A swinging pendulum, a bouncing car suspension, a plucked guitar string, a vibrating atom — all of these are oscillations, and the simplest, most useful model for describing them is simple harmonic motion (SHM). This workbook builds SHM from the ground up: what conditions produce it, how to describe it mathematically, and how energy moves back and forth as it happens. Along the way you will meet the same angular frequency, ω, that you used to describe circular motion in the Gravitation and circular motion workbook — it turns out oscillations and circles are two views of exactly the same mathematics.
By the end of this workbook you should be able to:
state the conditions that lead to simple harmonic motion and use the defining equation a = −ω²x
describe an oscillation using time period T, frequency f, angular frequency ω, amplitude, equilibrium position and displacement, and use T = 1/f = 2π/ω
use the time period of a mass–spring system, T = 2π√(m/k)
use the time period of a simple pendulum, T = 2π√(l/g)
describe, qualitatively, the energy changes that occur during one cycle of an oscillation
describe an oscillator using phase angle, φ
solve problems using the full set of SHM equations listed in the data booklet
1. Conditions for simple harmonic motion
An oscillation is a repeated, back-and-forth movement about a fixed central point. Planets orbiting, hearts beating, guitar strings vibrating, atoms jiggling in a solid — oscillations are everywhere in physics. Simple harmonic motion (SHM) is an idealised model of oscillation: no energy is lost, so the amplitude and time period stay constant forever. Very few real oscillators are perfect examples of SHM, but many are good approximations, which is why the model is so useful.
The restoring force
An oscillator has an equilibrium position: the point where the resultant force on it is zero, and where it would stay at rest if undisturbed. If the oscillator is displaced away from equilibrium, a restoring force pulls (or pushes) it back. This restoring force is what makes oscillation possible at all — without it, a displaced object would simply stay displaced.
For a mass on a spring, the restoring force follows Hooke's law:
F = −kx
The minus sign is essential: it shows that the force, and therefore the acceleration, always acts in the opposite direction to the displacement, x, always pushing or pulling the mass back towards equilibrium, never away from it.
Live simulation: force and displacement in a spring
Stretch or compress the spring and watch how the restoring force changes. Check that the force is always directed back towards the natural length, and that doubling the displacement doubles the force.
Drag the mass to stretch or compress the spring, and read off the applied force and displacement. Try the "Energy" and "Lab" tabs too, once you have explored the basic spring.
Interactive simulation — open the online version of this workbook to launch it.
A restoring force proportional to displacement is the key condition for SHM:
Key idea. Simple harmonic motion occurs whenever the restoring force (and so the acceleration) on an object is proportional to its displacement from equilibrium, and always acts in the opposite direction to that displacement.
Since F = ma, and mass m is constant, a force proportional to −x means the acceleration is also proportional to −x:
a ∝ −x
To turn this into an equation we need a constant of proportionality. That constant turns out to be ω² (you will meet ω, the angular frequency, properly in the next section) — giving the defining equation of SHM:
defining equation of SHM, a = −ω²x
Fig. 1.1 Acceleration–displacement graph for SHM. The straight line through the origin, with a negative gradient of −ω², shows a ∝ −x: the further the object is displaced, the greater its acceleration back towards equilibrium.
Why the model applies so widely: Many very different physical systems — springs, pendulums, vibrating molecules, alternating current — all obey a ∝ −x for small displacements, even though the actual restoring forces involved look completely different. This is why the single SHM model is worth studying in such detail: understanding it once means understanding the mathematics behind a huge range of real oscillators.
Worked example 1.1
A mass oscillates horizontally between two springs with an angular frequency of 8.8 rad s−1. Calculate its acceleration when its displacement is 1.0 cm to the right of equilibrium.
Working:
a = −ω²x = −(8.8)² × (+0.010) ... (complete the calculation, in m s−2, and state its direction)
Check your understanding
1State the two conditions that must both be true of the resultant force (or acceleration) on an object for it to undergo simple harmonic motion.
