An ambulance moves toward you at 30 m/s emitting sound at 500 Hz (speed of sound = 340 m/s). A classmate says: 'The sound waves reach you faster because the source is moving toward you, which raises the frequency.' What is wrong with this explanation?
ANothing — the wave speed increases when the source moves toward the observer
BThe frequency is actually lower when the source approaches, not higher
CWave speed in air is fixed by the medium and unchanged by source motion — the higher frequency arises from compressed wavelength, not faster waves
DThe classmate is correct for sound but wrong for light, where the frequency shift involves time dilation instead
Wave speed in a medium is determined by the medium's properties (density and elasticity for sound) — not by source motion. What changes when the source approaches is the wavelength: each successive wavefront is emitted closer to the previous one, packing crests together. More crests arrive per second not because they travel faster, but because they are spaced more closely. The correct formula gives f' = 500 × 340/310 ≈ 548 Hz, but the mechanism is wavelength compression, never speed increase. Option A states the specific misconception being addressed here.
Question 2 Multiple Choice
If a source moves away from a stationary observer at speed v_s, the denominator in the Doppler formula becomes (v_wave + v_s), giving a lower frequency. Which explanation correctly identifies why the denominator increases?
AThe wave speed decreases when the source moves away from the observer
BThe source emits wavefronts less frequently when receding
CEach successive wavefront is emitted farther from the previous one, stretching the wavelength behind the source
DThe medium becomes less dense behind the source, which slows the waves
When the source recedes, it travels v_s·T away from the observer between emitting successive crests (T = 1/f is the emission period). Each new crest is launched v_s·T farther from the previous one, stretching the spacing to λ' = (v_wave + v_s)/f. The crests still propagate at v_wave, but they arrive less frequently: f' = v_wave/λ' = f·v_wave/(v_wave + v_s). The emission rate (option B) does not change — the source vibrates at its own frequency regardless of motion. Options A and D both incorrectly attribute the effect to wave speed changes.
Question 3 True / False
The Doppler effect for a moving source changes the wavelength of sound as measured in the medium, but does not change the speed at which those waves travel through the medium.
TTrue
FFalse
Answer: True
Wave speed in air (~340 m/s) is set by air's density and bulk modulus — properties of the medium that are unaffected by source motion. What changes is the spatial spacing of wavefronts: ahead of an approaching source the crests bunch together (shorter λ), and behind it they spread apart (longer λ). Both the compressed and stretched wavefronts still travel at the same speed v_wave. The observer measures more or fewer arrivals per second — a frequency shift — entirely because of the changed wavelength, with no change in propagation speed.
Question 4 True / False
The Doppler formula f' = f·v_wave/(v_wave − v_s) applies equally to a moving source and a moving observer, as long as v_s is interpreted as the relative speed between source and observer.
TTrue
FFalse
Answer: False
This is a common but incorrect shortcut. The formula f' = f·v_wave/(v_wave − v_s) applies specifically to a stationary observer with a moving source. For a moving observer approaching a stationary source, the correct formula is f' = f·(v_wave + v_obs)/v_wave — a different expression, not equivalent by substituting relative speed. The two cases differ physically: a moving source changes the wavelength deposited in the medium, while a moving observer sweeps through a fixed wavelength at a different rate. These mechanisms are distinct, and in classical wave theory 'who is moving' relative to the medium matters.
Question 5 Short Answer
Explain the physical mechanism by which a moving source produces a higher observed frequency when approaching, without invoking any change in wave speed.
Think about your answer, then reveal below.
Model answer: The source emits one wavefront per period T = 1/f. When stationary, each successive crest is launched from the same position, so crests are spaced λ = v_wave·T apart. When the source moves toward the observer at speed v_s, it travels a distance v_s·T toward the observer between emitting one crest and the next. Each new crest is therefore launched v_s·T closer to the previous one, compressing the spacing to λ' = (v_wave − v_s)·T = (v_wave − v_s)/f. The crests still propagate at v_wave but arrive at the observer more frequently because they are packed together: f' = v_wave/λ' = f·v_wave/(v_wave − v_s) > f.
The mechanism is entirely geometric: source motion between successive emissions changes where each new crest originates, compressing or stretching wavefront spacing. Wave speed is unchanged; only the spacing — and hence the arrival rate — changes.