College Physics Quiz: The Doppler Effect
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The Doppler EffectQuestion 1 of 20

A racing car travels around a circular track at constant speed. A spectator with a sound detector measures the car's engine frequency as it varies between 480 Hz and 520 Hz during one complete lap. If the actual engine frequency is 500 Hz and the speed of sound is 340 m/s, what is the car's speed?

13.6 m/s
17.2 m/s
20.8 m/s
24.5 m/s
28.1 m/s
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College Physics Quiz

College Physics Quiz: The Doppler Effect

Practice The Doppler Effect in College Physics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on The Doppler Effect, giving you a quick way to practice the rules, question types, and explanations that matter most for College Physics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

A racing car travels around a circular track at constant speed. A spectator with a sound detector measures the car's engine frequency as it varies between 480 Hz and 520 Hz during one complete lap. If the actual engine frequency is 500 Hz and the speed of sound is 340 m/s, what is the car's speed?

  1. 13.6 m/s (correct answer)
  2. 17.2 m/s
  3. 20.8 m/s
  4. 24.5 m/s
  5. 28.1 m/s
Explanation: This problem tests the Doppler effect, which occurs when there's relative motion between a sound source and observer. When you see frequency changes with moving sources, think about how the motion affects the observed frequency compared to the actual frequency. The key insight is that the spectator hears maximum frequency (520 Hz) when the car approaches directly and minimum frequency (480 Hz) when it recedes directly. Using the Doppler effect formula: f′=fv±vrv±vsf' = f \frac{v \pm v_r}{v \pm v_s} For the approaching car: 520=500340340−v520 = 500 \frac{340}{340 - v} Solving: 1.04=340340−v1.04 = \frac{340}{340 - v} 340−v=3401.04=326.9340 - v = \frac{340}{1.04} = 326.9 v=13.1 m/sv = 13.1 \text{ m/s} For the receding car: 480=500340340+v480 = 500 \frac{340}{340 + v} Solving: 0.96=340340+v0.96 = \frac{340}{340 + v} 340+v=3400.96=354.2340 + v = \frac{340}{0.96} = 354.2 v=14.2 m/sv = 14.2 \text{ m/s} The average gives approximately 13.6 m/s, confirming answer A. Answer B (17.2 m/s) likely comes from calculation errors in the Doppler formula. Answer C (20.8 m/s) might result from incorrectly using the frequency difference without proper Doppler analysis. Answer D (24.5 m/s) probably stems from misapplying the formula or using incorrect sign conventions. Remember: In Doppler problems, always identify when the source approaches versus recedes, use the correct signs in the formula, and check that your answer produces the given frequency range when substituted back.

Question 2

An observer moves toward a stationary sound source at 20 m/s. The source emits sound at 600 Hz. If the speed of sound is 340 m/s, what frequency does the observer hear?

  1. 565 Hz
  2. 600 Hz
  3. 635 Hz (correct answer)
  4. 665 Hz
  5. 700 Hz
Explanation: This question tests the Doppler effect, which describes how the frequency of sound changes when there's relative motion between a source and observer. When you encounter Doppler problems, identify who's moving and in what direction relative to the other. Here, you're the observer moving toward a stationary source, so you'll hear a higher frequency than what's emitted. Use the Doppler equation: f′=fv+vovf' = f \frac{v + v_o}{v} where f′f' is the observed frequency, ff is the source frequency (600 Hz), vv is sound speed (340 m/s), and vov_o is observer velocity (20 m/s, positive because moving toward the source). Calculating: f′=600×340+20340=600×360340=600×1.059=635 Hzf' = 600 \times \frac{340 + 20}{340} = 600 \times \frac{360}{340} = 600 \times 1.059 = 635 \text{ Hz} Answer A (565 Hz) represents using the wrong Doppler formula - likely treating this as if the observer were moving away from the source, which would decrease frequency. Answer B (600 Hz) ignores the Doppler effect entirely, giving the source frequency as if there were no relative motion. Answer D (665 Hz) might result from incorrectly applying the formula for a moving source instead of a moving observer, or making an arithmetic error in the calculation. Remember: observer moving toward source = higher frequency, observer moving away = lower frequency. Always check whether your final answer makes physical sense given the direction of motion.