The force (and so acceleration) must be proportional to the displacement from equilibrium, and it must always act in the opposite direction to that displacement (i.e. always directed back towards equilibrium).
2A student says: "A ball rolling at constant speed in a straight line has zero resultant force, so it must be in equilibrium and therefore undergoing SHM." Explain the error in this statement.
Being in equilibrium (resultant force zero) is different from oscillating about an equilibrium position. SHM requires the object to move away from and then be pulled back towards a fixed central point.
Equilibrium (zero resultant force) is a necessary feature of the central point of an SHM oscillator, but the ball is not oscillating — it is not repeatedly displaced from and restored to a fixed point. SHM requires a restoring force that appears specifically when the object is displaced from equilibrium; the ball experiences no such force at all, so it is not oscillating and is not an example of SHM.
3An oscillator has a defining equation a = −(4.0)x, where a is in m s−2 and x is in m. Determine its angular frequency, and calculate its acceleration at a displacement of 0.25 m.
Compare the given equation directly with a = −ω²x to identify ω² first.
Comparing with a = −ω²x: ω² = 4.0, so ω = 2.0 rad s−1. Then a = −(4.0) × (0.25) ... (complete the calculation, in m s−2)
2. Describing SHM
To describe any particular SHM oscillator precisely, physicists use a small, standard set of quantities. Most of these — time period, frequency and angular frequency — should already be familiar from your work on circular motion.
Displacement, x: the distance of the oscillator from its equilibrium position, in a specified direction, at a given instant. Displacement is a vector and varies continuously as the oscillator moves.
Amplitude, x₀: the maximum displacement of the oscillator from equilibrium. Unlike displacement, amplitude is a single fixed value for a given oscillation (assuming no energy loss).
Equilibrium position: the position at which the resultant force on the oscillator is zero — the central point about which it oscillates.
Time period, T: the time taken for one complete oscillation (cycle). Unit: s.
Frequency, f: the number of complete oscillations per unit time. Unit: hertz, Hz (1 Hz = 1 oscillation per second).
Angular frequency, ω: a way of describing oscillation frequency in radians per second, borrowed directly from circular motion (see the Gravitation and circular motion workbook). Unit: rad s−1.
Frequency and time period carry exactly the same information, just expressed differently — we simply use whichever is more convenient:
time period, T = 1/f = 2π/ω
Fig. 2.1 A displacement–time graph for SHM is sinusoidal. The amplitude, x₀, is the maximum height of the curve above (or below) the equilibrium line; the period, T, is the time for one complete cycle.
Live simulation: SHM and circular motion
This applet shows the same connection you met in the Gravitation and circular motion workbook: a point moving at constant speed around a circle, and its shadow oscillating back and forth along a diameter with SHM. Change the frequency and amplitude, and watch how the displacement–time graph responds.
Press start, then try the "Show Vertical" and "Show Horizontal" options to see both projections of the circular motion. Adjust the frequency and amplitude sliders and watch how the graph changes.
Interactive simulation — open the online version of this workbook to launch it.
GeoGebra applet "C1. UCM & SHM" by Learn IB Physics, via geogebra.org.
Worked example 2.1
A mass oscillates between two identical springs. The distance between its two extreme positions is 18 cm, and it completes one full oscillation every 1.5 s.
Working:
Amplitude: the distance between the extremes is twice the amplitude, so x₀ = 18/2 = 9 cm.
Frequency: f = 1/T = 1/1.5 ... (complete the calculation, in Hz)
Angular frequency: ω = 2πf ... (use your answer above to complete this, in rad s−1)
Common mistake The amplitude of an oscillation is not the distance between its two extreme positions — that distance equals twice the amplitude. A single movement from one extreme to the other is also not a complete oscillation — it is only half of one.
Check your understanding
4A loudspeaker cone vibrates with a frequency of 250 Hz. Calculate its time period and its angular frequency.
