplitude appears
True for small angles only

A SMALL SWING

C.1 SIMPLE HARMONIC MOTION · SL + HL

Why the mass cancels

The force pulling the bob back towards the center is a component of its weight, so it is proportional to the mass. The inertia that resists the acceleration is also proportional to the mass. Double the mass and both double, leaving the acceleration, and the period, unchanged. It is the same reason all bodies fall with the same acceleration in A.1.

Why only small angles

The restoring force is \(mg \sin \theta\), where \(\theta\) is the angle of the string to the vertical. Simple harmonic motion needs a force proportional to the displacement, which would be \(mg \theta\). For small angles in radians, \(\sin \theta\) and \(\theta\) are almost equal: at 10° they differ by half a per cent. At larger angles the force falls short of what is needed, and the period grows with amplitude.

\(l\) is measured to the center of the bob, from the point of suspension.

Keep the swing below about 10° in any experiment. The period of a 1 m pendulum is then within 0.2% of the formula.

The story · The Hague, 1656

Background story · not assessed

Huygens puts a pendulum in a clock.

Painted portrait of Christiaan Huygens, a man with long dark curled hair and a lace collar — **Christiaan Huygens**, painted by Caspar Netscher in 1671

Engraving of the mechanism of a clock seen from the side, with gear wheels, hanging weights on pulleys and a long pendulum ending in a round bob — His **first pendulum clock**, as he drew it in 1658

About 1602, Galileo had found that a pendulum’s period does not depend on the size of its swing
On 25 December 1656 Huygens invented a clock regulated by a pendulum
The best clocks went from errors of about 15 minutes a day to about 15 seconds
He also found that wide swings spoil the timekeeping

Pendulum clocks were the most precise timekeepers in the world until the 1930s.

Portrait: Caspar Netscher, 1671, public domain. Drawing: Christiaan Huygens, 1658, public domain.

C.1 SIMPLE HARMONIC MOTION · SL + HL

Worked example

A pendulum that ticks seconds.

Find the length for a period of 2.0 s \(g = 9.8 \,\mathrm{m} \,\mathrm{s}^{-2}\) \(T^{2} = \dfrac{4 \pi^{2} l}{g}\) so \(l = \dfrac{g T^{2}}{4 \pi^{2}}\) \(l = \dfrac{9.8 \times 2.0^{2}}{4 \pi^{2}}\) \(l = \class{res}{0.99 \,\mathrm{m}}\)

On the Moon: \(g = 1.6 \,\mathrm{m} \,\mathrm{s}^{-2}\) \(T = 2 \pi \sqrt{\dfrac{l}{g}} = 2 \pi \sqrt{\dfrac{0.99}{1.6}}\) \(T = \class{res}{4.9 \,\mathrm{s}}\) A spring oscillator would not change

C.1 SIMPLE HARMONIC MOTION · SL + HL

The seconds pendulum

A pendulum with a period of 2.0 s swings from one side to the other in exactly one second, so each tick or tock of the clock marks a second. Its length, just under a meter, is why grandfather clocks are the height they are.

The two oscillators compared

The pendulum’s restoring force comes from gravity, so its period depends on \(g\) and changes from place to place. The spring’s restoring force comes from the spring, so its period depends on \(k\) and \(m\) and is the same on the Earth, on the Moon and in orbit. A pendulum clock taken up a mountain runs slow; a watch regulated by a spring does not.

\(T \propto \sqrt{l}\): to double the period, make the pendulum four times as long.

A classic experiment

Measuring \(g\) with a pendulum.

\(T^{2} = \dfrac{4 \pi^{2}}{g} \, l\)

Time 20 swings for each of several lengths
Plot \(T^{2}\) against \(l\): a straight line through the origin
\(g = \dfrac{4 \pi^{2}}{\text{gradient}}\)

PERIOD SQUARED AGAINST LENGTH

C.1 SIMPLE HARMONIC MOTION · SL + HL

Why square the period

A graph of \(T\) against \(l\) is a curve, and a curve cannot be tested by eye or used to find a constant. Squaring the equation gives \(T^{2}\) proportional to \(l\): a straight line, whose gradient contains \(g\). Choosing what to plot so as to get a straight line is the standard way to test a relationship.

