Doppler effect practice questions.

A siren moves at constant speed directly towards a stationary observer. Which statement explains why the observer hears a higher frequency than the frequency emitted?

• The speed of sound in the air increases
• Wavefronts ahead of the source are compressed, so the wavelength reaching the observer is shorter
• The amplitude of the wave increases as the source approaches
• The moving source actually emits a much greater number of wavefronts every second than it does at rest

Answer: Wavefronts ahead of the source are compressed, so the wavelength reaching the observer is shorter

The moving source partly catches up with each wavefront it emits, reducing the wavelength ahead of it; since the wave speed in air is unchanged, f = v/λ increases.

A hydrogen line emitted at a wavelength of 656 nm is observed at 662 nm in the light from a distant galaxy. Determine the speed at which the galaxy is receding from Earth.

• 2.7×10⁵ m s⁻¹
• 5.4×10⁶ m s⁻¹
• 2.7×10⁷ m s⁻¹
• 2.7×10⁶ m s⁻¹

Answer: 2.7×10⁶ m s⁻¹

v ≈ c Δλ/λ = 3.0×10⁸ × (6/656) ≈ 2.7×10⁶ m s⁻¹; the shift to longer wavelength (redshift) shows the galaxy is receding.

A sound source moves directly away from a stationary observer at constant speed. Compared with the same source at rest, what does the observer receive?

• A longer wavelength, a lower frequency and an unchanged wave speed
• A longer wavelength, a lower frequency and a lower wave speed
• A shorter wavelength, a higher frequency and an unchanged wave speed
• An unchanged wavelength, a lower frequency and a lower wave speed

Answer: A longer wavelength, a lower frequency and an unchanged wave speed

Behind a receding source the wavefronts are stretched apart, so λ increases and f decreases; the wave speed is fixed by the medium and does not change.

A point source of sound moves in a straight line at a constant speed less than the speed of sound. Which statement describes the pattern of the wavefronts it emits?

• The wavefronts are equally spaced circles centred on the source
• The wavefronts are closer together in front of the source and further apart behind it
• The wavefronts are further apart in front of the source and closer together behind it
• Wavefronts form only behind the moving source

Answer: The wavefronts are closer together in front of the source and further apart behind it

Each wavefront spreads from the point where it was emitted; because the source advances between emissions, the fronts bunch together ahead of it and spread out behind.

A satellite transmits a radio signal of frequency 8.0×10⁹ Hz while moving directly towards a ground station at 7.5×10³ m s⁻¹. Estimate the Doppler shift in the frequency received by the station.

• 2.0×10⁴ Hz
• 1.0×10⁵ Hz
• 2.0×10⁵ Hz
• 2.0×10⁶ Hz

Answer: 2.0×10⁵ Hz

For electromagnetic waves with v << c, Δf ≈ f v/c = 8.0×10⁹ × 7.5×10³/3.0×10⁸ = 2.0×10⁵ Hz, an increase because the satellite approaches.

An observer moves at constant speed towards a stationary sound source. Which statement explains why the observed frequency is higher than the emitted frequency?

• The wavelength of the sound in the air decreases
• The speed of sound relative to the air increases
• The source emits more wavefronts per second
• The observer crosses the wavefronts at a greater rate

Answer: The observer crosses the wavefronts at a greater rate

With a moving observer the wavelength in the medium is unchanged; the observer simply meets wavefronts more often per second, so the observed frequency rises.