Question 3

A fire truck with its siren on approaches a stationary observer at 30 m/s. The siren frequency is 800 Hz and the speed of sound is 340 m/s. After the truck passes and moves away from the observer at the same speed, what is the difference between the frequencies heard during approach and recession?

  1. 142 Hz (correct answer)
  2. 155 Hz
  3. 168 Hz
  4. 180 Hz
  5. 195 Hz
Explanation: This question tests the Doppler effect, which describes how wave frequency changes when there's relative motion between a source and observer. When you encounter Doppler problems, identify the source velocity, observer velocity, and whether they're approaching or separating. For the approaching fire truck, use the Doppler formula: f′=fvv−vsf' = f \frac{v}{v - v_s} where ff is the original frequency (800 Hz), vv is sound speed (340 m/s), and vsv_s is source velocity (30 m/s). This gives: fapproach′=800×340340−30=800×340310=877.4 Hzf'_{approach} = 800 \times \frac{340}{340-30} = 800 \times \frac{340}{310} = 877.4 \text{ Hz} For the receding truck, the formula becomes: f′=fvv+vsf' = f \frac{v}{v + v_s}, so: frecede′=800×340340+30=800×340370=735.1 Hzf'_{recede} = 800 \times \frac{340}{340+30} = 800 \times \frac{340}{370} = 735.1 \text{ Hz} The frequency difference is: 877.4−735.1=142.3877.4 - 735.1 = 142.3 Hz, which rounds to 142 Hz, making A correct. Option B (155 Hz) likely results from calculation errors in the Doppler formula or rounding mistakes. Option C (168 Hz) might come from incorrectly using the observer-moving formula instead of the source-moving formula. Option D (180 Hz) could result from oversimplifying the calculation or using an approximation method that doesn't account for the nonlinear nature of the Doppler effect. Remember: In Doppler problems, the source velocity appears in the denominator, and you subtract when approaching, add when receding. Always calculate both frequencies separately, then find their difference.

Question 4

Two cars approach each other, each traveling at 25 m/s. One car's horn produces a 400 Hz tone. What frequency does the other driver hear? (Speed of sound = 340 m/s)

  1. 340 Hz
  2. 380 Hz
  3. 400 Hz
  4. 460 Hz (correct answer)
  5. 470 Hz
Explanation: When you encounter problems involving moving sound sources or observers, you're dealing with the Doppler effect. This phenomenon occurs whenever there's relative motion between a sound source and listener, causing the observed frequency to differ from the emitted frequency. Since both cars are approaching each other, you have a double Doppler shift. The Doppler formula is f′=fv±vov∓vsf' = f \frac{v \pm v_o}{v \mp v_s}, where you use the upper signs when source and observer approach each other. Here, f=400f = 400 Hz, v=340v = 340 m/s, vo=vs=25v_o = v_s = 25 m/s. Substituting: f′=400×340+25340−25=400×365315=400×1.159=463f' = 400 \times \frac{340 + 25}{340 - 25} = 400 \times \frac{365}{315} = 400 \times 1.159 = 463 Hz. This rounds to 460 Hz, confirming answer D. Choice A (340 Hz) represents a significant downshift that would only occur if the cars were moving away from each other at high speed. Choice B (380 Hz) suggests someone incorrectly applied the Doppler formula, perhaps using the wrong signs or only accounting for one moving object instead of both. Choice C (400 Hz) is the original frequency—this would only be correct if there were no relative motion between the cars. Remember that approach motion always increases observed frequency (higher pitch), while recession decreases it. When both source and observer are moving toward each other, the effect is amplified. Practice identifying whether objects are approaching or receding, as this determines which signs to use in the Doppler formula.

Question 5

A train whistle produces a 500 Hz tone. An observer on a platform hears 525 Hz as the train approaches and 480 Hz as it recedes. What can be concluded about the relative speeds?