Getting a good value

Timing twenty swings and dividing by twenty cuts the effect of your reaction time by a factor of twenty. Use small swings. Measure the length to the center of the bob. A careful school experiment gives \(g\) to about 1%.

If the line misses the origin

A small positive or negative intercept usually means that every length was measured to the wrong point on the bob, by the same amount. The gradient, and so the value of \(g\), is not affected: one more advantage of using a graph.

For a gradient of 4.03 s² m⁻¹: \(g = \frac{4 \pi^{2}}{4.03} = 9.8\) m s⁻².

Quick check

Two oscillators.

01The mass on a spring is made four times as large. The period?

It doubles

02The amplitude of a mass on a spring is doubled. The period?

Unchanged

03A pendulum’s bob is replaced by a heavier one. The period?

Unchanged

04A pendulum is taken to the Moon. The period?

Longer: \(g\) is smaller

05Why must a pendulum’s swings be small?

So the restoring force is proportional to displacement

06Period of 0.50 kg on a spring of 200 N m⁻¹?

\(2 \pi \sqrt{\frac{0.50}{200}} = 0.31 \,\mathrm{s}\)

PRESS → TO REVEAL EACH ANSWER

Part 03

Energy.

An oscillation as energy passing back and forth between two forms

C.1 SIMPLE HARMONIC MOTION · SL + HL

Kinetic and potential

Energy during one cycle.

At each end: at rest, so all potential
At the center: fastest, so all kinetic
In between: some of each
With no resistive forces the total stays the same

ENERGY PASSING BACK AND FORTH

C.1 SIMPLE HARMONIC MOTION · SL + HL

What the potential energy is

For the mass on a spring it is the elastic potential energy of the stretched or compressed spring. For a pendulum it is gravitational potential energy: the bob is highest at each end of its swing and lowest at the center. In both cases the energy is stored when the body is displaced and returned as the body comes back.

Why an oscillation happens at all

At the center there is no force, but the body has kinetic energy and overshoots. Moving outwards, it does work against the restoring force and slows, until all the kinetic energy has been stored. Then the stored energy is handed back and the body returns. Inertia carries it through the center; the restoring force brings it back. Take away either and there is no oscillation.

At SL you describe these changes in words and with sketch graphs. Calculations of the energy are for HL, in Part 4.

Where the energy is

Energy against displacement.

Potential: zero at the center, greatest at \(\pm x_{0}\)
Kinetic: the reverse
The curves are parabolas
Total: a horizontal line

ONE OSCILLATOR, THREE CURVES

C.1 SIMPLE HARMONIC MOTION · SL + HL

Why parabolas

The energy stored in a spring is \(\frac{1}{2} k x^{2}\), from A.3: proportional to the square of the displacement. That is the potential energy curve. The kinetic energy is whatever is left of the total, so its curve is the same parabola turned upside down.

HL only Where the two are equal

The curves cross where each is half the total. That is not half way out. Half the greatest potential energy needs \(x^{2} = \frac{1}{2} x_{0}^{2}\), so \(x = \frac{x_{0}}{\sqrt{2}} = 0.71 x_{0}\): about seven tenths of the way to the end. For most of the distance the energy is mainly kinetic.

Doubling the amplitude quadruples the total energy, since the potential energy at the end is proportional to \(x_{0}^{2}\).

Drawing the curves as straight lines, or as sine curves. Against displacement they are parabolas. Sine-like curves appear only when energy is plotted against time.

When the energy is where

Energy against time.

Each energy rises and falls twice in every period
Never negative
When one is greatest the other is zero
Their sum is constant

ONE PERIOD, STARTING AT AN END

C.1 SIMPLE HARMONIC MOTION · SL + HL

Why twice

In one period the body visits both ends. The potential energy is at its greatest at each of them, whichever side it is on, because energy depends on the square of the displacement and does not care about the sign. So the potential energy peaks twice in a period. The kinetic energy peaks each time the body passes the center, which also happens twice.

The frequency of the energy is therefore double the frequency of the oscillation. An oscillator of period 2.0 s has kinetic energy that goes through a full cycle every 1.0 s.

A sketch needs: both curves between zero and the same greatest value, never below the axis, two humps of each per period, and the peaks of one above the zeros of the other.

Quick check

Energy.

01Where is the kinetic energy greatest?

At the equilibrium position

02Where is the potential energy greatest?