T = 1/f = 1/250 ... (complete the calculation, in s). ω = 2πf = 2π × 250 ... (complete the calculation, in rad s−1)
5An oscillator has an angular frequency of 12.0 rad s−1. Calculate its time period and frequency.
Rearrange T = 2π/ω first, then use f = 1/T (or f = ω/2π directly).
T = 2π/ω = 2π/12.0 ... (complete the calculation, in s). f = 1/T (use your answer above to complete this, in Hz)
6State what is meant by the equilibrium position of an oscillator, and explain why the resultant force there must be zero.
The equilibrium position is the central point about which the oscillator moves, where it would remain at rest if left undisturbed. If the resultant force there were not zero, the object would accelerate away from that point rather than staying there, so by definition the resultant force at equilibrium must be zero.
3. Time period of a mass–spring system
You have already met the equation that defines SHM, a = −ω²x, and the restoring force behind it, Hooke's law, F = −kx. Combining these two equations lets us predict exactly how quickly a mass on a spring will oscillate — without needing to measure it directly.
Deriving the equation
Your task
Starting with Hooke's law, F = −kx, and incorporating Newton's second law, the defining equation of SHM, and the equation relating time period to angular frequency, derive the expression for the time period of a mass–spring system:
T = 2π√(m/k)
Use the space below (or your notebook) to show every step of your working.
For a mass on a spring, the restoring force follows Hooke's law: F = −kx
Newton's second law gives F = ma, so: ma = −kx
For any object undergoing SHM, a = −ω²x
Substituting this expression for a into the line above: m(−ω²x) = −kx
The factor of x and the minus sign cancel from both sides, leaving: mω² = k, so ω² = k/m
Since T = 2π/ω, substituting ω = √(k/m) gives the time period of a mass–spring system: T = 2π√(m/k)
T = 2π√(m/k)
Notice: a bigger mass gives a bigger time period — a heavier object oscillates more slowly. A bigger spring constant (a stiffer spring) gives a smaller time period — a stiffer spring oscillates faster. Neither the amplitude nor g appears in the equation at all: for an ideal spring, the time period of a mass–spring system does not depend on how far you pull it before releasing it.
Worked example 3.1
A trolley of mass 0.40 kg is attached to a horizontal spring of spring constant 25 N m−1 and set oscillating. Calculate the time period of the oscillation.
Working:
T = 2π√(m/k) = 2π√(0.40/25) ... (complete the calculation, in s)
Worked example 3.2
A mass of 150 g hangs from a vertical spring and, when disturbed, oscillates with a time period of 0.55 s. Calculate the spring constant of the spring.
Working:
Start from T = 2π√(m/k) and rearrange to make k the subject: square both sides, T² = 4π²m/k, so k = 4π²m/T².
k = 4π² × 0.150 / 0.55² ... (complete the calculation, in N m−1, remembering to convert the mass to kg first)
Check your understanding
7A block of mass 0.25 kg is attached to a spring of spring constant 40 N m−1 and oscillates horizontally on a frictionless surface. Calculate the angular frequency and the time period of the oscillation.
ω = √(k/m) = √(40/0.25) ... (complete the calculation, in rad s−1). T = 2π/ω (use your answer above to complete this, in s) — or equivalently T = 2π√(m/k) = 2π√(0.25/40).
8A student keeps the same spring but doubles the mass attached to it. State and explain what happens to the time period of the oscillation.
Look at how m appears in T = 2π√(m/k) — it is under a square root, not in direct proportion.
Because m is under a square root in T = 2π√(m/k), doubling m does not double T. Instead, T increases by a factor of √2 ≈ 1.41, so the oscillation becomes slower, but only by about 41%, not 100%.
9To find the spring constant of a spring experimentally, a student measures the time period T for several different masses m and plots a graph of T² on the y-axis against m on the x-axis. Show that this graph should be a straight line through the origin, and find an expression for its gradient in terms of k.
Start from T = 2π√(m/k) and square both sides so that T² is written as (something) × m.
