A source emits sound at 440 Hz while stationary. In scenario A, the source moves toward a stationary observer at 20 m/s. In scenario B, the source is stationary and the observer moves toward it at 20 m/s. Do both scenarios produce the same observed frequency?
AYes — by symmetry, the observed shift depends only on the closing speed, not which party is moving
BNo — equal source and observer speeds produce different observed frequencies because the mechanisms differ
CYes — the Doppler formula f' = f(v + 20)/v applies equally to both cases
DNo — observer motion always produces a larger frequency shift than source motion at the same speed
Source motion compresses the wavefronts themselves, changing the wavelength: f' = f · v_wave / (v_wave − v_s). Observer motion changes how fast the observer sweeps through fixed-wavelength wavefronts: f' = f · (v_wave + v_o) / v_wave. At 20 m/s approach and v_wave = 343 m/s, source motion gives f' = 440 × 343/323 ≈ 467 Hz; observer motion gives f' = 440 × 363/343 ≈ 466 Hz — close but not identical, and diverging as speeds increase. Option A is the classic misconception: the scenarios look symmetric but are mechanistically distinct.
Question 2 Multiple Choice
What physically changes when an observer moves toward a stationary sound source?
AThe wavelength of the sound decreases as the observer approaches the source
BThe speed of sound through the air increases in the direction of observer motion
CThe observer's velocity relative to the wavefronts increases, so more pressure crests are encountered per second
DThe source emits higher-frequency waves in response to the approaching observer
The wavefronts are equally spaced — the source hasn't moved, so wavelength is unchanged. The observer's effective speed relative to the wavefronts is v_wave + v_o, meaning they sweep through more crests per second, increasing observed frequency. The speed of sound in air is determined by the medium, not relative motion (option B). Source motion (not observer motion) changes wavelength (option A). The source is unaware of the observer (option D).
Question 3 True / False
When an observer moves away from a stationary sound source, the speed of sound through the air decreases.
TTrue
FFalse
Answer: False
The speed of sound is a property of the medium — temperature and density of air — and is unaffected by the motion of either source or observer. What changes is the observer's speed *relative to the wavefronts*: v_wave − v_o. This reduces the rate of crest encounters, lowering observed frequency, but the wave itself still travels through air at the same v_wave.
Question 4 True / False
For equal closing speeds, a moving observer produces a different observed frequency than a moving source.
TTrue
FFalse
Answer: True
The formulas are structurally different: source motion f' = f · v_wave/(v_wave − v_s); observer motion f' = f · (v_wave + v_o)/v_wave. At the same speed these yield slightly different values because source motion physically alters the spacing of wavefronts (wavelength), while observer motion leaves the wavelength intact and only changes the encounter rate. The asymmetry grows with speed.
Question 5 Short Answer
Why does the Doppler formula for a moving observer differ structurally from the formula for a moving source, even though both change the observed frequency?
Think about your answer, then reveal below.
Model answer: Because the physical mechanisms are different. Source motion compresses or stretches the spacing between wavefronts — changing the wavelength itself — before the wave even reaches the observer. Observer motion leaves the wavelength unchanged; it only alters the rate at which the observer sweeps through fixed wavefronts. These distinct mechanisms produce formulas with different mathematical structure: source motion appears in the denominator (affecting wavelength), observer motion appears in the numerator (affecting encounter rate).
The key is to ask: what is actually different about the wave? For source motion, the wave pattern itself is distorted. For observer motion, the wave is identical to the stationary case — only the observer's relationship to it changes. Both produce a frequency shift, but tracing the physics to its origin reveals why the formulas look similar but are not interchangeable.