At the greatest displacement

03What form is the potential energy of a pendulum?

Gravitational

04What shape is the graph of potential energy against displacement?

A parabola

05An oscillator has period 4.0 s. Period of its kinetic energy?

2.0 s

06What happens to the total energy if there are no resistive forces?

It stays constant

PRESS → TO REVEAL EACH ANSWER

Part 04 · Additional higher level

The equations of the motion.

Displacement and velocity as functions of time · phase angle
Calculating the energy

C.1 SIMPLE HARMONIC MOTION · ADDITIONAL HL

Where it is and how fast, at any time

Displacement and velocity.

\(x = x_{0} \sin (\omega t + \phi)\)

\(v = \omega x_{0} \cos (\omega t + \phi)\)

Greatest speed
\(\omega x_{0}\)At the equilibrium position
Greatest acceleration
\(\omega^{2} x_{0}\)At each end
Angles
In radians\(\omega t + \phi\) is an angle: set the calculator to radians

C.1 SIMPLE HARMONIC MOTION · ADDITIONAL HL

Checking against the defining equation

Velocity is the rate of change of displacement, and the gradient of a sine curve is a cosine curve, scaled by \(\omega\): that gives \(v\). The gradient of the cosine curve is minus a sine curve, scaled by \(\omega\) again: \(a = -\omega^{2} x_{0} \sin (\omega t + \phi)\), which is \(-\omega^{2} x\). The sine function is the solution of \(a = -\omega^{2} x\).

The circle again

\(\omega t + \phi\) is the angle the rotating radius of Part 1 has turned through. \(\phi\) is where it started. The sine of that angle, times the radius, is the height of the shadow.

The greatest values

Sine and cosine never exceed 1, so the greatest displacement is \(x_{0}\) and the greatest speed is \(\omega x_{0}\). A larger amplitude or a higher frequency both mean a faster passage through the center.

Working in degrees. \(\omega t\) is in radians: with \(\omega = 7.85\) rad s⁻¹ and \(t = 0.10\) s the angle is 0.785 rad, not 0.785°.

Where in the cycle it starts

Phase angle.

\(\phi\) sets the displacement at \(t = 0\): \(x = x_{0} \sin \phi\)
\(\phi = 0\): starts at the center, moving in the positive direction
\(\phi = \dfrac{\pi}{2}\): starts at \(+x_{0}\), and \(x = x_{0} \cos \omega t\)

A whole cycle is \(2 \pi\) rad.

THE SAME OSCILLATION, STARTED EARLIER

C.1 SIMPLE HARMONIC MOTION · ADDITIONAL HL

Phase difference between two oscillators

Two oscillations of the same frequency keep a fixed relationship: one always reaches its peak a certain fraction of a cycle before the other. That fraction, as an angle, is their phase difference. Half a cycle apart is \(\pi\) rad: when one is at \(+x_{0}\) the other is at \(-x_{0}\), and they are said to be in antiphase. A whole cycle apart is \(2 \pi\) rad, which is the same as being in phase.

From a time to an angle

A time difference \(\Delta t\) is the fraction \(\frac{\Delta t}{T}\) of a cycle, so the phase difference is \(\frac{\Delta t}{T} \times 2 \pi\). For two oscillators of period 0.80 s, one 0.20 s ahead of the other, it is \(\frac{0.20}{0.80} \times 2 \pi = \frac{\pi}{2}\) rad.

Within one oscillator

Velocity leads displacement by \(\frac{\pi}{2}\), and acceleration is \(\pi\) out of phase with displacement: the three graphs of Part 1, restated as phase angles.

The phase angle matters again in C.3 and C.4: waves that meet in phase reinforce, and waves that meet in antiphase cancel.

Worked example

Position and velocity at a given time.

\(x_{0} = 0.050 \,\mathrm{m}\) · \(T = 0.80 \,\mathrm{s}\) · \(\phi = 0\) \(\omega = \dfrac{2 \pi}{T} = \dfrac{2 \pi}{0.80} = 7.85 \,\mathrm{rad} \,\mathrm{s}^{-1}\) \(\omega t = 7.85 \times 0.10 = 0.785 \,\mathrm{rad}\) \(x = x_{0} \sin \omega t = 0.050 \times \sin 0.785\) \(x = \class{res}{0.035 \,\mathrm{m}}\)

\(v = \omega x_{0} \cos \omega t\) \(v = 7.85 \times 0.050 \times \cos 0.785\) \(v = \class{res}{0.28 \,\mathrm{m} \,\mathrm{s}^{-1}}\) Greatest speed: \(\omega x_{0} = 0.39 \,\mathrm{m} \,\mathrm{s}^{-1}\)

C.1 SIMPLE HARMONIC MOTION · ADDITIONAL HL

Does the answer make sense?

