All questions
Question 1
Light from a distant galaxy has an observed hydrogen-alpha spectral line at 659.3 nm. The rest wavelength of this line is 656.3 nm. What is the approximate speed of the galaxy relative to Earth? (The speed of light, c = 3.00 × 10⁸ m s⁻¹)
- 1.37 × 10⁶ m s⁻¹ away from Earth. (correct answer)
- 1.37 × 10⁶ m s⁻¹ towards Earth.
- 4.57 × 10⁵ m s⁻¹ away from Earth.
- 4.57 × 10⁵ m s⁻¹ towards Earth.
Explanation: The observed wavelength (659.3 nm) is longer than the rest wavelength (656.3 nm), which indicates a redshift. This means the galaxy is moving away from Earth. The change in wavelength is Δλ = 659.3 nm - 656.3 nm = 3.0 nm. For speeds much less than c, the Doppler formula for light is Δλ/λ ≈ v/c. Rearranging for v gives v ≈ c(Δλ/λ) = (3.00 × 10⁸ m s⁻¹) × (3.0 nm / 656.3 nm) ≈ 1.37 × 10⁶ m s⁻¹.
Question 2
A stationary ambulance emits a sound from its siren at 800 Hz. A person on a bicycle travels towards the ambulance at 20.0 m s⁻¹. What frequency does the cyclist hear? (The speed of sound in air = 340 m s⁻¹)
- 753 Hz
- 756 Hz
- 847 Hz (correct answer)
- 850 Hz
Explanation: This is a case of a moving observer approaching a stationary source. The formula is f' = f ((v + uₒ) / v), where v is the speed of sound and uₒ is the speed of the observer. Plugging in the values: f' = 800 Hz × ((340 m s⁻¹ + 20.0 m s⁻¹) / 340 m s⁻¹) = 800 Hz × (360 / 340) ≈ 847 Hz. Distractor D (850 Hz) is the frequency that would be heard if the source were moving towards a stationary observer at the same speed.
Question 3
The spectral lines from a distant, rotating star are observed. How does a single absorption line from the star's spectrum appear compared to the same line from a stationary laboratory source?
- As a single, sharp line shifted to a shorter wavelength.
- As a single, sharp line shifted to a longer wavelength.
- Broader than the corresponding laboratory spectral line. (correct answer)
- Split into two distinct, sharp spectral lines.
Explanation: One edge of the rotating star is moving towards the observer (blueshift), while the other edge is moving away (redshift). The center of the star has no radial velocity due to rotation. The observed spectral line is the sum of light from all parts of the star's disk. This combination of blueshifted, redshifted, and unshifted light causes the spectral line to appear smeared out, or broadened.
Question 4
A stationary police radar gun emits microwaves of frequency f. The waves reflect from a car approaching at speed u. The reflected waves are detected by the gun. The speed of the waves is c. Given that u ≪ c, what is the approximate frequency difference between the emitted and detected waves?
- f u / c
- 2 f u / c (correct answer)
- f u / (c - u)
- f u / (c + u)
Explanation: This is a two-step Doppler shift. First, the car acts as a moving observer approaching a stationary source, so it observes a frequency f' ≈ f(1 + u/c). Second, the car acts as a moving source reflecting this frequency f' as it approaches the stationary gun. The gun detects a frequency f'' ≈ f'(1 + u/c). Substituting for f', we get f'' ≈ f(1 + u/c)² = f(1 + 2u/c + u²/c²). Since u ≪ c, the u²/c² term is negligible. So, f'' ≈ f(1 + 2u/c). The frequency difference is Δf = f'' - f ≈ f(1 + 2u/c) - f = 2fu/c.
Question 5
A small sound source is attached to the edge of a rotating turntable, moving in a horizontal circle at a constant speed. A stationary observer is located far away, in the same plane as the turntable. How does the frequency heard by the observer vary?
- It is always equal to the source frequency.
- It is always higher than the source frequency.
- It switches abruptly between a single high frequency and a single low frequency.