Squaring T = 2π√(m/k) gives T² = (4π²/k) m. This has the form y = (gradient) x + 0 with y = T² and x = m, so a graph of T² against m is indeed a straight line through the origin, with gradient 4π²/k. Once the gradient is measured from the graph, k can be found from k = 4π²/gradient.
4. The simple pendulum
A simple pendulum — a small bob on a light string, swinging through small angles — is another classic example of SHM. The restoring force here is a component of gravity rather than a spring, but the same defining equation, a = −ω²x, still applies.
Deriving the equation
Deriving the time period of a simple pendulum
For a pendulum of length L displaced through a small angle θ, the restoring force is the component of gravity along the direction of motion: F = −mg sinθ
Newton's second law gives F = ma, so: ma = −mg sinθ
For a small angle (measured in radians), sinθ ≈ θ, so: ma = −mgθ
The angle θ and the arc-length displacement x are related by θ = x/L, so: ma = −mg(x/L)
Dividing both sides by m: a = −(g/L)x
Comparing this with the defining equation of SHM, a = −ω²x, shows that ω² = g/L
Since T = 2π/ω, substituting ω = √(g/L) gives the time period of a simple pendulum:
T = 2π√(l/g)
Small-angle approximation: the derivation above only works because sinθ ≈ θ when θ is small (roughly less than about 10°, measured in radians). Swing a pendulum through a large angle and its motion is still periodic, but it is no longer simple harmonic, and this equation for T no longer applies accurately.
Notice: just as with the mass–spring system, the mass of the bob does not appear anywhere in T = 2π√(l/g). A heavy bob and a light bob on strings of the same length, swinging through small angles, have the same time period.
Live simulation: the pendulum lab
Set a single pendulum swinging with a small angle and use the built-in stopwatch and ruler to test how the time period depends on length, mass, and amplitude. Try changing one variable at a time.
Interactive simulation — open the online version of this workbook to launch it.
A simple pendulum on the Earth's surface (g = 9.81 m s−2) has a length of 0.80 m. Calculate its time period.
Working:
T = 2π√(l/g) = 2π√(0.80/9.81) ... (complete the calculation, in s)
Worked example 4.2
The same pendulum is taken to the Moon, where gMoon = 1.62 m s−2. Without recalculating T from scratch, find the ratio TMoon / TEarth.
Working:
Since T ∝ 1/√g at constant length, TMoon / TEarth = √(gEarth / gMoon) = √(9.81/1.62) ... (complete the calculation — and state whether the pendulum swings faster or slower on the Moon)
Check your understanding
10A simple pendulum has a time period of 1.8 s on Earth. Calculate its length.
Rearrange T = 2π√(l/g) to make l the subject before substituting numbers.
Rearranging T = 2π√(l/g): T² = 4π²l/g, so l = gT²/4π² = 9.81 × 1.8² / 4π² ... (complete the calculation, in m)
11Explain why a pendulum clock loses time if its bob is accidentally set swinging through a much larger angle than usual, even though its length has not changed.
The equation T = 2π√(l/g) relies on the small-angle approximation sinθ ≈ θ. At a large swing angle this approximation breaks down, the motion is no longer simple harmonic, and the true time period becomes slightly longer than the small-angle formula predicts — so a clock calibrated for small swings runs slow.
12Two simple pendulums, X and Y, are the same length, but bob X has twice the mass of bob Y. Compare their time periods, and explain your reasoning.
Look for mass in the equation T = 2π√(l/g) — is it there at all?
The two pendulums have the same time period. Mass does not appear in T = 2π√(l/g), because although a heavier bob needs a larger force to accelerate it (F = ma), gravity also pulls on it proportionally harder (F = mg sinθ) — the mass cancels out of the equation of motion entirely.
5. Energy changes during SHM
As an oscillator moves through one cycle, energy is continuously exchanged between two forms: potential energy (elastic, for a spring, or gravitational, for a pendulum) and kinetic energy. Provided there is no friction or air resistance, the total energy of the oscillator stays constant throughout.
