Oscillations and Waves — SHM, Superposition, Beats and Doppler

One condition defines this entire chapter: a restoring force proportional to displacement and directed opposite to it, F = −kx. Everything else — the sinusoidal motion, the period, the energy exchange, the pendulum, the spring — is a consequence. That is worth stating up front because it tells you how to identify SHM in an unfamiliar system rather than memorising a list of cases: find the restoring force, check it is linear in the displacement, and read ω² off the coefficient. Waves then extend the same mathematics from one oscillating particle to a chain of them, and the second half of the chapter is what happens when two such disturbances meet.

Simple Harmonic Motion

From F = −kx and F = ma comes a = −(k/m)x, which is solved by x = A sin(ωt + φ) with ω = √(k/m). Differentiating gives v = Aω cos(ωt + φ) and a = −Aω²sin(ωt + φ) = −ω²x. The useful non-time form is v = ω√(A² − x²), which lets you get the speed at any position without knowing when.

What happens at the two extremes
QuantityAt the mean position (x = 0)At the extreme (x = ±A)
SpeedMaximum, AωZero
AccelerationZeroMaximum, Aω²
Restoring forceZeroMaximum
Kinetic energyMaximum, ½kA²Zero
Potential energyZeroMaximum, ½kA²
⚠️ Energy oscillates at twice the frequency of the motion
The total energy E = ½kA² is constant, but the kinetic and potential parts trade back and forth. Because KE ∝ v² ∝ cos²(ωt), and cos² completes a full cycle twice as often as cos does, the energy varies with frequency 2f while the displacement varies with f. That is why the kinetic energy is maximum twice in every oscillation — once passing the centre going right, once going left. Two related results that get asked: KE = PE when x = ±A/√2, and the average kinetic energy over a full cycle equals the average potential energy, each being half the total.
The standard oscillators
SystemPeriodDepends on
Spring–massT = 2π√(m/k)Mass and stiffness; unaffected by gravity, so it is the same vertical or horizontal
Simple pendulumT = 2π√(l/g)Length and g only — not on mass or (for small angles) amplitude
Springs in series1/k_eff = 1/k₁ + 1/k₂Softer than either — like resistors in parallel
Springs in parallelk_eff = k₁ + k₂Stiffer than either
🎯 Why a pendulum in a lift changes period but a spring does not
The pendulum's restoring force is the tangential component of gravity, so anything that changes the effective g changes the period: in a lift accelerating upward T = 2π√(l/(g+a)) and the clock runs fast, accelerating downward it runs slow, and in free fall the pendulum stops oscillating altogether because there is no restoring force left. The spring is different — its restoring force comes from the spring itself, and gravity only shifts the equilibrium position downward without changing k. So T = 2π√(m/k) holds in a lift, on the Moon and in orbit alike. This pair is asked almost every year, and the "pendulum in free fall" case is the memorable one: T becomes infinite.

Waves, Speed and Standing Waves

A progressive wave is written y = A sin(ωt − kx), where k = 2π/λ is the wave number, and its speed is v = ω/k = fλ. Two facts about speed do most of the work: it is fixed by the medium, not by the source, and it therefore stays constant when the frequency changes.

Wave speed in different media
WaveSpeed
Transverse wave on a stretched stringv = √(T/μ), where μ is mass per unit length
Sound in a gasv = √(γP/ρ) = √(γRT/M) — so v ∝ √T
Sound in a solidv = √(Y/ρ) with Y the Young's modulus
⚠️ The speed of sound does not depend on pressure
The formula v = √(γP/ρ) looks as though raising the pressure should raise the speed, and this is the most reliably set trap in the topic. But for a gas at fixed temperature, raising P raises ρ in exactly the same proportion, so P/ρ is unchanged and v is unchanged. What genuinely changes v is temperature, through v ∝ √T, which is why sound travels faster on a hot day. Two more genuine dependencies: v ∝ 1/√M, so sound is faster in a lighter gas — about three times faster in hydrogen than in oxygen — and humidity raises the speed slightly, because water vapour is lighter than dry air.
Standing waves in pipes and strings
SystemFundamentalHarmonics present
String fixed at both endsf₁ = v/2LAll integer multiples: f₁, 2f₁, 3f₁ …
Open organ pipef₁ = v/2LAll integer multiples
Closed organ pipef₁ = v/4L — an octave lowerOdd multiples only: f₁, 3f₁, 5f₁ …
🧠 Why a closed pipe sounds an octave lower and skips the even harmonics
A closed end must be a node (the air cannot move there) and an open end an antinode. The shortest pattern that fits node-to-antinode is a quarter of a wavelength, so L = λ/4 and f₁ = v/4L — half the frequency of an open pipe of the same length, which is exactly one octave lower. Continuing the pattern, only odd quarter-wavelengths fit, so only odd harmonics exist. This is not merely a formula: it is why a flute (effectively open) and a closed pipe of the same length have audibly different timbres, since the missing even harmonics change the tone quality. In the exam the giveaway is the word "closed at one end", and the two errors to avoid are using v/2L and including 2f₁.

