Understand Doppler Effect - IB Physics

A fire truck with siren frequency $1000$ Hz travels at $25$ m/s on a circular track of radius $50$ m. An observer stands at the center of the track. At the instant when the truck is moving tangentially past a point closest to the observer, what frequency does the observer hear? Take the speed of sound as $340$ m/s.
Answer: $1000$ Hz At the instant when the truck is moving tangentially past the closest point, its velocity is perpendicular to the line connecting it to the observer at the center. Since there is no component of velocity along the line of sight between source and observer, there is no Doppler shift at this instant. The frequency heard is exactly 1000 Hz. Choice B assumes the truck is moving directly toward the observer. Choice C assumes the truck is moving directly away. Choice D represents a compromise that incorrectly averages approaching and receding frequencies.
A source S emits sound of frequency $f$ . An observer O moves away from the stationary source at speed $u$ . In a second experiment, the observer O is stationary and the source S moves away from O at the same speed $u$ . Let $f_O$ be the frequency observed in the first experiment and $f_S$ be the frequency observed in the second. How do $f_O$ and $f_S$ compare?
Answer: $f_O < f_S$ For the moving observer: $f_O = f \left( \frac{v-u}{v} \right) = f(1 - u/v)$ . For the moving source: $f_S = f \left( \frac{v}{v+u} \right) = f(1/(1+u/v))$ . We need to compare $1 - u/v$ with $1/(1+u/v)$ . Let $x = u/v$ . For any $x > 0$ , the value of $1-x$ is always less than $1/(1+x)$ . For example, if $x=0.1$ , $1-0.1=0.9$ while $1/1.1 \approx 0.909$ . Therefore, $f_O < f_S$ . This highlights a key difference between the Doppler effect for sound (medium-dependent) and light (relative velocity only).
A source of sound moves towards a stationary observer at a speed equal to half the speed of sound. What is the ratio of the observed frequency to the source frequency, $f'/f$ ?
Answer: 2.0 The formula for the observed frequency $f'$ from a source moving towards a stationary observer is $f' = f \left( \frac{v}{v-v_s} \right)$ , where $v$ is the speed of sound and $v_s$ is the speed of the source. Given $v_s = 0.5v$ , we can substitute this into the equation: $f' = f \left( \frac{v}{v-0.5v} \right) = f \left( \frac{v}{0.5v} \right) = f \left( \frac{1}{0.5} \right) = 2f$ . The ratio $f'/f$ is therefore 2.0.
All radio signals from the Voyager 1 space probe, now in interstellar space, are received on Earth with a slightly lower frequency than they were transmitted with. What is the best explanation for this phenomenon?
Answer: The probe is moving away from Earth, causing a Doppler redshift in the radio signals. A lower received frequency corresponds to a longer wavelength, which is a redshift. For electromagnetic waves like radio signals, a redshift is caused by the source moving away from the observer (or vice versa). Voyager 1 is travelling away from the solar system and thus away from Earth. Distractor A is a description of the discredited 'tired light' hypothesis. Distractor C describes a real but much smaller effect. Distractor D describes atmospheric attenuation, which affects signal strength (amplitude), not frequency.
A police car with its siren operating at frequency $f_0 = 1200$ Hz is moving at $30$ m/s toward a stationary observer. After passing the observer, the car continues at the same speed away from the observer. 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?
Answer: $212$ Hz When approaching: $f_1 = f_0 \frac{v}{v - v_s} = 1200 \frac{340}{340 - 30} = 1200 \frac{340}{310} = 1316$ Hz. When receding: $f_2 = f_0 \frac{v}{v + v_s} = 1200 \frac{340}{340 + 30} = 1200 \frac{340}{370} = 1103$ Hz. The difference is $1316 - 1103 = 213$ Hz, which rounds to 212 Hz. Choice A uses an incorrect formula. Choice C incorrectly adds the frequencies. Choice D represents only half the actual difference.
A car travels at a constant speed $u$ in a circular path of radius $R$ . The car's horn emits a sound of constant frequency $f$ . A stationary observer is located in the plane of the circle at a distance $D \gg R$ from the center of the circle. What is the maximum observed frequency $f_{max}$ ? The speed of sound is $v$ .