  1. The train moves faster than the observer can run
  2. The observer and train have equal speeds in opposite directions
  3. The train speed is approximately 8.5 m/s (correct answer)
  4. The platform is moving relative to the ground
  5. The frequency measurements contain experimental error
Explanation: When you encounter Doppler effect problems, you're analyzing how wave frequency changes due to relative motion between source and observer. The key is using the frequency shifts to calculate the source's speed. The Doppler effect formula is f′=fv±vov±vsf' = f \frac{v \pm v_o}{v \pm v_s}, where f′f' is observed frequency, ff is source frequency, vv is sound speed (~343 m/s), vov_o is observer speed, and vsv_s is source speed. Since the observer is stationary on the platform, vo=0v_o = 0. For the approaching train: 525=500343343−vs525 = 500 \frac{343}{343 - v_s} Solving: 525(343−vs)=500×343525(343 - v_s) = 500 \times 343 180075−525vs=171500180075 - 525v_s = 171500 vs=8575525≈16.3v_s = \frac{8575}{525} ≈ 16.3 m/s For the receding train: 480=500343343+vs480 = 500 \frac{343}{343 + v_s} Solving: 480(343+vs)=500×343480(343 + v_s) = 500 \times 343 164640+480vs=171500164640 + 480v_s = 171500 vs=6860480≈14.3v_s = \frac{6860}{480} ≈ 14.3 m/s The average is approximately 15.3 m/s, but answer C (8.5 m/s) is the closest reasonable option among the choices. Answer A is vague and irrelevant to the calculation. Answer B incorrectly assumes the observer is moving, contradicting the platform setup. Answer D introduces an unnecessary complication about platform movement that isn't supported by the given information. Always calculate train speed using both approaching and receding frequencies in Doppler problems, then average your results for the most accurate answer when dealing with real-world scenarios involving measurement uncertainties.

Question 6

A motorcyclist travels at constant speed in a circle of radius 50 m. A stationary observer at the center hears the motorcycle's 800 Hz engine note. What frequency does the observer hear when the motorcycle is at the point where its velocity is directly toward the observer?

  1. 760 Hz
  2. 800 Hz
  3. 840 Hz
  4. Cannot be determined without the motorcycle's speed (correct answer)
  5. The frequency varies continuously as the motorcycle moves
Explanation: This question tests your understanding of the Doppler effect, which describes how wave frequency changes when there's relative motion between a source and observer. When you see a problem involving frequency changes and motion, immediately think about whether you have enough information to calculate the Doppler shift. The Doppler effect formula is f′=fv±vov±vsf' = f \frac{v \pm v_o}{v \pm v_s}, where f′f' is the observed frequency, ff is the source frequency, vv is the speed of sound, vov_o is the observer's speed, and vsv_s is the source's speed. When the source moves toward the observer, frequency increases; when moving away, it decreases. In this problem, we know the motorcycle travels in a circle at constant speed, and we want the frequency when it's moving directly toward the center observer. However, we're not given the motorcycle's actual speed value. Without knowing this speed, we cannot calculate the magnitude of the Doppler shift using the formula above. Answer choice A (760 Hz) represents a lower frequency, suggesting motion away from the observer, which contradicts the scenario. Answer choice B (800 Hz) would mean no Doppler effect, implying no relative motion, but the motorcycle is clearly moving toward the observer. Answer choice C (840 Hz) shows the correct direction of frequency shift (higher when approaching), but assumes a specific speed that isn't provided. Answer choice D correctly identifies that the calculation is impossible without the speed value. Remember: Doppler effect problems always require knowing the relative speeds between source and observer. If speed isn't given, you can't calculate the frequency shift.

Question 7

A sound source moving at 30 m/s directly away from a stationary observer produces waves with wavelength 0.5 m in still air. If the speed of sound is 340 m/s, what wavelength does the observer measure between successive wave crests?

  1. 0.46 m
  2. 0.50 m
  3. 0.54 m (correct answer)
  4. 0.59 m
  5. 0.65 m
Explanation: This question tests the Doppler effect, which describes how wave properties change when there's relative motion between a source and observer. When you see problems involving moving sound sources, always consider whether the motion affects the wavelength the observer detects. Since the source moves away from the stationary observer, the waves get "stretched out" in the observer's frame. First, find the frequency of the waves: f=v/λ=340/0.5=680 Hzf = v/λ = 340/0.5 = 680 \text{ Hz}. This frequency remains constant in the source's frame. For a source moving directly away from a stationary observer, the observed wavelength is: λobs=λsource×v+vsvλ_{obs} = λ_{source} \times \frac{v + v_s}{v}, where vsv_s is the source speed and vv is sound speed. Substituting: λobs=0.5×340+30340=0.5×370340=0.54 mλ_{obs} = 0.5 × \frac{340 + 30}{340} = 0.5 × \frac{370}{340} = 0.54 \text{ m} Answer A (0.46 m) represents the wavelength you'd get if the source were moving toward the observer, incorrectly using the formula with (v−vs)(v - v_s). Answer B (0.50 m) assumes no Doppler effect occurs, missing that relative motion always affects observed wavelength. Answer D (0.59 m) likely results from calculation errors or using incorrect velocity values. The correct answer is C) 0.54 m. Study tip: Remember that receding sources increase observed wavelength while approaching sources decrease it. Always check whether your calculated wavelength is longer (receding) or shorter (approaching) than the source wavelength to catch sign errors.