- It varies continuously, being highest when the source moves directly towards the observer. (correct answer)
Explanation: The observer detects a Doppler shift based on the component of the source's velocity directed towards them. As the source rotates, this radial velocity component changes continuously. It is maximum positive (moving directly towards) at one point, zero at two points (moving tangentially), and maximum negative (moving directly away) at another. This sinusoidal variation in radial velocity results in a smooth, continuous variation in the observed frequency, which is highest when the source's velocity is directly towards the observer.
Question 6
A bat flies toward a stationary wall at speed vₑ. It emits an ultrasound pulse of frequency f₀. The pulse reflects off the wall and returns to the bat. What is the frequency fₑ of the echo detected by the bat? (The speed of sound is vₛ).
- f₀ (vₛ / (vₛ - vₑ))
- f₀ ( (vₛ + vₑ) / vₛ )
- f₀ ( (vₛ + vₑ) / (vₛ - vₑ) ) (correct answer)
- f₀ ( (vₛ - vₑ) / (vₛ + vₑ) )
Explanation: This is a two-step Doppler shift problem. Step 1: The bat is a moving source approaching a stationary observer (the wall). The frequency received by the wall is f_wall = f₀ (vₛ / (vₛ - vₑ)). Step 2: The wall acts as a stationary source emitting at frequency f_wall. The bat is a moving observer approaching this source. The frequency detected by the bat is fₑ = f_wall ((vₛ + vₑ) / vₛ). Substituting the expression for f_wall into the second equation gives fₑ = f₀ (vₛ / (vₛ - vₑ)) × ((vₛ + vₑ) / vₛ) = f₀ ((vₛ + vₑ) / (vₛ - vₑ)).
Question 7
A car's horn emits a sound at 425 Hz. When the car is moving directly away from a stationary observer, the observer detects a frequency of 400 Hz. What is the speed of the car? (The speed of sound in air = 340 m s⁻¹)
- 20.0 m s⁻¹
- 21.3 m s⁻¹ (correct answer)
- 22.7 m s⁻¹
- 25.0 m s⁻¹
Explanation: The formula for a moving source receding from a stationary observer is f' = f(v / (v + uₛ)). We need to solve for the source speed, uₛ. Rearranging gives v + uₛ = v(f / f'), so uₛ = v(f / f') - v. Plugging in the values: uₛ = 340 m s⁻¹ × (425 Hz / 400 Hz) - 340 m s⁻¹ = 340 × 1.0625 - 340 = 361.25 - 340 = 21.25 m s⁻¹. This is approximately 21.3 m s⁻¹. Distractor A (20.0 m s⁻¹) is the result of using the approximation uₛ ≈ v(Δf/f).
Question 8
Two identical tuning forks produce a sound of 440 Hz. One tuning fork is stationary. The other is moved away from a stationary observer at a speed of 3.0 m s⁻¹. What is the approximate beat frequency heard by the observer? (The speed of sound in air = 340 m s⁻¹)
- 0 Hz
- 2 Hz
- 4 Hz (correct answer)
- 8 Hz
Explanation: The observer hears a frequency of f₁ = 440 Hz from the stationary fork. From the moving fork, the observer hears a lower frequency, f₂. Using the moving source formula: f₂ = 440 Hz × (340 / (340 + 3.0)) ≈ 436.15 Hz. The beat frequency is the difference between the two frequencies: f_beat = |f₁ - f₂| = |440 - 436.15| = 3.85 Hz, which is approximately 4 Hz.
Question 9
A stationary observer hears a sound from a source moving towards them at a constant speed, which is less than the speed of sound. Which statement correctly describes the properties of the sound waves between the source and the observer, compared to waves from a stationary source?
- The wavelength is decreased and the wave speed is increased.
- The wavelength is increased and the wave speed is unchanged.
- The wavelength is unchanged and the wave speed is increased.
- The wavelength is decreased and the wave speed is unchanged. (correct answer)
Explanation: The speed of a wave is determined by the properties of the medium through which it travels, not by the motion of the source. Therefore, the wave speed remains unchanged. As the source moves towards the observer, it emits successive wave crests from positions closer to the observer. This motion effectively 'compresses' the waves in the forward direction, resulting in a decreased wavelength. The increased frequency heard by the observer is a consequence of these shorter-wavelength crests arriving at the observer's location at an unchanged speed (f' = v/λ').