At the extremes (x = ±x₀): the oscillator is momentarily at rest (v = 0), so all of its energy is potential energy. Kinetic energy is zero here.
At equilibrium (x = 0): the oscillator is moving at its maximum speed, so all of its energy is kinetic energy. Potential energy is zero here (measured relative to the equilibrium position).
In between: the oscillator has some of each — as it moves away from equilibrium, kinetic energy is continuously transferred to potential energy, and as it moves back towards equilibrium, that potential energy is transferred back to kinetic energy.
Fig. 5.1 Energy–displacement graph for an undamped SHM oscillator. Potential energy EP (teal) is zero at equilibrium and maximum at the extremes; kinetic energy EK (amber) is maximum at equilibrium and zero at the extremes. Their sum, the total energy ET (dashed line), stays constant throughout the cycle.
Live simulation: mass on a spring
Set the spring oscillating and watch the potential, kinetic and total energy bars change as the mass moves. Pause at the extremes and at equilibrium to check your answers to the questions below.
Interactive simulation — open the online version of this workbook to launch it.
A mass oscillates on a spring with SHM. State the type(s) of energy it possesses (i) at maximum displacement, (ii) at equilibrium, and (iii) at a point in between.
Working:
(i) At maximum displacement the mass is momentarily at rest, so it has only potential energy (elastic potential energy stored in the spring).
(ii) At equilibrium the mass is moving at its maximum speed and the spring is at its natural length, so it has only kinetic energy.
(iii) In between, the mass has both potential and kinetic energy, with their sum always equal to the total energy of the system.
Check your understanding
13Describe how the kinetic energy and the potential energy of an oscillator change as it moves from its equilibrium position to its maximum displacement.
As the oscillator moves from equilibrium to maximum displacement, its speed continuously decreases, so its kinetic energy continuously decreases from a maximum to zero. At the same time its potential energy continuously increases from zero to a maximum, as work is done against the restoring force.
14Using Fig. 5.1, explain why the total energy line ET is horizontal (constant) even though EP and EK both vary with displacement.
Think about what physical principle must hold if there is no friction or air resistance acting on the oscillator.
By conservation of energy, in the absence of friction or air resistance no energy is lost from the system, only transferred between potential and kinetic forms. At every displacement, EP + EK must add up to the same fixed total, ET, so the graph of ET against x is a horizontal line.
15A pendulum bob undergoing SHM passes through a point that is exactly halfway between equilibrium and its maximum displacement. Is its kinetic energy at this point exactly half of its maximum kinetic energy? Explain your reasoning, referring to the shape of the graph in Fig. 5.1.
Look carefully at the curves in Fig. 5.1 — are they straight lines or curves? What does that tell you about proportionality?
No. The EK and EP curves in Fig. 5.1 are parabolas, not straight lines, so energy does not vary in direct proportion to displacement. At the halfway point the bob still has most of its speed (and so most of its kinetic energy), because the potential energy curve rises slowly near equilibrium and only rises steeply near the extremes — so the kinetic energy at the halfway point is considerably more than half of its maximum value.
6. Phase angle
Two oscillators can have exactly the same amplitude and angular frequency, yet be at different points in their cycle at the same instant — one might be at its maximum displacement while the other is passing through equilibrium, for example. The quantity that describes this difference is called the phase angle.
Phase angle, φ: a measure, in radians, of how far through a cycle one oscillator is compared with a reference oscillator (or compared with the point where its displacement would otherwise have been zero and increasing). One full cycle corresponds to a phase angle of 2π rad.
If a reference oscillator has displacement x = x₀ sin(ωt), then a second oscillator with the same amplitude and angular frequency, but shifted in time, has displacement:
x = x₀ sin(ωt + φ)
A positive phase angle means the oscillator leads the reference — it reaches every point in its cycle earlier in time. A negative phase angle means it lags the reference — it reaches every point later.