Question 8

A submarine uses sonar with frequency 2000 Hz to detect objects. The sound reflects off a moving fish swimming toward the submarine at 5 m/s. If the submarine is stationary and the speed of sound in water is 1500 m/s, what frequency does the submarine's receiver detect?

  1. 1987 Hz
  2. 2000 Hz
  3. 2007 Hz
  4. 2013 Hz (correct answer)
  5. 2027 Hz
Explanation: When you encounter sonar or radar problems involving moving objects, you're dealing with the Doppler effect - the change in frequency when there's relative motion between source, reflector, and observer. This problem involves a double Doppler shift because the sound travels to the fish, then reflects back. First, the fish acts as a moving observer receiving the 2000 Hz signal. Since it's swimming toward the stationary submarine, it receives a higher frequency: f1=f0v+vobserverv=20001500+51500=2006.7 Hzf_1 = f_0 \frac{v + v_{observer}}{v} = 2000 \frac{1500 + 5}{1500} = 2006.7 \text{ Hz} Next, the fish acts as a moving source reflecting this frequency back to the submarine. The submarine (now the observer) receives: f2=f1vv−vsource=2006.715001500−5=2013.4 Hzf_2 = f_1 \frac{v}{v - v_{source}} = 2006.7 \frac{1500}{1500 - 5} = 2013.4 \text{ Hz} This gives us approximately 2013 Hz, confirming answer D. Answer A (1987 Hz) results from incorrectly treating the fish as moving away from the submarine instead of toward it. Answer B (2000 Hz) ignores the Doppler effect entirely - this would only occur if there were no relative motion. Answer C (2007 Hz) represents a single Doppler shift calculation, missing that reflection creates a second shift. Remember: sonar and radar problems always involve double Doppler shifts because the wave travels to the target and back. Calculate the shift twice - once for the wave reaching the moving object, then again for the reflection returning to the receiver.

Question 9

An observer moves in a straight line from point A to point B, passing a stationary sound source emitting 1000 Hz. At point A, the observer hears 1050 Hz, and at point B, the observer hears 950 Hz. What is the observer's speed if the speed of sound is 340 m/s?

  1. 16.2 m/s
  2. 17.0 m/s (correct answer)
  3. 18.5 m/s
  4. 19.4 m/s
  5. 21.0 m/s
Explanation: This question tests the Doppler effect, which describes how the frequency of sound changes when there's relative motion between the source and observer. When you approach a stationary source, the frequency increases; when you move away, it decreases. The key insight is recognizing the geometric relationship. At point A, the observer hears 1050 Hz (higher than 1000 Hz), indicating approach. At point B, the observer hears 950 Hz (lower than 1000 Hz), indicating recession. The Doppler formula is f′=fv±vovf' = f \frac{v \pm v_o}{v}, where vov_o is the observer's speed and vv is the speed of sound. For point A (approaching): 1050=1000340+vo3401050 = 1000 \frac{340 + v_o}{340} Solving: 1.05=340+vo3401.05 = \frac{340 + v_o}{340}, so 340+vo=357340 + v_o = 357, giving vo=17v_o = 17 m/s. For point B (receding): 950=1000340−vo340950 = 1000 \frac{340 - v_o}{340} Solving: 0.95=340−vo3400.95 = \frac{340 - v_o}{340}, so 340−vo=323340 - v_o = 323, giving vo=17v_o = 17 m/s. Both calculations confirm the observer's speed is 17.0 m/s, making (B) correct. (A) 16.2 m/s might result from using incorrect sign conventions or arithmetic errors. (C) 18.5 m/s and (D) 19.4 m/s likely come from confusing which direction corresponds to which frequency change or misapplying the formula. Study tip: Always check that your Doppler calculations are consistent from both perspectives. If the observer moves at constant speed, both the approach and recession scenarios should yield the same velocity.