Beats and the Doppler Effect

Two waves of nearly equal frequency superpose to give a slow rise and fall in loudness, and the beat frequency is simply |f₁ − f₂|. Beats are how a musician tunes an instrument: as the two frequencies approach each other the beats slow, and at perfect tuning they vanish.

🧠 Beats tell you the difference but not which is higher — until you change one
A tuning fork of 256 Hz sounded with an unknown string gives 4 beats per second. The string is either 252 or 260 Hz, and the beat frequency alone cannot distinguish them, since it is an absolute difference. The standard method is to change one of them slightly and see what happens: if you load the fork with wax its frequency drops, and if the beats then get slower the string must be below the fork, whereas if they get faster the string is above. Similarly, tightening a string raises its frequency. Almost every beats question is really testing whether you know that the sign of the difference has to be found by a second experiment.

The Doppler effect for sound gives the observed frequency as f′ = f (v ± v_o)/(v ∓ v_s), where v is the speed of sound, v_o the observer's speed and v_s the source's. Rather than memorising the signs, remember the physical rule: approaching raises the frequency, receding lowers it, and then choose the signs so the algebra agrees with that.

⚠️ Doppler shift needs motion along the line joining source and observer
Only the component of velocity along the line of sight produces a shift. So a source moving in a circle around a stationary observer at the centre produces no Doppler shift at all, however fast it moves, because its distance is never changing. Likewise, at the instant a passing train is exactly level with you the shift is momentarily zero, which is why the pitch drops through the true frequency as it goes by rather than jumping. Two further points that get examined: wind changes v, not v_s or v_o — you add the wind speed to the speed of sound along the direction of propagation — and if the source and observer move with the same velocity there is no shift, since their separation is constant.
ℹ️ What superposition gives you: interference, nodes and reflection
Superposition simply adds displacements. Waves in phase add to give constructive interference with amplitude A₁ + A₂ and intensity ∝ (A₁+A₂)²; waves in antiphase give destructive interference. Because intensity goes as the square of amplitude, two equal waves in phase give four times the intensity of one, not twice — a favourite question. Path difference converts to phase difference through Δφ = 2πΔx/λ, so constructive interference needs Δx = nλ and destructive needs (n + ½)λ. One more result worth holding: a wave reflecting from a denser medium or a rigid end suffers a phase reversal of π, while reflection from a rarer or free end does not — which is why a string fixed at its end has a node there.

Key takeaways

  • SHM is defined by F = −kx; then ω = √(k/m), a = −ω²x and v = ω√(A² − x²).
  • At the mean position speed is maximum and acceleration zero; at the extremes the reverse.
  • Energy oscillates at 2f while displacement oscillates at f; KE = PE at x = ±A/√2.
  • Pendulum T = 2π√(l/g) — mass-independent, and infinite in free fall. Spring T = 2π√(m/k) — unaffected by g.
  • v = fλ, set by the medium: √(T/μ) on a string, √(γRT/M) in a gas.
  • The speed of sound is independent of pressure but goes as √T and as 1/√M.
  • Closed pipe: f₁ = v/4L with odd harmonics only. Open pipe and string: v/2L with all harmonics.
  • Beat frequency = |f₁ − f₂|, so which is higher needs a second test (wax lowers a fork).