Answer: $f \frac{v}{v-u}$ The maximum observed frequency occurs when the component of the car's velocity along the line of sight to the observer is at its maximum. Because the observer is very far away (D >> R), the lines of sight are nearly parallel. The maximum component of velocity towards the observer is the car's full speed, $u$ . This corresponds to a moving source approaching a stationary observer. The formula is $f' = f \frac{v}{v-v_s}$ . Therefore, the maximum frequency is $f_{max} = f \frac{v}{v-u}$ .
An ambulance travels at $25$ m/s toward a building while its siren emits sound at $800$ Hz. The sound reflects off the building and returns to the ambulance driver. Taking the speed of sound as $340$ m/s, what frequency does the driver hear from the reflected sound?
Answer: $930$ Hz This involves two Doppler shifts. First, the ambulance acts as a moving source toward a stationary 'observer' (the building): $f_1 = 800 \frac{340}{340 - 25} = 864$ Hz. Then this frequency reflects and the ambulance (now moving toward the source of reflection) acts as a moving observer: $f_2 = 864 \frac{340 + 25}{340} = 930$ Hz. Choice A represents only the first Doppler shift. Choice B incorrectly treats the ambulance as moving away. Choice D ignores the Doppler effect entirely.
An astronomer observes the light from a star. They find that the spectral lines in the star's spectrum are shifted towards the blue end of the spectrum compared to the same lines from a laboratory source. What can be deduced about the star?
Answer: The star is moving towards the Earth. A shift towards the blue end of the spectrum means the observed wavelengths are shorter and the frequencies are higher. This is known as a blueshift. A blueshift occurs when the source of the waves and the observer are moving closer to each other. Therefore, the star is moving towards the Earth. Rapid rotation would cause spectral line broadening, and temperature affects the overall continuous spectrum, not the shift of discrete lines.
A train sounding its whistle at a constant frequency approaches a stationary observer. The train is accelerating. How does the pitch of the whistle as perceived by the observer change as the train approaches?
Answer: The pitch continuously increases. The observed frequency for an approaching source is given by $f' = f(v/(v-v_s))$ , where $v$ is the speed of sound and $v_s$ is the speed of the source. Since the train is accelerating as it approaches, its speed $v_s$ is increasing. As $v_s$ increases, the denominator $(v-v_s)$ decreases, which causes the observed frequency $f'$ (the pitch) to continuously increase.
A train whistle emits sound at $500$ Hz. An observer on a platform measures the frequency to be $485$ Hz. If the speed of sound is $343$ m/s, what can be concluded about the train's motion?
Answer: The train is moving away from the observer at approximately $10.6$ m/s Since the observed frequency (485 Hz) is lower than the source frequency (500 Hz), the train must be moving away from the observer. Using $f' = f_0 \frac{v}{v + v_s}$ : $485 = 500 \frac{343}{343 + v_s}$ . Solving: $\frac{485}{500} = \frac{343}{343 + v_s}$ , so $343 + v_s = \frac{500 \times 343}{485} = 353.6$ , giving $v_s = 10.6$ m/s. Choices A and D incorrectly suggest motion toward the observer. Choice C uses an incorrect calculation method.
An aircraft flying at $200$ m/s emits a radio signal at $100$ MHz toward the ground. The signal reflects off the ground and returns to the aircraft. If the aircraft maintains constant velocity throughout this process, what is the frequency of the reflected signal received by the aircraft?
Answer: $100.13$ MHz When you encounter problems involving moving sources or observers with wave phenomena, you're dealing with the Doppler effect. This question involves a "double Doppler shift" because the aircraft acts as both source and observer, with the ground serving as a reflector. Let's work through this systematically. The Doppler shift formula is $f' = f \frac{v \pm v_o}{v \pm v_s}$ , where $v = 3 \times 10^8$ m/s (speed of light), $v_o$ is observer velocity, and $v_s$ is source velocity. First shift occurs when the signal travels from aircraft to ground. The aircraft is a moving source ( $v_s = 200$ m/s) approaching a stationary observer (ground), so: $f_1 = 100 \times \frac{3 \times 10^8}{3 \times 10^8 - 200} = 100 \times \frac{3 \times 10^8}{2.9998 \times 10^8}$ Second shift occurs when the reflected signal returns. Now the ground acts as a source and the aircraft as a moving observer approaching the source: $f_2 = f_1 \times \frac{3 \times 10^8 + 200}{3 \times 10^8}$ Combining both shifts: $f_2 = 100 \times \frac{3 \times 10^8 + 200}{3 \times 10^8 - 200} = 100 \times \frac{3.0002 \times 10^8}{2.9998 \times 10^8} \approx 100.13$ MHz Answer D (100.13 MHz) correctly accounts for both Doppler shifts. Answer A (100.00 MHz) ignores the Doppler effect entirely. Answers B and C represent incorrect applications of the formula or only considering one shift instead of both. Remember: radar problems always involve double Doppler shifts—one for transmission and one for reception. Calculate each shift separately, then combine them.