Fig. 6.1 Displacement–time graphs for two oscillators with the same amplitude and angular frequency, but different phase. Oscillator B (amber) reaches each point in its cycle earlier than oscillator A (teal) — it has a positive phase angle φ relative to A, marked as the small horizontal gap between their peaks.
Live simulation: phase angle explorer
Phase angle explorer
xA = x₀ sin(ωt) xB = x₀ sin(ωt + φ)
Worked example 6.1
Oscillator A has displacement xA = x₀ sin(ωt). Oscillator B has the same amplitude and angular frequency, but reaches its maximum displacement exactly one-sixth of a period before oscillator A does. Determine the phase angle of B relative to A, in radians.
Working:
One full cycle corresponds to a phase angle of 2π rad, so one-sixth of a period corresponds to a phase angle of (1/6) × 2π = π/3 rad.
Because B reaches each point in its cycle earlier than A, it leads A, so its phase angle is positive: φ = +π/3 rad ... (convert to a decimal, to 2 s.f., if required)
Check your understanding
16Two identical oscillators, P and Q, have the same amplitude and angular frequency. Q reaches every point in its cycle exactly a quarter of a period after P does. State the phase angle of Q relative to P, in radians, and state whether Q leads or lags P.
A quarter of a period is a quarter of a full cycle. Convert that fraction of a cycle into radians using 2π rad = one full cycle, then decide on the sign.
A quarter of a period corresponds to a phase angle of (1/4) × 2π = π/2 rad. Because Q reaches each point later than P, Q lags P, so its phase angle relative to P is negative: φ = −π/2 rad.
17Using Fig. 6.1, describe how you could estimate the phase angle φ between oscillators A and B directly from the graph.
Measure the horizontal (time) gap between two corresponding points on the two curves, such as their peaks. Express this gap as a fraction of the time period T (the horizontal distance for one complete cycle of either curve), then multiply that fraction by 2π rad to convert it into a phase angle.
18State what is meant by two oscillators being in phase, and what is meant by two oscillators being in antiphase. Give one real example of two objects that could reasonably be described as oscillating in antiphase.
Antiphase means the phase difference is exactly half a cycle. Think of two objects that always move in exactly opposite directions to each other.
Two oscillators are in phase when their phase difference is zero (or a whole number of cycles, 2π rad) — they reach maximum displacement, equilibrium, and minimum displacement at exactly the same times. They are in antiphase when their phase difference is exactly half a cycle (π rad) — one is at its positive maximum exactly when the other is at its negative maximum. Example: two pistons in a horizontally-opposed ("boxer") engine, or two ends of a simple seesaw, moving in exactly opposite directions at every instant.
7. Solving problems with SHM data-booklet equations
You now have all the equations needed to fully describe and solve problems on any SHM oscillator. Two further equations complete the set: one connecting velocity directly to displacement (without needing time), and a trio of energy equations built from the ones you met qualitatively in Section 5. All of these appear in the physics data booklet, so you do not need to memorise them — but you do need to know when and how to use each one.
Quantity
Equation
Notes
Displacement
x = x₀ sin(ωt + φ)
φ = 0 if the oscillator starts at x = 0 moving in the positive direction
Velocity (vs. time)
v = ωx₀ cos(ωt + φ)
found by differentiating x with respect to t
Velocity (vs. displacement)
v = ±ω√(x₀² − x²)
useful when time is not known or not needed; ± because the oscillator passes through each x twice per cycle, moving in opposite directions
Acceleration
a = −ω²x
the defining equation of SHM (Section 1)
Total energy
ET = ½mω²x₀²
constant throughout the cycle
Potential energy
EP = ½mω²x²
maximum at x = ±x₀, zero at x = 0
Kinetic energy
EK = ET − EP EK = ½mω²(x₀² − x²)
maximum at x = 0, zero at x = ±x₀
Maximum speed: setting x = 0 in v = ±ω√(x₀² − x²) gives the maximum speed of any SHM oscillator: vmax = ωx₀. This is the speed as the oscillator passes through equilibrium.