Question 10
A sound source moving at speed u approaches a stationary observer, resulting in an observed frequency f_approach. The same source then moves away from the observer at the same speed u, resulting in f_recede. The source frequency is f₀. How does the magnitude of the frequency increase, Δf_approach = f_approach - f₀, compare to the magnitude of the frequency decrease, Δf_recede = f₀ - f_recede?
- Δf_approach is greater than Δf_recede. (correct answer)
- Δf_approach is less than Δf_recede.
- Δf_approach is equal to Δf_recede.
- The relationship depends on the value of the source frequency f₀.
Explanation: For a moving source, f_approach = f₀(v/(v-u)) and f_recede = f₀(v/(v+u)). The shifts are Δf_approach = f₀(v/(v-u) - 1) = f₀(u/(v-u)) and Δf_recede = f₀(1 - v/(v+u)) = f₀(u/(v+u)). Since (v-u) is smaller than (v+u), the fraction 1/(v-u) is larger than 1/(v+u). Therefore, the frequency shift upon approach is greater in magnitude than the shift upon recession.
Question 11
The light from Galaxy A has a redshift z = 0.01. The light from Galaxy B has a redshift z = 0.02. Redshift z is defined as Δλ/λ₀. What can be deduced about the galaxies, assuming the standard cosmological model?
- Galaxy B is moving away from us at half the speed of Galaxy A.
- Both galaxies are moving away from us at the same speed.
- Galaxy A is moving away from us, and Galaxy B is moving towards us.
- Galaxy B is moving away from us at twice the speed of Galaxy A. (correct answer)
Explanation: Redshift indicates that an object is moving away from the observer. Since both galaxies have a positive redshift, they are both moving away from us. For small redshifts (z ≪ 1), the recessional velocity v is approximately v ≈ zc. Since the redshift of Galaxy B (0.02) is twice the redshift of Galaxy A (0.01), Galaxy B is moving away from us at approximately twice the speed of Galaxy A.
Question 12
A sound source S emits frequency f. The speed of sound is v. In Case 1, the source moves towards a stationary observer at speed u. In Case 2, the observer moves towards the stationary source at the same speed u. Let the observed frequencies be f₁ and f₂ respectively. Which statement is correct for all speeds 0 < u < v?
- f₁ = f₂
- f₁ > f₂ (correct answer)
- f₁ < f₂
- The relationship depends on whether u > v/2.
Explanation: In Case 1 (moving source), f₁ = f(v / (v - u)). In Case 2 (moving observer), f₂ = f((v + u) / v). To compare them, we can compare the factors multiplying f. We need to compare v/(v - u) with (v + u)/v. This is equivalent to comparing v² with (v + u)(v - u) = v² - u². Since v² > v² - u² for any u > 0, it follows that v/(v - u) > (v + u)/v. Therefore, f₁ > f₂.
Question 13
A police car's siren has a frequency of 600 Hz. The police car is travelling at 40 m s⁻¹ and chasing a truck travelling in the same direction at 30 m s⁻¹. What frequency does the driver of the truck hear? (The speed of sound in air = 340 m s⁻¹)
- 581 Hz
- 618 Hz
- 620 Hz (correct answer)
- 622 Hz
Explanation: This situation involves both a moving source and a moving observer. The general Doppler effect formula is f' = f ((v ± uₒ) / (v ∓ uₛ)). The police car (source) is moving towards the truck (observer), so the sign in the denominator is negative. The truck (observer) is moving away from the police car (source), so the sign in the numerator is negative. Therefore, f' = f ((v - uₒ) / (v - uₛ)) = 600 Hz × ((340 - 30) / (340 - 40)) = 600 Hz × (310 / 300) = 620 Hz. Distractor B is the result of incorrectly using the relative speed (10 m s⁻¹) in the moving source formula.
Question 14
A weather radar system operating at 10 GHz detects precipitation moving toward the station. If the received signal shows a frequency shift of +67 Hz, and electromagnetic waves travel at 3.0×108 m/s, what is the speed of the precipitation?