A police car travels at 30 m s⁻¹ towards a stationary observer. Its siren emits a sound of frequency 500 Hz. A second car travels at 20 m s⁻¹ in the same direction, away from the observer and behind the police car. What is the frequency of the siren as heard by the driver of the second car? The speed of sound in air is 340 m s⁻¹.
Answer: 516 Hz This HL problem involves a moving source (police car) and a moving observer (second car). Using the general Doppler effect formula $f' = f \frac{v \pm v_o}{v \mp v_s}$ . The source (police car, $v_s = 30$ ) is moving towards the observer (second car). The observer (second car, $v_o = 20$ ) is moving away from the source. The signs are chosen to reflect these motions: observer moving away is $v - v_o$ in the numerator, and source moving towards is $v - v_s$ in the denominator. So, $f' = 500 \times \frac{340 - 20}{340 - 30} = 500 \times \frac{320}{310} \approx 516$ Hz.
A submarine uses sonar with frequency $2000$ Hz while moving at $8$ m/s toward a stationary underwater cliff. The speed of sound in water is $1500$ m/s. After the sonar pulse reflects from the cliff, what frequency does the submarine's hydrophone detect?
Answer: $2021$ Hz When you encounter sonar problems involving moving sources and reflectors, you're dealing with a double Doppler shift - the frequency changes twice during the round trip. First, analyze the outbound journey. The submarine (source) moves toward the stationary cliff at 8 m/s, so the cliff receives a higher frequency. Using the Doppler formula where the source approaches a stationary observer: $f' = f \frac{v}{v - v_s} = 2000 \times \frac{1500}{1500 - 8} = 2000 \times \frac{1500}{1492} = 2010.7$ Hz. Next, consider the return journey. The cliff now acts as a source reflecting at 2010.7 Hz, and the submarine (moving toward this "source") receives an even higher frequency. For an observer approaching a stationary source: $f'' = f' \frac{v + v_o}{v} = 2010.7 \times \frac{1500 + 8}{1500} = 2010.7 \times 1.0053 = 2021.4$ Hz, which rounds to 2021 Hz. Answer A (2011 Hz) represents only applying one Doppler shift instead of recognizing the double shift nature of sonar reflection. Answer B (2043 Hz) likely results from incorrectly adding the velocity effects rather than using the proper Doppler formulas. Answer C (2032 Hz) suggests using an incorrect combined formula or making sign errors in the velocity directions. The key strategy for sonar problems is remembering that reflection creates two separate Doppler shifts: source-to-reflector, then reflector-to-detector. Always calculate both shifts sequentially, being careful about which object is moving relative to which in each leg of the journey.
A stationary observer stands near a large flat wall. A car approaches the wall at a speed of 10 m s⁻¹, sounding a horn with a frequency of 400 Hz. The observer is positioned such that the car is moving away from them. What is the approximate beat frequency heard by the observer from the direct and reflected sounds? The speed of sound is 340 m s⁻¹.
Answer: 24 Hz The observer hears two sounds. 1) The direct sound: the source (car) moves away, so $f_{direct} = 400 \times (340/(340+10)) = 388.6$ Hz. 2) The reflected sound: the car moves towards the wall, so the sound hits the wall at $f_{wall} = 400 \times (340/(340-10)) = 412.1$ Hz. The wall reflects this frequency as a stationary source, so the observer hears $f_{reflected} = 412.1$ Hz. The beat frequency is the difference: $f_{beat} = |f_{reflected} - f_{direct}| = |412.1 - 388.6| = 23.5$ Hz, which is approximately 24 Hz.
A stationary bat emits an ultrasound pulse of frequency $f_0$ . The pulse reflects from an insect moving directly away from the bat with speed $v_i$ . The bat detects the reflected pulse. The speed of sound is $v_s$ . Which expression gives the frequency $f_d$ of the detected pulse?