0.10 s is an eighth of the 0.80 s period, so the oscillator is an eighth of the way round its cycle: half way, in time, from the center to the end. Yet it has already covered 0.035 m of the 0.050 m, seven tenths of the distance. It moves fastest near the center and slows as it approaches the end, so it covers most of the ground early.

The time to reach a given place

Rearranged: \(t = \frac{1}{\omega} \sin^{-1} \frac{x}{x_{0}}\). To reach half the amplitude takes \(\frac{1}{7.85} \sin^{-1} 0.5 = \frac{0.524}{7.85} = 0.067\) s: a twelfth of a period, not an eighth.

\(0.785 = \frac{\pi}{4}\). Whenever \(t\) is a simple fraction of \(T\), the angle is the same fraction of \(2 \pi\).

Speed at a place, without the time

Velocity and displacement.

\(v = \pm \omega \sqrt{x_{0}^{2} - x^{2}}\)

At \(x = 0\): \(v = \pm \omega x_{0}\), the greatest speed
At \(x = \pm x_{0}\): \(v = 0\)
\(\pm\): at each place it may be moving either way

\(\omega = 7.85 \,\mathrm{rad} \,\mathrm{s}^{-1}\) · \(x_{0} = 0.050 \,\mathrm{m}\) · find the speed at \(x = 0.030 \,\mathrm{m}\) \(v = \omega \sqrt{x_{0}^{2} - x^{2}}\) \(v = 7.85 \sqrt{0.050^{2} - 0.030^{2}}\) \(v = 7.85 \times 0.040 = \class{res}{0.31 \,\mathrm{m} \,\mathrm{s}^{-1}}\)

C.1 SIMPLE HARMONIC MOTION · ADDITIONAL HL

Where the equation comes from

Take \(x = x_{0} \sin \theta\) and \(v = \omega x_{0} \cos \theta\), where \(\theta\) stands for \(\omega t + \phi\). Since \(\sin^{2} \theta + \cos^{2} \theta = 1\), \(\left(\frac{x}{x_{0}}\right)^{2} + \left(\frac{v}{\omega x_{0}}\right)^{2} = 1\). Rearranging for \(v\) gives the equation on the slide. The time has been eliminated.

The same thing from energy

Kinetic energy = total energy − potential energy: \(\frac{1}{2} m v^{2} = \frac{1}{2} m \omega^{2} x_{0}^{2} - \frac{1}{2} m \omega^{2} x^{2}\). Cancel \(\frac{1}{2} m\) and take the square root. The velocity equation is conservation of energy in disguise.

Subtracting before squaring: \(\sqrt{x_{0}^{2} - x^{2}}\) is not \(x_{0} - x\). Here it is \(\sqrt{0.0025 - 0.0009} = 0.040\), not 0.020.

Putting numbers to Part 3

Calculating the energy.

\(E_\mathrm{T} = \dfrac{1}{2} m \omega^{2} x_{0}^{2}\) \(E_\mathrm{p} = \dfrac{1}{2} m \omega^{2} x^{2}\) \(E_\mathrm{k} = E_\mathrm{T} - E_\mathrm{p}\)

Total energy \(\propto (\text{amplitude})^{2}\) and \(\propto (\text{frequency})^{2}\).

ENERGY AGAINST DISPLACEMENT

C.1 SIMPLE HARMONIC MOTION · ADDITIONAL HL

Where the total comes from

At the center all the energy is kinetic and the speed is \(\omega x_{0}\), so \(E_\mathrm{T} = \frac{1}{2} m (\omega x_{0})^{2}\). The potential energy at displacement \(x\) is the same expression with \(x\) in place of \(x_{0}\): for a spring, \(m \omega^{2} = k\), and it is the familiar \(\frac{1}{2} k x^{2}\).