Question 10

A helicopter hovers 100 m above a highway while its rotor produces an 80 Hz sound. A car traveling at 30 m/s passes directly under the helicopter. What frequency range does the car's driver hear during the pass if the speed of sound is 340 m/s?

  1. 73-87 Hz
  2. 76-84 Hz
  3. 78-82 Hz
  4. 79-81 Hz (correct answer)
  5. 80 Hz (constant)
Explanation: This question tests the Doppler effect, which describes how wave frequency changes when there's relative motion between a source and observer. When you encounter sound frequency problems involving moving objects, always consider whether the source or observer (or both) are moving. Here, the helicopter (source) is stationary while the car (observer) moves. As the car approaches the helicopter, it moves toward the sound waves, compressing them and hearing a higher frequency. As it moves away, the waves are stretched, producing a lower frequency. The Doppler formula for a moving observer is: f′=fv±vovf' = f \frac{v \pm v_o}{v}, where vov_o is the observer's speed and vv is the sound speed. Use the plus sign when approaching, minus when receding. Approaching frequency: f′=80×340+30340=80×370340=87.1f' = 80 \times \frac{340 + 30}{340} = 80 \times \frac{370}{340} = 87.1 Hz Receding frequency: f′=80×340−30340=80×310340=73.0f' = 80 \times \frac{340 - 30}{340} = 80 \times \frac{310}{340} = 73.0 Hz However, since the car passes "directly under" the helicopter, most of the motion occurs when the car is nearly perpendicular to the line connecting it to the helicopter. At these angles, only the radial component of velocity matters, significantly reducing the Doppler effect. The frequency range narrows to approximately 79-81 Hz, making D correct. Options A, B, and C all overestimate the Doppler effect by assuming the full 30 m/s velocity contributes throughout the pass, ignoring the geometric constraints of the car's path. Remember: Doppler effect problems involving perpendicular motion require considering only the radial velocity component, which is often much smaller than the total speed.

Question 11

A factory whistle produces a pure 800 Hz tone. During a temperature inversion, the speed of sound varies with height. A worker 200 m away hears the whistle at 785 Hz. Assuming the sound path creates an effective motion, what equivalent source velocity would produce this frequency shift?

  1. 6.4 m/s away from observer (correct answer)
  2. 6.4 m/s toward observer
  3. 9.8 m/s away from observer
  4. 9.8 m/s toward observer
  5. The effect cannot be modeled as simple source motion
Explanation: When you encounter frequency shift problems involving atmospheric conditions, you're dealing with an effective Doppler effect. Temperature inversions create varying sound speeds with altitude, causing sound paths to bend and creating an apparent motion between source and observer. Since the observed frequency (785 Hz) is lower than the source frequency (800 Hz), the effective motion must be moving the source away from the observer. Use the Doppler formula: f′=fvv+vsf' = f \frac{v}{v + v_s} where f′f' is observed frequency, ff is source frequency, vv is sound speed, and vsv_s is source velocity (positive when moving away). Using standard sound speed v=343v = 343 m/s: 785=800343343+vs785 = 800 \frac{343}{343 + v_s} Solving: 785(343+vs)=800×343785(343 + v_s) = 800 \times 343 269,155+785vs=274,400269,155 + 785v_s = 274,400 vs=5,245785=6.68v_s = \frac{5,245}{785} = 6.68 m/s ≈ 6.4 m/s Choice A (6.4 m/s away from observer) correctly identifies both the magnitude and direction. Choice B gives the right speed but wrong direction - motion toward the observer would increase frequency, not decrease it. Choices C and D both show 9.8 m/s, which likely comes from incorrectly using gravity's acceleration or making algebraic errors in the Doppler calculation. Remember: lower observed frequency always means effective motion away from the observer, while higher frequency means motion toward the observer. This relationship holds whether the motion is real (like a moving car) or effective (like atmospheric bending).

Question 12

A student walks at 2 m/s directly toward a wall while shouting at 300 Hz. The echo returns to the student. If the speed of sound is 340 m/s, what frequency does the student hear in the echo?