- 1.0 m/s using single Doppler shift analysis (correct answer)
- 2.0 m/s accounting for round-trip electromagnetic propagation
- 0.5 m/s based on the standard weather radar equation
- 4.0 m/s considering both transmission and reflection components
Explanation: For radar, the frequency shift involves a double Doppler effect: Δf=c2vrf0 where vr is the radial velocity. Solving: vr=2f0Δf⋅c=2×101067×3.0×108=1.005 m/s. The factor of 2 in the denominator already accounts for the round-trip nature. Choice B incorrectly applies an additional factor of 2. Choice C uses half the correct calculation. Choice D applies an extra factor of 4. Question 15
A police car traveling at 30 m/s approaches a stationary observer while its siren emits sound at 800 Hz. After passing the observer, the car continues at the same speed. If the speed of sound is 340 m/s, what is the difference between the frequencies heard by the observer before and after the car passes?
- 142 Hz with the higher frequency occurring as the car approaches (correct answer)
- 71 Hz with the higher frequency occurring as the car approaches
- 154 Hz with the higher frequency occurring as the car recedes
- 77 Hz with the higher frequency occurring as the car recedes
Explanation: Using the Doppler effect formula: approaching frequency = f0v−vsv=800×340−30340=877.4 Hz; receding frequency = f0v+vsv=800×340+30340=735.1 Hz. The difference is 877.4−735.1=142.3 Hz, with the higher frequency when approaching. Choice B uses half the correct calculation. Choices C and D incorrectly identify which case produces the higher frequency. Question 16
Two identical sound sources each emit 500 Hz tones. Source A moves toward a stationary observer at 20 m/s while Source B moves away from the same observer at 20 m/s. If the speed of sound is 340 m/s and the sources are positioned so their waves interfere at the observer's location, what phenomenon occurs?
- Constructive interference producing a steady tone at 531 Hz
- Beat frequency of 59 Hz due to the frequency difference between sources (correct answer)
- Destructive interference resulting in complete silence at the observer
- Beat frequency of 30 Hz based on the average Doppler shift
Explanation: Source A produces frequency fA=500×340−20340=531.25 Hz. Source B produces fB=500×340+20340=472.22 Hz. The beat frequency is ∣fA−fB∣=∣531.25−472.22∣=59.03 Hz. Choice A incorrectly assumes the sources combine to one frequency. Choice C is wrong because the amplitudes aren't necessarily equal and opposite. Choice D incorrectly calculates the beat frequency. Question 17
A train whistle emits sound at 600 Hz. An observer on a platform hears 648 Hz as the train approaches and 558 Hz as it recedes. Using these measurements, what can be determined about the speeds involved?
- Train speed is 35 m/s and sound speed is 350 m/s simultaneously
- Sound speed is 335 m/s assuming train speed is 20 m/s
- Train speed is 25 m/s assuming sound speed is 340 m/s (correct answer)
- Observer speed is 15 m/s with train stationary relative to platform
Explanation: This question tests the Doppler effect, which occurs when there's relative motion between a sound source and observer. When you encounter Doppler problems, you need to identify what's known and what you're solving for, then apply the appropriate equation.
The Doppler effect equation is f′=fv∓vsv±vo, where f′ is observed frequency, f is source frequency, v is sound speed, vo is observer speed, and vs is source speed. For a stationary observer and moving source, this simplifies to f′=fv∓vsv.
Using the approaching case: 648=600v−vsv, so v−vsv=1.08
Using the receding case: 558=600v+vsv, so v+vsv=0.93
Solving these equations simultaneously gives vs=25 m/s when v=340 m/s. Option C is correct.
Option A incorrectly assumes both speeds can be determined independently from the given data - you need one known value to find the other. Option B uses an incorrect assumption about train speed and yields an unrealistic sound speed for air at room temperature. Option D misidentifies the moving object; if the observer were moving and the train stationary, the frequency shifts would be different due to the asymmetric nature of the Doppler effect.
Remember: Doppler problems typically require assuming a standard value (like sound speed ≈ 340 m/s in air) to solve for the unknown quantity. Question 18
An ambulance with a siren frequency of 1200 Hz is moving toward a building at 25 m/s. The sound reflects off the building and returns to a stationary observer positioned between the ambulance and building. If the speed of sound is 340 m/s, what frequency does the observer hear from the reflected sound?
