5.5 - Oscillations

Syllabus
2021
Topic
5.5
Level
A2

Learning objectives

5.5.143Condition for simple harmonic motionUnderstand that the condition for simple harmonic motion is F = − kx, and hence understand how to identify situations in which simple harmonic motion will occur5.5.144SHM displacement, velocity and acceleration equationsUse a = −ω²x, x = A cos ωt, v = −Aω sin ωt, a = −Aω² cos ωt, T = 1/f = 2π/ω, and ω = 2πf for simple harmonic motion.5.5.145SHM period equationsUse T = 2π√(m/k) for a mass–spring oscillator and T = 2π√(l/g) for a simple pendulum.5.5.146Displacement-time graphs for oscillationsBe able to draw and interpret a displacement-time graph for an object oscillating and know that the gradient at a point gives the velocity at that point5.5.147Velocity-time graphs for oscillationsBe able to draw and interpret a velocity-time graph for an oscillating object and know that the gradient at a point gives the acceleration at that point5.5.148ResonanceUnderstand what is meant by resonance5.5.149Core Practical 16 - unknown mass by resonanceCORE PRACTICAL 16: Determine the value of an unknown mass using the resonant frequencies of the oscillation of known masses5.5.150Energy conservation in oscillating systemsUnderstand how to apply conservation of energy to damped and undamped oscillating systems5.5.151Free and forced oscillationsUnderstand the distinction between free and forced oscillations5.5.152Resonance amplitude and dampingUnderstand how the amplitude of a forced oscillation changes at and around the natural frequency of a system and know, qualitatively, how damping affects resonance5.5.153Damping and plastic deformationUnderstand how damping and the plastic deformation of ductile materials reduce the amplitude of oscillation.

Simple harmonic motion needs a linear restoring force

F=kxsoa=kmx=ω2xF=-kx\qquad\text{so}\qquad a=-\frac{k}{m}x=-\omega^2x

Displacement xx is measured from equilibrium. The minus sign means force and acceleration always point back towards equilibrium; proportionality means their magnitudes grow linearly with distance from equilibrium.

Evidence SHM verdict
FF-against-xx is a straight line through the origin with negative gradient satisfies the SHM condition
restoring force is curved or not proportional to xx not exact SHM over that range
force points away from equilibrium unstable motion, not SHM

For a mass on an ideal spring, extension about the equilibrium position produces restoring force kx-kx. A simple pendulum is approximately SHM only at small angles, where the restoring component is approximately proportional to displacement.

A repeated or sinusoidal-looking motion is not sufficient evidence by itself. The defining test is acceleration proportional to and opposite displacement about a stable equilibrium; the constant kk here is the positive proportionality constant.

SHM equations encode amplitude, phase and extrema

a=ω2x,x=Acosωt,v=Aωsinωt,a=Aω2cosωt,T=1f=2πωa=-\omega^2x,\quad x=A\cos\omega t,\quad v=-A\omega\sin\omega t,\quad a=-A\omega^2\cos\omega t,\quad T=\frac1f=\frac{2\pi}{\omega}

Position Speed Acceleration
x=+Ax=+A zero maximum towards equilibrium
x=0x=0 maximum, AωA\omega zero
x=Ax=-A zero maximum towards equilibrium

The cosine form chooses t=0t=0 at positive maximum displacement. A different initial condition requires a phase constant, but the relationships between xx, vv and aa remain: velocity is one quarter-cycle out of phase with displacement, and acceleration is in antiphase with displacement.

For A=0.040mA=0.040\,\mathrm{m} and f=2.0Hzf=2.0\,\mathrm{Hz}, ω=2πf=12.6rads1\omega=2\pi f=12.6\,\mathrm{rad\,s^{-1}}, so vmax=Aω=0.50ms1v_{max}=A\omega=0.50\,\mathrm{m\,s^{-1}} and amax=Aω2=6.3ms2a_{max}=A\omega^2=6.3\,\mathrm{m\,s^{-2}}.

Do not use degrees inside these time equations: ωt\omega t is in radians. The sign of vv depends on direction of travel, while speed is its magnitude; maximum acceleration occurs at the extremes, not at equilibrium.