The kinetic energy in full

\(E_\mathrm{k} = \frac{1}{2} m \omega^{2} (x_{0}^{2} - x^{2})\). It is not in the data booklet, because it is just the difference of the two that are.

The squares matter

Double the amplitude and the energy is four times as great. The same is true of frequency. This is why the energy carried by a wave, in C.2, is proportional to the square of its amplitude.

Fractions are quick here: at \(x = \frac{1}{2} x_{0}\) the potential energy is \(\frac{1}{4}\) of the total, so the kinetic energy is \(\frac{3}{4}\).

Worked example

The energy of an oscillator.

\(m = 0.20 \,\mathrm{kg}\) · \(\omega = 7.85 \,\mathrm{rad} \,\mathrm{s}^{-1}\) · \(x_{0} = 0.050 \,\mathrm{m}\) \(E_\mathrm{T} = \dfrac{1}{2} m \omega^{2} x_{0}^{2}\) \(E_\mathrm{T} = \dfrac{1}{2} \times 0.20 \times 7.85^{2} \times 0.050^{2}\) \(E_\mathrm{T} = \class{res}{1.5 \times 10^{-2} \,\mathrm{J}}\)

At \(x = 0.030 \,\mathrm{m}\) \(E_\mathrm{p} = \dfrac{1}{2} m \omega^{2} x^{2} = \dfrac{1}{2} \times 0.20 \times 7.85^{2} \times 0.030^{2}\) \(E_\mathrm{p} = \class{res}{5.5 \times 10^{-3} \,\mathrm{J}}\) \(E_\mathrm{k} = E_\mathrm{T} - E_\mathrm{p} = \class{res}{9.9 \times 10^{-3} \,\mathrm{J}}\)

C.1 SIMPLE HARMONIC MOTION · ADDITIONAL HL

Quick check · HL

The equations of the motion.

01What is the greatest speed of an oscillator, in terms of \(\omega\) and \(x_{0}\)?

\(\omega x_{0}\)

02What phase angle makes \(x = x_{0} \sin (\omega t + \phi)\) start at \(+x_{0}\)?

\(\frac{\pi}{2}\) rad

03Two oscillators are half a cycle apart. Phase difference?

\(\pi\) rad

04The amplitude is tripled. The total energy?

Nine times as great

05At \(x = \frac{1}{2} x_{0}\), what fraction of the energy is kinetic?

\(\frac{3}{4}\)

06\(\omega\) = 10 rad s⁻¹, \(x_{0}\) = 0.10 m. Speed at \(x\) = 0.060 m?

\(10 \sqrt{0.10^{2} - 0.060^{2}} = 0.80 \,\mathrm{m} \,\mathrm{s}^{-1}\)

PRESS → TO REVEAL EACH ANSWER

Summary

Simple harmonic motion on one slide.

Defining equation

\(a = -\omega^{2} x\)

Proportional to displacement, towards equilibrium

Period

\(T = \frac{1}{f} = \frac{2 \pi}{\omega}\)

Independent of amplitude

Two oscillators

\(2 \pi \sqrt{\frac{m}{k}}\) · \(2 \pi \sqrt{\frac{l}{g}}\)

Mass on a spring; pendulum at small angles

Energy

Kinetic ⇌ potential

All potential at the ends, all kinetic at the center

Motion in time

\(x = x_{0} \sin (\omega t + \phi)\)

\(v = \omega x_{0} \cos (\omega t + \phi)\)
\(v = \pm \omega \sqrt{x_{0}^{2} - x^{2}}\)

Energy calculations

\(E_\mathrm{T} = \frac{1}{2} m \omega^{2} x_{0}^{2}\)

\(E_\mathrm{p} = \frac{1}{2} m \omega^{2} x^{2}\)

C.1 SIMPLE HARMONIC MOTION · SL + HL

Summary

Key words.

Equilibrium positionWhere the resultant force is zero

DisplacementDistance from equilibrium, with a direction

AmplitudeThe greatest displacement

PeriodTime for one complete oscillation

FrequencyOscillations per second

Angular frequency\(\omega = 2 \pi f\), in rad s⁻¹

Restoring forceDirected towards equilibrium

IsochronousPeriod independent of amplitude

Phase angleWhere in its cycle an oscillation starts, in radians

C.1 SIMPLE HARMONIC MOTION · SL + HL