- 1286 Hz because the building acts as a moving source
- 1372 Hz due to double Doppler shift with the building as stationary reflector (correct answer)
- 1158 Hz because the reflected wave appears to come from a receding source
- 1200 Hz since reflection preserves the original frequency unchanged
Explanation: This involves a double Doppler shift. First, the building receives sound at frequency f1=1200×340−25340=1295 Hz. Then this frequency is reflected back, and since the ambulance is moving away from the observer relative to the reflection point, the observer hears f2=1295×340−25340=1372 Hz. Choice A applies only one Doppler shift. Choice C incorrectly treats it as a receding source. Choice D ignores the Doppler effect entirely. Question 19
A bat emits ultrasonic calls at 50 kHz while flying at 15 m/s toward a moth that is flying away at 5 m/s. The sound reflects off the moth and returns to the bat. If the speed of sound is 340 m/s, what frequency does the bat detect in the returning echo?
- 54.8 kHz considering the combined relative velocity of 20 m/s
- 51.5 kHz accounting for relative motion during transmission only
- 53.2 kHz using sequential Doppler shifts for approach and reflection (correct answer)
- 52.4 kHz applying double Doppler effect with moving source and receiver
Explanation: When you encounter Doppler effect problems involving reflection, you must analyze the sound's journey in two separate stages, as the roles of source and observer change at each stage.
First, consider the bat's call traveling to the moth. The bat (source) moves toward the stationary reference frame at 15 m/s, while the moth (observer) moves away at 5 m/s. Using the Doppler formula: f1=f0v+vsv+vo=50,000340−15340−5=51,538 Hz
Second, analyze the reflected sound returning to the bat. Now the moth acts as the source (emitting 51,538 Hz) moving away at 5 m/s, while the bat becomes the observer moving toward the source at 15 m/s: f2=f1v+vsv+vo=51,538340+5340+15=53,200 Hz≈53.2 kHz
Answer A incorrectly treats this as a single Doppler shift with combined velocities, ignoring that reflection creates two distinct interactions. Answer B only accounts for the outbound journey, missing the return trip's additional frequency shift. Answer D mentions "double Doppler effect" but arrives at an incorrect frequency, likely from mathematical errors in the sequential calculation.
Study tip: For any Doppler problem involving reflection, always break it into two separate calculations—sound going out, then sound coming back. The frequency shifts compound, and you cannot simply add velocities. Question 20
A star emits light with a characteristic wavelength of 656 nm in its rest frame. An astronomer observes this wavelength to be 659 nm. If this shift is entirely due to the star's motion relative to Earth, what can be concluded about the star's radial velocity?
- The star moves toward Earth at 1.37×106 m/s based on classical Doppler theory
- The star moves away from Earth at 4.57×105 m/s based on redshift analysis
- The star moves toward Earth at 4.57×105 m/s correcting for relativistic effects
- The star moves away from Earth at 1.37×106 m/s using the non-relativistic approximation (correct answer)
Explanation: When you encounter wavelength shifts from distant stars, you're dealing with the Doppler effect - the change in observed frequency (or wavelength) when there's relative motion between source and observer.
First, determine the direction of motion. Since the observed wavelength (659 nm) is longer than the emitted wavelength (656 nm), this is a redshift, meaning the star is moving away from Earth. This immediately eliminates options A and C, which claim the star moves toward us.
For the velocity calculation, use the non-relativistic Doppler formula since stellar velocities are typically much less than the speed of light:
λ0Δλ=cv
Where Δλ=659−656=3 nm and λ0=656 nm.
v=λ0Δλ×c=6563×3.00×108=1.37×106 m/s
Option B gives the wrong velocity value (4.57×105 m/s) despite correctly identifying the direction. Option A has this same incorrect velocity but wrong direction. Option C uses the incorrect velocity and wrong direction, despite mentioning "relativistic effects" which aren't needed here since v<<c.
Option D correctly identifies both the direction (away from Earth) and the proper velocity using non-relativistic approximation.
Study tip: For Doppler shift problems, always check the direction first (redshift = moving away, blueshift = moving toward), then apply the appropriate formula. Most stellar motion problems use non-relativistic approximations.