Apply Doppler Effect - IB Physics

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).
Answer: f₀ ( (vₛ + vₑ) / (vₛ - vₑ) ) 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)).
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)
Answer: 620 Hz 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.
A weather radar system operating at $10 \text{ GHz}$ detects precipitation moving toward the station. If the received signal shows a frequency shift of $+67 \text{ Hz}$ , and electromagnetic waves travel at $3.0 \times 10^8 \text{ m/s}$ , what is the speed of the precipitation?
Answer: $1.0 \text{ m/s}$ using single Doppler shift analysis For radar, the frequency shift involves a double Doppler effect: $\Delta f = \frac{2v_r f_0}{c}$ where $v_r$ is the radial velocity. Solving: $v_r = \frac{\Delta f \cdot c}{2f_0} = \frac{67 \times 3.0 \times 10^8}{2 \times 10^{10}} = 1.005 \text{ m/s}$ . 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.
A siren on a car moves at a constant velocity along a straight road, passing a stationary observer. At the instant the car is closest to the observer, what is the observed frequency of the siren?
Answer: The same as the emitted frequency. The Doppler shift is determined by the component of the source's velocity along the line connecting the source to the observer. At the instant of closest approach, the car's velocity vector is perpendicular to the line of sight to the observer. At this moment, the radial component of the velocity is zero. Therefore, there is no Doppler shift at this specific instant, and the observed frequency is equal to the emitted frequency.
A train sounds its 500 Hz horn as it moves away from a stationary observer at 30.0 m s⁻¹. What is the frequency heard by the observer? (The speed of sound in air = 340 m s⁻¹)
Answer: 459 Hz This is a case of a moving source receding from a stationary observer. The formula is f' = f (v / (v + uₛ)), where v is the speed of sound and uₛ is the speed of the source. Plugging in the values: f' = 500 Hz × (340 m s⁻¹ / (340 m s⁻¹ + 30.0 m s⁻¹)) = 500 Hz × (340 / 370) ≈ 459 Hz.
A sound source and an observer are both in motion relative to the ground. The Doppler effect will NOT be observed if:
Answer: The observer and source move with the same velocity. The Doppler effect for sound depends on the relative motion between the source and the observer along the line connecting them. If the source and observer are moving with the same velocity (i.e., same speed and same direction), the distance between them remains constant. There is no relative motion between them, so the observed frequency will be the same as the emitted frequency. All other scenarios involve a changing distance or a non-zero radial velocity component.
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?
Answer: f₁ > f₂ 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.
Light from a star is observed to be blueshifted. The fractional change in its wavelength, |Δλ|/λ, is 0.02%. What is the approximate fractional change in its frequency, |Δf|/f?
Answer: 0.02% For electromagnetic waves when the relative speed v is much less than the speed of light c, the approximate Doppler shift formulas are |Δλ|/λ ≈ v/c and |Δf|/f ≈ v/c. Therefore, the magnitude of the fractional change in frequency is approximately equal to the magnitude of the fractional change in wavelength. A 0.02% change in wavelength corresponds to a 0.02% change in frequency.
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?
Answer: Δf_approach is greater than Δf_recede. 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.
A police car traveling at $30 \text{ m/s}$ approaches a stationary observer while its siren emits sound at $800 \text{ Hz}$ . After passing the observer, the car continues at the same speed. If the speed of sound is $340 \text{ m/s}$ , what is the difference between the frequencies heard by the observer before and after the car passes?
Answer: $142 \text{ Hz}$ with the higher frequency occurring as the car approaches Using the Doppler effect formula: approaching frequency = $f_0 \frac{v}{v - v_s} = 800 \times \frac{340}{340-30} = 877.4 \text{ Hz}$ ; receding frequency = $f_0 \frac{v}{v + v_s} = 800 \times \frac{340}{340+30} = 735.1 \text{ Hz}$ . The difference is $877.4 - 735.1 = 142.3 \text{ Hz}$ , 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.
An ambulance with a siren frequency of $1200 \text{ Hz}$ is moving toward a building at $25 \text{ m/s}$ . The sound reflects off the building and returns to a stationary observer positioned between the ambulance and building. If the speed of sound is $340 \text{ m/s}$ , what frequency does the observer hear from the reflected sound?
Answer: $1372 \text{ Hz}$ due to double Doppler shift with the building as stationary reflector This involves a double Doppler shift. First, the building receives sound at frequency $f_1 = 1200 \times \frac{340}{340-25} = 1295 \text{ Hz}$ . Then this frequency is reflected back, and since the ambulance is moving away from the observer relative to the reflection point, the observer hears $f_2 = 1295 \times \frac{340}{340-25} = 1372 \text{ Hz}$ . Choice A applies only one Doppler shift. Choice C incorrectly treats it as a receding source. Choice D ignores the Doppler effect entirely.
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?
Answer: Broader than the corresponding laboratory spectral line. 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.
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)
Answer: 847 Hz 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.
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)
Answer: 4 Hz 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.
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?
Answer: It varies continuously, being highest when the source moves directly towards the observer. 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.
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?
Answer: The wavelength is decreased and the wave speed is unchanged. 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').
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?
Answer: Galaxy B is moving away from us at twice the speed of Galaxy A. 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.
A star emits light with a characteristic wavelength of $656 \text{ nm}$ in its rest frame. An astronomer observes this wavelength to be $659 \text{ nm}$ . If this shift is entirely due to the star's motion relative to Earth, what can be concluded about the star's radial velocity?
Answer: The star moves away from Earth at $1.37 \times 10^6 \text{ m/s}$ using the non-relativistic approximation 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: $\frac{\Delta \lambda}{\lambda_0} = \frac{v}{c}$ Where $\Delta \lambda = 659 - 656 = 3 \text{ nm}$ and $\lambda_0 = 656 \text{ nm}$ . $v = \frac{\Delta \lambda}{\lambda_0} \times c = \frac{3}{656} \times 3.00 \times 10^8 = 1.37 \times 10^6 \text{ m/s}$ Option B gives the wrong velocity value ( $4.57 \times 10^5 \text{ m/s}$ ) 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.
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?
Answer: 2 f u / c 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.
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)
Answer: 1.37 × 10⁶ m s⁻¹ away from Earth. 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.

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