Answer: $f_d = f_0 \left( \frac{v_s - v_i}{v_s + v_i} \right)$ This is a double Doppler shift. First, the insect (a moving observer) receives the pulse at frequency $f_{insect} = f_0 ((v_s - v_i)/v_s)$ since it's moving away. Then, the insect acts as a moving source, reflecting this pulse back to the stationary bat. The insect is moving away from the bat, so the detected frequency is $f_d = f_{insect} (v_s/(v_s + v_i))$ . Substituting the first equation into the second gives $f_d = f_0 \left( \frac{v_s - v_i}{v_s} \right) \left( \frac{v_s}{v_s + v_i} \right) = f_0 \left( \frac{v_s - v_i}{v_s + v_i} \right)$ .
Which statement correctly distinguishes the Doppler effect for sound waves from the Doppler effect for electromagnetic waves?
Answer: The observed frequency shift for sound depends on who is moving (source or observer), whereas for light it depends only on their relative velocity. For sound waves, which travel in a medium, the frequency shift depends on the velocity of the source and the velocity of the observer relative to the medium. The formulas are different for a moving source vs. a moving observer, even if their relative speed is the same. For electromagnetic waves in a vacuum, there is no medium, and the principles of special relativity show that the shift depends only on the relative velocity between the source and the observer.
A source of sound moves with constant velocity directly towards a stationary observer, passes the observer, and then moves directly away. Which of the following describes the frequency measured by the observer?
Answer: A high constant frequency, which then abruptly drops to a low constant frequency. While the source approaches with constant velocity, the observed frequency is constant and higher than the source frequency. While it recedes with constant velocity, the observed frequency is constant and lower than the source frequency. The transition from the high pitch to the low pitch occurs abruptly at the moment the source passes the observer.
A distant galaxy is receding from Earth at a speed of $6.0 \times 10^6 \, \text{m s}^{-1}$ . A spectral line of hydrogen, which has a wavelength of 656 nm when measured in a laboratory on Earth, is observed in the galaxy's spectrum. What is the observed wavelength of this spectral line? The speed of light is $c = 3.0 \times 10^8 \, \text{m s}^{-1}$ .
Answer: 669 nm For light and speeds much less than $c$ , the Doppler shift is given by $\Delta\lambda / \lambda_0 \approx v/c$ . The fractional speed is $v/c = (6.0 \times 10^6) / (3.0 \times 10^8) = 0.02$ . The change in wavelength is $\Delta\lambda = \lambda_0 \times (v/c) = 656 \, \text{nm} \times 0.02 = 13.12 \, \text{nm}$ . Since the galaxy is receding, this is a redshift, meaning the wavelength increases. The observed wavelength is $\lambda = \lambda_0 + \Delta\lambda = 656 + 13.12 = 669.12 \, \text{nm}$ .
A weather monitoring station uses radar operating at $3.0 \times 10^9$ Hz to track a storm system moving directly toward the station at $15$ m/s. The electromagnetic waves travel at $3.0 \times 10^8$ m/s. What is the frequency shift of the reflected signal received by the station?
Answer: $300$ Hz For electromagnetic waves reflecting off a moving target, the total frequency shift is $\Delta f = \frac{2v_{target}f_0}{c}$ where the factor of 2 accounts for the round trip. Here: $\Delta f = \frac{2 \times 15 \times 3.0 \times 10^9}{3.0 \times 10^8} = \frac{90 \times 10^9}{3.0 \times 10^8} = 300$ Hz. Choice A omits the factor of 2 for the round trip. Choice C incorrectly uses $1.5 \times$ the correct formula. Choice D doubles the correct answer unnecessarily.
An ambulance with its siren emitting a sound of constant frequency moves directly away from a stationary observer. What is the primary reason the observer perceives a lower frequency?
Answer: The time interval between successive wavefronts arriving at the observer is increased. The Doppler effect is caused by the change in the arrival time of successive wavefronts. As the source moves away, each wavefront is emitted from a position further from the observer. This increases the path length and thus the travel time for each successive wavefront, leading to a longer time interval between their arrivals and hence a lower perceived frequency. A and D describe changes in amplitude (loudness), not frequency. B is a common misconception; the speed of sound in a medium is independent of the source's motion.

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