Oscillations and Waves

1What makes a motion simple harmonic

A motion is simple harmonic when the restoring force is proportional to the displacement and points back towards the mean position — equivalently, when a = −ω²x. That one line is the entire definition; everything else in the chapter is its solution. Integrating it gives x = A sin(ωt + φ), and with it v = ωA cos(ωt + φ) = ω√(A² − x²). The speed is greatest at the mean position and zero at the extremes; the acceleration does exactly the opposite.

Figure. Two full periods. Velocity leads displacement by a quarter of a period: the particle is fastest exactly where x = 0 and momentarily at rest at each extreme. The velocity curve is drawn as v/ω so that both fit one axis — its true height is ωA.

How to recognise SHM in a problem

1. Find the mean positionLocate where the net force vanishes and measure x from there — not from the spring's natural length or the pivot.
2. Linearise the restoring forceShow that F = −(constant) × x for small x. That constant divided by m is ω².
3. Read ω off the coefficientω comes from the equation, never from the amplitude, so T = 2π/ω does not care how hard you pushed.

2Energy in SHM

The total energy E = ½mω²A² = ½kA² never changes; only its division does. Potential energy ½kx² is largest at the extremes, kinetic energy ½k(A² − x²) at the mean position, and each is the other's mirror image in the same parabola.

Because both go as the square of a sinusoid, the exchange runs at twice the frequency of the motion itself: the particle crosses the centre twice per period, so kinetic energy peaks twice per period while the displacement peaks once.

Figure. Two parabolas that sum to a constant at every displacement. They cross at x = ±A/√2, where each holds half the energy; at x = ±A/2 the kinetic share is still three-quarters. Nothing here oscillates in time — this is the energy as a function of where the particle is.

How it works

1. Fix E from the amplitudeE = ½kA² = ½mω²A², settled the moment the oscillator and the amplitude are settled.
2. Split it with the displacementU = ½kx² and KE = E − U = ½k(A² − x²). No integration is needed anywhere.
3. Average only over a whole cyclesin² and cos² each average to ½, so ⟨KE⟩ = ⟨U⟩ = E/2 — as averages, not as an instantaneous truth.

3The spring oscillator, and springs in combination

Hooke's law F = −kx is already in the form a = −(k/m)x, so ω = √(k/m) and T = 2π√(m/k). Gravity never appears: hanging the mass vertically moves the mean position down by mg/k and leaves the period untouched, which is why one formula serves a horizontal glider and a vertical spring balance alike.

Combinations follow from what is shared. Springs in parallel share the displacement and add their forces, so stiffness adds: k_eq = Σkᵢ. Springs in series share the force and add their extensions, so compliance adds: 1/k_eq = Σ1/kᵢ. Cutting a spring into two equal halves doubles the stiffness of each half, because k is inversely proportional to the natural length.

Figure. Parallel springs share the displacement → k_eff adds. Series springs share the force → reciprocals add. Period follows √(m/k_eff), so doubling k halves T².

How it works

1. Displacement shared, so add kParallel springs stretch by the same x, so their forces add: k_eq = k₁ + k₂.
2. Force shared, so add 1/kSeries springs carry the same tension, so their extensions add: 1/k_eq = 1/k₁ + 1/k₂.
3. Then one formula covers itFeed k_eq into T = 2π√(m/k_eq). Nothing else about the arrangement matters.

4The simple pendulum

Resolve the bob's weight along and across the string: mg cosθ is balanced by the tension, and mg sinθ is left over as the restoring force. For small θ, sinθ ≈ θ = s/L, so the restoring force is −(mg/L)s — Hooke's law with k = mg/L — and T = 2π√(L/g). The mass cancels, the amplitude cancels, and only the length and the local g survive.

Anything that changes the apparent gravity changes the period through g_eff. In a lift accelerating upward at a, g_eff = g + a; accelerating downward, g − a; in free fall g_eff = 0 and the pendulum does not oscillate at all. The same substitution handles a pendulum in a car rounding a bend, where g_eff is the vector sum of g and the centripetal acceleration.