Spring and pendulum periods depend on different system properties

Tspring=2πmkTpendulum=2πlgT_{spring}=2\pi\sqrt{\frac{m}{k}}\qquad T_{pendulum}=2\pi\sqrt{\frac{l}{g}}

Oscillator Longer period when... Independent of, in the model
mass-spring mm increases or kk decreases amplitude for an ideal linear spring
simple pendulum ll increases or gg decreases bob mass and small amplitude

A 0.20kg0.20\,\mathrm{kg} mass on a 50Nm150\,\mathrm{N\,m^{-1}} spring has T=2π0.20/50=0.40sT=2\pi\sqrt{0.20/50}=0.40\,\mathrm{s}. Quadrupling the mass doubles the period because TmT\propto\sqrt m.

Use the oscillating mass that moves with the spring. Pendulum length is from pivot to the bob's centre of mass, and the pendulum equation assumes small angular displacement, negligible air resistance and a light inextensible support.

Do not interchange kk and gg, or infer a linear period-mass relationship. Squaring gives useful straight-line forms: T2=4π2m/kT^2=4\pi^2m/k and T2=4π2l/gT^2=4\pi^2l/g.

A displacement-time graph reveals velocity through its gradient

For SHM, displacement varies sinusoidally between +A+A and A-A about the equilibrium line x=0x=0. The period TT is the time between equivalent points moving in the same direction; amplitude is the maximum distance from equilibrium.

Point on xx-tt graph Gradient and motion
positive or negative extreme gradient zero, so velocity zero and direction reverses
crossing equilibrium upwards steep positive gradient, maximum positive velocity
crossing equilibrium downwards steep negative gradient, maximum negative velocity
between these points tangent gradient gives instantaneous velocity

Fix the axes and units, mark ±A\pm A, and place repeated peaks exactly one period apart. If initial displacement is +A+A with release from rest, begin at a maximum and follow a cosine curve; other initial states shift the phase without changing AA or TT.

The graph is displacement against time, not the path through space. Its height gives position and its gradient gives velocity; a zero crossing therefore does not mean the object has stopped.

A velocity-time graph reveals acceleration through its gradient

SHM velocity is sinusoidal with the same period as displacement and amplitude vmax=Aωv_{max}=A\omega. The gradient of a velocity-time graph is instantaneous acceleration.

Point on vv-tt graph Physical state
v=+vmaxv=+v_{max} or vmax-v_{max} object crosses equilibrium; gradient and acceleration are zero
v=0v=0 with negative gradient object is at positive extreme; acceleration is most negative
v=0v=0 with positive gradient object is at negative extreme; acceleration is most positive

If x=Acosωtx=A\cos\omega t, then v=Aωsinωtv=-A\omega\sin\omega t. Velocity therefore reaches each corresponding feature one quarter-period after displacement, and its sign records direction. Draw peaks separated by TT and label the vertical scale using AωA\omega.

Do not read acceleration from the graph's height; use its tangent gradient. Zero velocity occurs at the displacement extremes, where acceleration is largest, while maximum speed occurs at equilibrium, where acceleration is zero.

Resonance is maximum energy transfer at the natural frequency

Resonance occurs when a periodic driving force acts at the natural frequency, or very close to it in a damped real system. The driver then transfers energy efficiently on successive cycles, producing the largest steady oscillation amplitude.

Step Resonant response
1 a system has its own natural frequency for free oscillation
2 an external force supplies periodic energy
3 matched timing makes energy additions reinforce the motion
4 amplitude grows until energy input per cycle balances losses

Away from the natural frequency, the driver and oscillator do not stay in the most effective timing relationship, so less net energy is transferred and the response amplitude is smaller.

Resonance does not mean the amplitude grows without limit: damping and non-linear effects restrict it. Merely vibrating at a high frequency is not resonance; the relevant comparison is driving frequency with the system's natural frequency.

Core Practical 16: find an unknown mass from resonance

Attach the same spring to a vibration generator and add a known mass. Drive it sinusoidally, vary frequency in small steps, and record the resonant frequency at the maximum steady amplitude. Repeat the frequency sweep from both directions and average the peak value.