Worked example 7.1
An oscillator has amplitude 0.15 m and angular frequency 4.0 rad s−1, and starts at x = 0 moving in the positive direction. Calculate its displacement and velocity at t = 0.30 s.
Working:
Since the oscillator starts at x = 0 moving in the positive direction, φ = 0, so x = x₀ sin(ωt) = 0.15 × sin(4.0 × 0.30) ... (complete the calculation, in m, with the angle in radians)
v = ωx₀ cos(ωt) = 4.0 × 0.15 × cos(4.0 × 0.30) ... (complete the calculation, in m s−1)
Worked example 7.2
A mass of 0.20 kg oscillates with SHM with amplitude 0.080 m and angular frequency 6.0 rad s−1. Calculate the total energy of the oscillation, and the kinetic energy when its displacement is 0.050 m.
Working:
ET = ½mω²x₀² = ½ × 0.20 × 6.0² × 0.080² ... (complete the calculation, in J)
EP at x = 0.050 m: EP = ½mω²x² = ½ × 0.20 × 6.0² × 0.050² ... (complete the calculation, in J)
EK = ET − EP (use your two answers above to complete this, in J)
Check your understanding
19An oscillator has amplitude 0.24 m and angular frequency 3.5 rad s−1. Calculate its maximum speed.
vmax = ωx₀ = 3.5 × 0.24 ... (complete the calculation, in m s−1)
20An oscillator has amplitude 0.10 m and angular frequency 2.0 rad s−1. Calculate its speed when its displacement is 0.060 m.
Use v = ±ω√(x₀² − x²) — you are given x directly, so there is no need to work through t.
v = ±ω√(x₀² − x²) = ±2.0 × √(0.10² − 0.060²) ... (complete the calculation, in m s−1)
21A mass of 0.50 kg on a spring has amplitude 0.12 m and time period 0.80 s. Calculate the angular frequency, then the total energy of the oscillation.
Find ω from T first (ω = 2π/T), then substitute it into the total energy equation.
ω = 2π/T = 2π/0.80 ... (complete the calculation, in rad s−1). ET = ½mω²x₀² = ½ × 0.50 × ω² × 0.12² (use your value of ω to complete this, in J)
22An oscillator has amplitude x₀ and you are told its kinetic energy equals its potential energy at some instant. Show that this occurs when x = x₀/√2.
Set EK = EP using the equations EK = ½mω²(x₀² − x²) and EP = ½mω²x², then cancel the common factors before rearranging for x.
Setting EK = EP: ½mω²(x₀² − x²) = ½mω²x². The common factor ½mω² cancels from both sides, leaving x₀² − x² = x², so x₀² = 2x², giving x² = x₀²/2 and therefore x = x₀/√2, as required.
Glossary
Oscillation
one complete repetition of a periodic to-and-fro motion about a central point.
Simple harmonic motion (SHM)
oscillation in which the acceleration is proportional to the displacement from equilibrium and always directed towards it: a = −ω²x.
Restoring force
the resultant force on an oscillator that always acts to push or pull it back towards its equilibrium position.
Equilibrium position
the central position about which an oscillator moves, where the resultant force on it is zero.
Displacement, x
the distance (and direction) of an oscillator from its equilibrium position at a given instant.
Amplitude, x₀
the maximum displacement of an oscillator from its equilibrium position.
Time period, T
the time taken for one complete oscillation.
Frequency, f
the number of complete oscillations per unit time, measured in hertz (Hz).
Angular frequency, ω
oscillation frequency expressed in radians per second; ω = 2πf = 2π/T.
Phase angle, φ
a measure, in radians, of how far through its cycle one oscillator is compared with a reference oscillator of the same amplitude and frequency.
In phase
describes two oscillators with a phase difference of zero (or a whole number of cycles) — they reach corresponding points in their cycles at exactly the same times.
Antiphase
describes two oscillators with a phase difference of exactly half a cycle (π rad) — one is at its positive maximum exactly when the other is at its negative maximum.