Figure. Plot T² instead of T and the law straightens into a line through the origin of slope 4π²/g — which is exactly how g is measured in a school laboratory. Halve g and the line doubles in steepness, so the period at any given length grows by √2.

How it works

1. Resolve, do not follow the arcOnly mg sinθ restores; mg cosθ goes into the tension and does no restoring work.
2. Linearise, and know the costsinθ ≈ θ is what makes the motion harmonic at all; at θ = 20° the true period is already about 0.8% longer.
3. Swap g for g_effAny steady acceleration of the support adds vectorially to g. Use the magnitude of that sum, never g itself.

5Damping, forced oscillation and resonance

Every real oscillator loses energy. Add a drag force −bv to Hooke's law and the amplitude decays as A₀e^(−bt/2m), while the oscillation slows a little to ω′ = √(ω₀² − (b/2m)²); the energy, going as amplitude squared, falls faster, as e^(−bt/m). For light damping ω′ is barely below ω₀, so the visible effect is the fading, not any change of pitch.

Drive that oscillator with a periodic force and, once the transients die, it settles at the driving frequency, not its own. Its steady-state amplitude, though, depends sharply on how close the driving frequency is to ω₀: it is largest at ω_drive ≈ ω₀, and there its size is limited only by the damping. Light damping gives a tall, narrow resonance peak; heavy damping a low, broad one. This single idea drives a child's swing, tunes a radio's LCR circuit and, at its most destructive, shook apart the Tacoma Narrows bridge.

How it works

1. Damping fades and slightly slowsAmplitude decays as e^(−bt/2m); ω′ = √(ω₀² − (b/2m)²), only just below ω₀ for light damping.
2. A driven system obeys the driverIn the steady state it oscillates at ω_drive, whatever its own natural ω₀.
3. Resonance at a matchThe response amplitude peaks at ω_drive ≈ ω₀; the lighter the damping, the taller and sharper that peak.

6Travelling waves

y = A sin(kx − ωt) is a shape moving in the +x direction at speed v = ω/k = fλ; make the two signs agree and it moves in −x instead. Two quite different speeds live inside that one equation. The wave speed v is how fast the pattern advances; the particle speed ∂y/∂t = −ωA cos(kx − ωt) is how fast a piece of the medium moves up and down, and it never exceeds ωA.

The two are tied together by the shape of the snapshot: particle velocity = −(wave speed) × (slope of the y–x curve). A steep stretch of the wave is a fast-moving stretch of the medium, and a crest — where the curve is flat — is momentarily at rest.

How to read a wave function

1. Read k and ω off the argumentk = 2π/λ in rad m⁻¹ and ω = 2πf in rad s⁻¹. Both are coefficients, not measurements.
2. Divide for the wave speedv = ω/k = fλ. This belongs to the medium and does not contain A anywhere.
3. Differentiate for the particle speedThe greatest particle speed is ωA = kAv — large only when the wave is steep.

7Wave speed belongs to the medium

A mechanical wave's speed is set by the medium's stiffness and its inertia, never by the source. On a stretched string v = √(T/μ), with T the tension and μ the mass per unit length. In a gas v = √(γP/ρ), and putting P/ρ = RT/M turns that into v = √(γRT/M) — the useful form, which says the speed of sound depends on the absolute temperature and on the gas, and not at all on the pressure. Because v ∝ √T, sound gains about 0.61 m s⁻¹ per kelvin near 0 °C: 331 m s⁻¹ at 273 K becomes 347 m s⁻¹ at 300 K. (The letter T is doing triple duty in this chapter — tension, absolute temperature, period. Read it from its neighbours every time.)

Frequency, by contrast, belongs to the source. Send a wave from a light string into a heavy one and the particle at the junction is shared by both, so it can only vibrate at one frequency: f cannot change. The speed does change, so λ = v/f absorbs the entire difference.

Figure. Both curves are square roots, so they flatten as you pull harder: near 100 N another 10 N of tension buys only about 5 m s⁻¹. The heavier string, at four times the mass per unit length, runs at exactly half the speed of the lighter one at every tension — the two curves are the same curve scaled by ½.

