Lesson 18: Sound, Standing Waves, and the Doppler EffectAP

Sound is a longitudinal pressure wave in a medium — in air at room temperature it travels at about 343 m/s. This lesson (CED Unit 9) treats the three sound topics the exam rotates: standing waves in pipes and on strings, the Doppler effect, and beats. All three are interference in different costumes: standing waves are interference of a wave with its own reflection; beats are interference of two waves in time; the Doppler effect is what a moving source or observer does to the wave pattern.

Standing waves: strings and pipes

A string fixed at both ends can only sustain patterns with nodes at both ends, so L = nλ/2 and the harmonics are f_n = nv/(2L) for n = 1, 2, 3, … — all integer multiples of the fundamental. Pipes behave by their ends: an open end is a displacement antinode, a closed end is a node.

• Open–open pipe: same as the string: f_n = nv/(2L), all harmonics.
• Closed–open pipe: the pattern must fit odd quarter-wavelengths, L = (2n−1)λ/4, so f_n = nv/(4L) for odd n only (1, 3, 5…). The missing even harmonics are why a clarinet (closed) sounds hollow next to a flute (open) of the same length.

The Doppler effect

When a source moves toward an observer, each new wavefront is emitted from a position closer to the previous one: the wavelength ahead is compressed, and the observed frequency rises: f' = f · v_sound/(v_sound − v_source) for approach, with + for recession. A moving observer instead crosses wavefronts faster: f' = f (v_sound + v_observer)/v_sound approaching. AP Physics 1 treats these quantitatively for sound in air with the convention that speeds are relative to the air. The pitch drop of a passing siren — high approaching, low receding — is the signature.

Beats

Two tones of close frequency f_1 and f_2 alternate between constructive and destructive interference, producing throbbing at the beat frequency f_beat = |f_1 − f_2|. Piano tuners listen for beats vanishing against a reference fork — interference as a precision instrument.

Worked example 1 — closed pipe harmonics

An organ pipe closed at one end has length 0.85 m (v_sound = 343 m/s). Find (a) its fundamental frequency, (b) the next two resonant frequencies.

(a) f_1 = v/(4L) = 343/(4 × 0.85) = 343/3.4 ≈ 101 Hz
(b) Odd harmonics only: f_3 = 3 × 101 ≈ 303 Hz;  f_5 = 5 × 101 ≈ 505 Hz

Check: an open pipe of the same length would have fundamental v/(2L) ≈ 202 Hz — the closed pipe sounds an octave lower, and its even harmonic at 202 Hz is forbidden, matching the answers' spacing ✓.

Worked example 2 — the passing ambulance

An ambulance siren emits 400 Hz while approaching at 30 m/s (v_sound = 340 m/s). Find the frequency heard (a) as it approaches, (b) after it passes and recedes at the same speed.

(a) f' = f × v/(v − v_s) = 400 × 340/(340 − 30) = 400 × 340/310 ≈ 439 Hz
(b) f' = f × v/(v + v_s) = 400 × 340/370 ≈ 368 Hz

Check: approach raises the pitch, recession lowers it, and the two shifts are not symmetric (439 vs 368 around 400) because the source's motion divides by different denominators — a subtlety students often miss. (A quick beats check: this siren against a stationary 440 Hz tone would beat at ~1 Hz while approaching.)

Real-world application: Doppler in medicine and on the road

Doppler ultrasound measures blood flow: sound reflecting from moving red blood cells returns frequency-shifted, and the shift gives flow speed and direction — detecting valve leaks and blockages without surgery. Police radar and baseball radar guns do the same with electromagnetic waves, and weather radar maps storm rotation by the Doppler shifts of rain. In astronomy the effect becomes redshift, the evidence that distant galaxies recede.

Experimental design note

Find the speed of sound with a resonance tube: hold a tuning fork of known frequency over a tube whose water level adjusts the air column; resonances occur at lengths L ≈ λ/4, 3λ/4, 5λ/4… The spacing between successive resonances is λ/2, so two resonance positions give λ robustly (and cancel the end correction that spoils a single measurement) — v = fλ then follows. Repeating with a second fork frequency and checking that v agrees is the mark of a sound experimental design — pun intended by generations of teachers.

Common mistakes (from College Board scoring commentary)

• Giving a closed pipe even harmonics, or using f = nv/(2L) for it.
• Putting nodes where antinodes belong at a pipe's open end (displacement antinode; pressure node — keep to displacement in this course).
• Doppler sign errors: approach must raise f′; check your fraction exceeds 1 for approach before calculating further.
• Computing beats as the average (f_1 + f_2)/2 instead of the difference.

AP Exam Experience

MCQ 1. A pipe open at both ends has length 0.50 m (v_sound = 340 m/s). Its fundamental frequency is:

• (A) 170 Hz
• (B) 340 Hz
• (C) 510 Hz
• (D) 680 Hz

Answer: (B). f_1 = v/(2L) = 340/(1.0) = 340 Hz. (A) is the closed-pipe formula applied wrongly.