Practice questions (8)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

  1. A simple pendulum is taken inside a lift in free fall. Its time period becomes:

    1. Unchanged
    2. Halved
    3. Infinite — it stops oscillating
    4. Doubled
    Show answer

    Answer: C — Infinite — it stops oscillating

    The pendulum's restoring force is the tangential component of gravity, so with effective g = 0 there is no restoring force at all and T = 2π√(l/g) becomes infinite — the bob simply floats where it is put. Note that a spring–mass system would be completely unaffected, since its restoring force comes from the spring rather than from gravity.

  2. If the pressure of a gas is doubled at constant temperature, the speed of sound in it:

    1. Doubles
    2. Becomes √2 times
    3. Remains unchanged
    4. Halves
    Show answer

    Answer: C — Remains unchanged

    Although v = √(γP/ρ) appears to depend on pressure, doubling P at constant temperature doubles ρ in exactly the same proportion, so P/ρ and hence v are unchanged. What genuinely changes the speed is temperature, since v ∝ √T, and molar mass, since v ∝ 1/√M.

  3. A pipe closed at one end and an open pipe of the same length are sounded. The closed pipe's fundamental frequency is:

    1. The same as the open pipe's
    2. Twice the open pipe's
    3. Half the open pipe's
    4. Four times the open pipe's
    Show answer

    Answer: C — Half the open pipe's

    A closed end is a node and an open end an antinode, so the shortest fit is L = λ/4 and f₁ = v/4L, against v/2L for the open pipe — exactly half, an octave lower. The closed pipe also produces only odd harmonics, which is why its timbre differs audibly from an open pipe of the same pitch.

  4. A source of sound moves in a circle at constant speed with a stationary observer at the centre. The observer hears:

    1. A continuously rising pitch
    2. No change in frequency at all
    3. A pitch alternately rising and falling
    4. A frequency raised by the factor v/(v − v_s)
    Show answer

    Answer: B — No change in frequency at all

    Only the component of velocity along the line joining source and observer produces a Doppler shift, and for circular motion about the observer that component is always zero — the distance never changes. This is the standard test of whether a candidate has understood the effect or merely memorised the formula.

  5. In SHM, the kinetic energy equals the potential energy when the displacement is:

    1. A/2
    2. A/√2
    3. A
    4. Zero
    Show answer

    Answer: B — A/√2

    PE = ½kx² and total energy = ½kA², so PE equals half the total when ½kx² = ¼kA², giving x = A/√2. The tempting A/2 would make PE only a quarter of the total, since the energy depends on the square of the displacement rather than on the displacement itself.

  6. A tuning fork of 256 Hz gives 4 beats per second with a string. On loading the fork with wax, the beats become slower. The string's frequency is:

    1. 260 Hz
    2. 252 Hz
    3. 256 Hz
    4. 248 Hz
    Show answer

    Answer: B — 252 Hz

    Wax lowers the fork below 256 Hz. If the beats slow, the gap has narrowed, so the string must lie below the fork — at 252 Hz. Had the string been at 260 Hz the gap would have widened and the beats would have quickened. The beat frequency alone gives only |f₁ − f₂|, so a second observation is always needed to fix the sign.

  7. Two identical waves in phase superpose. Compared with one wave alone, the resultant intensity is:

    1. Twice
    2. Four times
    3. The same
    4. √2 times
    Show answer

    Answer: B — Four times

    Amplitudes add for waves in phase, giving 2A, and intensity goes as the square of amplitude, so I ∝ (2A)² = 4A² — four times. The tempting "twice" comes from adding intensities directly, which is only valid for incoherent sources where the waves have no fixed phase relationship.

  8. Two springs of stiffness k₁ and k₂ are joined in series and a mass is attached. The effective stiffness is:

    1. k₁ + k₂
    2. k₁k₂/(k₁ + k₂)
    3. (k₁ + k₂)/2
    4. √(k₁k₂)
    Show answer

    Answer: B — k₁k₂/(k₁ + k₂)

    In series the extensions add for the same tension, so 1/k_eff = 1/k₁ + 1/k₂, giving k₁k₂/(k₁+k₂) — softer than either spring, which is the useful sanity check. Option 1 is the parallel result, where the springs share the load and the arrangement is stiffer than either; the pattern is the reverse of resistors.