How it works

1. Stiffness upstairs, inertia downstairsEvery mechanical wave speed is √(restoring quantity / inertia): T over μ, γP over ρ.
2. Quadruple to doubleBoth are square roots, so a four-fold change in tension buys only a doubling of the speed.
3. At a junction, f survivesFrequency crosses a boundary unchanged because the junction particle is shared; the wavelength does the adjusting.

8Standing waves: strings and pipes

Superpose two identical waves running opposite ways and the x and t dependences separate: y = 2A sin kx · cos ωt. Nothing travels any more. The points where sin kx = 0 never move at all — nodes, spaced λ/2 apart — and halfway between them sit antinodes swinging through 2A. Energy sloshes between neighbouring segments but no longer flows along the medium.

Boundaries decide which wavelengths survive. A string clamped at both ends must carry a node at each, so L = nλ/2 and fₙ = nv/2L, with every harmonic allowed. A pipe open at both ends has a displacement antinode at each and gives the same series. A pipe closed at one end has a node there and an antinode at the mouth, so it fits only a quarter wavelength and its odd multiples: L = (2n − 1)λ/4, hence fₙ = (2n − 1)v/4L.

How it works

1. Put nodes where the medium is heldA clamped string end and the closed end of a pipe are displacement nodes; a free end or an open mouth is an antinode.
2. Fit whole quarters between themNode to node is λ/2, node to antinode is λ/4. Count how many of those fit into L.
3. Convert with f = v/λv is still √(T/μ) or the speed of sound. The boundary chose λ; it never touched v.

9Beats

Two waves of nearly equal frequency arriving at the same place add to 2A cos(π(f₁ − f₂)t) · sin(π(f₁ + f₂)t): a tone at the mean frequency whose amplitude is itself slowly oscillating. The envelope 2A cos(πΔf t) passes through zero twice in each of its own cycles, and loudness follows the size of the amplitude rather than its sign, so the ear counts f_beat = |f₁ − f₂| beats per second — twice the envelope's own frequency.

Beats are a difference measurement, and that is exactly their value and exactly their limitation: a 4 Hz beat against a 256 Hz standard pins the unknown to within 4 Hz but leaves two candidates, 252 and 260. Only a deliberate nudge tells you which.

Figure. Four hertz added to six hertz, watched for exactly one second. The carrier runs at the mean of 5 Hz; the dashed envelope 2A cos(2πt) completes just one cycle in that second, but it crosses zero twice — so the ear hears two beats, exactly |6 − 4|. That factor of two between envelope frequency and beat frequency is the whole trap.

How it works

1. Add with the sum-to-product identitysin a + sin b = 2 cos((a − b)/2) sin((a + b)/2) — a slow envelope multiplying a fast carrier.
2. Count loudness maxima, not envelope cyclesThe envelope's magnitude peaks twice per cosine period, so f_beat = |f₁ − f₂| and not half of it.
3. Break the tie with a known perturbationWax lowers a fork's frequency, filing the prongs raises it. Watch which way the beat rate then moves.

10The Doppler effect in sound

Motion along the line joining source and observer changes the rate at which wavefronts arrive: f′ = f(v ± v_o)/(v ∓ v_s), with the signs chosen so that anything closing the gap raises the pitch. Unlike light, sound travels in a medium, so the two motions are genuinely different physical processes — a moving source compresses the wavelength it emits, while a moving observer leaves the wavelength alone and simply meets the existing wavefronts sooner.

The asymmetry shows in the algebra. The observer factor (v ± v_o)/v is linear and stays finite at any speed; the source factor v/(v ∓ v_s) blows up as v_s approaches v. They agree only while both speeds are small compared with v. Motion perpendicular to the line of sight gives no shift at that instant, whichever body is moving.

How it works

1. Decide the direction of the shift firstAsk whether the pitch ought to rise or fall, then pick the signs that deliver it. Never the other way round.
2. Keep only the component along the lineResolve both velocities along the source–observer line; the perpendicular part contributes nothing.
3. Wind moves the medium, not the pitchReplace v by v + w everywhere, w measured from source towards observer. With both bodies at rest, f′ = f however hard it blows.