MCQ 2. A tuning fork of 512 Hz is sounded with an unknown fork, producing 4 beats per second. The unknown frequency could be:

• (A) 508 Hz or 516 Hz
• (B) 512 Hz
• (C) 256 Hz
• (D) 1024 Hz

Answer: (A). f_beat = |f_1 − f_2| = 4 Hz, so f_2 = 512 ± 4 = 508 or 516 Hz. (Waxing one fork to lower its frequency and re-listening resolves the ambiguity — a classic follow-up.)

MCQ 3. A source emitting frequency f moves toward a stationary observer at 0.20 × v_sound. The observed frequency is closest to:

• (A) 0.80f
• (B) 1.00f
• (C) 1.25f
• (D) 1.50f

Answer: (C). f′ = f × v/(v − 0.20v) = f/0.80 = 1.25f.

FRQ. A guitar string of length 0.65 m vibrates in its fundamental mode at 196 Hz. (a) Calculate the speed of waves on the string. (b) The player presses the string at a fret, shortening the vibrating length to 0.49 m (tension unchanged). Calculate the new fundamental frequency. (c) Two identical strings are tuned to 196 Hz and 198 Hz and played together. Calculate the beat frequency heard. (d) Explain, using wavefront spacing, why a listener hears a higher frequency as a sound source approaches than after it passes. Scoring guideline. (a) 1 point: fundamental λ = 2L = 1.30 m; v = fλ = 196 × 1.30 ≈ 255 m/s. (b) 1 point: v unchanged (same tension and string), so f = v/(2L') = 255/(0.98) ≈ 260 Hz. (c) 1 point: f_beat = |198 − 196| = 2 Hz. (d) 1 point: as the source advances between emitting successive wavefronts, the spacing ahead of it is compressed (λ shorter), and since v_sound is fixed by the air, f' = v/λ rises; behind the source the spacing stretches and the frequency falls.

Worked example 3 — the string's harmonic series

A violin string of vibrating length 0.33 m has fundamental frequency 440 Hz. Find (a) the wave speed on the string, (b) the first two overtones, (c) the wavelength of the fundamental sound in air (v_air = 343 m/s).

(a) v_string = 2Lf_1 = 2 × 0.33 × 440 ≈ 290 m/s
(b) Overtones: f_2 = 880 Hz, f_3 = 1,320 Hz (all harmonics exist on a string)
(c) The sound in air has the same frequency (440 Hz) but a different speed and wavelength:
    λ_air = 343/440 ≈ 0.78 m — versus λ_string = 0.66 m

Check: crossing a boundary (string → air) preserves frequency — the driver sets it — while speed and wavelength renegotiate with the new medium; this is exactly the v = fλ triangle of Lesson 17, and it is why a guitar does not change pitch when you open a window.

The moving observer: the other half of Doppler

Let the source rest and the observer move toward it at 20 m/s, for a 500 Hz tone in air (340 m/s). The observer crosses wavefronts sooner: f' = f (v + v_o)/v = 500 × 360/340 ≈ 529 Hz. Compare a source approaching at the same 20 m/s: f' = 500 × 340/320 ≈ 531 Hz. Close — but not identical, because a moving source physically compresses the wavelength in the air, while a moving observer merely samples an unchanged pattern faster. At everyday speeds the difference is small; the exam tests whether you know which formula belongs to which mover, and that both give f' > f on approach. Add the two effects for an ambulance passing a moving car, and always sanity-check the direction of the shift before the arithmetic.

Worked example 4 — both movers at once

An ambulance (siren 400 Hz) approaches at 25 m/s while a car ahead of it pulls away at 15 m/s in the same direction (v_sound = 340 m/s). Find the frequency heard by the car's driver. Method: combine the two Doppler factors — the moving source compresses the wavelength to v/(v − v_s)... apply observer motion to the wave speed relative to the driver:

f' = f × (v − v_o)/(v − v_s) = 400 × (340 − 15)/(340 − 25) = 400 × 325/315 ≈ 413 Hz

Check: the driver still hears a raised pitch (413 > 400) because the ambulance gains at 10 m/s net; if the car matched the ambulance's speed exactly, numerator and denominator would shift together in a way that leaves a smaller but nonzero rise — precisely the kind of limit case worth a moment's thought. (If the car drove toward the ambulance at 15 m/s, the sign flips: f' = 400 × 355/315 ≈ 451 Hz.)

Decibels in one paragraph

Loudness is logged: the sound level β = 10 log₁₀(I/I₀) decibels, with I₀ = 10⁻¹² W/m² at the threshold of hearing. Two consequences to carry: doubling intensity adds only ~3 dB (two identical speakers are barely louder), while ten times the intensity adds 10 dB and sounds roughly twice as loud. Combined with the inverse-square falloff of intensity with distance (Lesson 17), this explains why moving twice as far from a speaker costs 6 dB — noticeable, not catastrophic — and why hearing damage is a logarithmic negotiation with very large intensities.