Stage Control or analysis
calibration repeat for several known masses using the same spring, drive amplitude and geometry
graph plot 1/fr21/f_r^2 against known mass; 1/fr2=4π2m/k1/f_r^2=4\pi^2m/k gives a straight calibration
unknown measure its resonant frequency, calculate 1/fr21/f_r^2, and interpolate its mass
quality use small frequency intervals near each peak and wait for a steady response

Calibration can absorb a small constant effective contribution from the moving spring or holder: use the fitted relationship rather than forcing a line through the origin without evidence. Repeat the unknown measurement and propagate the frequency reading uncertainty through the graph.

Do not identify resonance from the first visible motion; locate the maximum steady amplitude. Changing the spring, attachment point, damping or drive amplitude between calibration and unknown invalidates the comparison.

Oscillator energy swaps form or leaves the system

Position in an ideal mass-spring cycle Kinetic energy Elastic potential energy
x=0x=0 maximum zero relative to equilibrium
0<x<A0<|x|<A shared shared
x=±Ax=\pm A zero maximum, 12kA2\tfrac12kA^2

For undamped SHM, E=12mv2+12kx2=12kA2E=\tfrac12mv^2+\tfrac12kx^2=\tfrac12kA^2 is constant. A pendulum similarly exchanges gravitational potential and kinetic energy while total mechanical energy remains constant.

With damping, resistive forces do negative work. Mechanical energy is transferred to thermal energy, sound or deformation in the surroundings, so the maximum energy and amplitude decrease on successive cycles even though total energy of the larger closed system is conserved.

Damping does not destroy energy and it does not simply convert potential energy into kinetic energy; it transfers mechanical energy out of the oscillator. At a turning point speed is zero, but energy is not zero.

Free oscillations use natural frequency; forced ones follow the driver

Feature Free oscillation Forced oscillation
cause initial displacement or impulse, then no continuing periodic drive continuing periodic external force
frequency system's natural frequency driving frequency in steady state
energy initial store is exchanged and may decay through damping driver continually supplies energy
amplitude fixed if undamped, otherwise decreases depends on driving frequency and damping

A sequence of impulses can maintain a forced oscillation when each arrives at the same effective phase, adding energy repeatedly. The drive need not move sinusoidally, but its timing must contain the relevant periodic component.

A forced oscillator does not automatically move at its natural frequency. It follows the driver; the natural frequency matters because response becomes especially large when the two frequencies match closely.

Damping lowers and broadens the resonance peak

For a fixed driving-force amplitude, the steady response is small far below the natural frequency, rises as the driving frequency approaches it, reaches a maximum near resonance, and falls again above it.

Damping Resonance curve Physical consequence
light high, narrow peak large amplitude over a small frequency range
stronger lower, broader peak smaller maximum response and less sharply selected resonance

At steady amplitude, energy supplied by the driver each cycle equals energy dissipated. Near resonance, the timing permits the greatest energy transfer; increasing damping removes more mechanical energy per cycle, so a smaller amplitude is enough to balance the input.

Damping does not eliminate the natural frequency or force the oscillator to stop while it is driven. It changes the amplitude-frequency response; a finite resonance peak is expected even for light damping.

Damping and plastic deformation remove recoverable mechanical energy

Damping forces oppose motion and do work on the surroundings. Fluid resistance, friction or a dashpot converts organised oscillation energy into thermal energy, so each cycle returns less mechanical energy and the amplitude falls.

Material response Energy on unloading Effect on oscillation
elastic deformation most stored strain energy is returned motion can continue with little material loss
plastic deformation in a ductile material some work remains as permanent deformation and heat less energy returns, so amplitude is reduced

A ductile component designed to yield during strong structural motion absorbs work over its force-extension cycle. Together with viscous dampers, this reduces energy available for the next swing and limits damaging amplitudes.

Plastic deformation is permanent; it is not the same as an elastic element temporarily storing energy. Both mechanisms obey energy conservation: the oscillator's mechanical energy becomes internal energy or irreversible deformation rather than disappearing.