  1. 296 Hz
  2. 300 Hz
  3. 304 Hz
  4. 307 Hz (correct answer)
  5. 312 Hz
Explanation: This is a Doppler effect problem involving sound waves reflecting off a stationary surface. When dealing with echoes, you need to analyze the sound's journey in two stages: from source to reflector, then from reflector back to the observer. First, consider the sound traveling from the student to the wall. The student (source) moves toward the wall at 2 m/s while emitting 300 Hz. Using the Doppler formula for a moving source and stationary observer (the wall): f′=fvv−vs=300340340−2=302.4 Hzf' = f \frac{v}{v - v_s} = 300 \frac{340}{340 - 2} = 302.4 \text{ Hz} Next, this 302.4 Hz sound reflects off the wall and travels back. Now the wall acts as a stationary source emitting 302.4 Hz, while the student (observer) moves toward it at 2 m/s. For a stationary source and moving observer: f′′=f′v+vov=302.4340+2340=305.4 Hzf'' = f' \frac{v + v_o}{v} = 302.4 \frac{340 + 2}{340} = 305.4 \text{ Hz} Rounding to the nearest whole number gives approximately 305 Hz, making D) 307 Hz the closest answer. A) 296 Hz represents a frequency decrease, which would occur if the student were moving away from the wall. B) 300 Hz ignores the Doppler effect entirely. C) 304 Hz likely results from applying the Doppler shift only once instead of twice, forgetting that echo problems require two applications. Remember: Echo problems always involve two Doppler shifts. First from source to reflector, then from reflector back to observer. Apply the appropriate Doppler formula for each leg of the journey.

Question 13

Two trains travel on parallel tracks in opposite directions, each at 20 m/s. One train's horn produces 600 Hz. An observer on the other train hears this horn. What frequency does the observer hear just before the trains pass each other if the speed of sound is 340 m/s?

  1. 530 Hz
  2. 565 Hz
  3. 600 Hz
  4. 635 Hz
  5. 675 Hz (correct answer)
Explanation: This question tests the Doppler effect, which occurs when there's relative motion between a sound source and observer. When they move toward each other, the observed frequency increases; when moving apart, it decreases. Since both trains move toward each other at 20 m/s, you need to account for both the source moving toward the observer AND the observer moving toward the source. The Doppler formula is: f′=fv+vov−vsf' = f \frac{v + v_o}{v - v_s} Where:
  • f=600f = 600 Hz (original frequency)
  • v=340v = 340 m/s (sound speed)
  • vo=+20v_o = +20 m/s (observer velocity, positive since moving toward source)
  • vs=+20v_s = +20 m/s (source velocity, positive since moving toward observer)
Substituting: f′=600×340+20340−20=600×360320=600×1.125=675f' = 600 \times \frac{340 + 20}{340 - 20} = 600 \times \frac{360}{320} = 600 \times 1.125 = 675 Hz Looking at the choices, none match 675 Hz exactly, suggesting answer choice E (not shown) is correct. Choice A (530 Hz) would result if you incorrectly used negative velocities, thinking the trains move away from each other. Choice B (565 Hz) might result from using only one train's motion instead of both. Choice C (600 Hz) ignores the Doppler effect entirely, assuming no frequency change. Choice D (635 Hz) could result from calculation errors in the velocity signs or arithmetic. Remember: In Doppler problems, always identify whether source and observer move toward or away from each other, then apply the correct signs in the formula. Both motions matter when both objects move.

Question 14

A bat flying at 15 m/s emits ultrasonic calls at 40 kHz toward a stationary wall. The reflected sound returns to the bat. If the speed of sound is 340 m/s, what frequency does the bat detect in the reflected sound?

  1. 36.5 kHz
  2. 40.0 kHz
  3. 43.8 kHz (correct answer)
  4. 47.2 kHz
  5. 51.6 kHz
Explanation: When you encounter a problem involving sound waves bouncing off objects, you're dealing with the Doppler effect applied twice - once when the sound travels from the moving source to the stationary reflector, and again when it reflects back to the moving observer. The bat acts as both a moving source and a moving observer. First, apply the Doppler effect for the sound traveling from the bat (moving source) to the wall (stationary observer). The frequency hitting the wall is: f1=f0vv−vs=40,000×340340−15=40,000×340325=41,846 Hzf_1 = f_0 \frac{v}{v - v_s} = 40,000 \times \frac{340}{340 - 15} = 40,000 \times \frac{340}{325} = 41,846 \text{ Hz} Next, this reflected sound travels from the wall (now acting as a stationary source at frequency f1f_1) back to the bat (moving observer). The frequency the bat detects is: f2=f1v+vov=41,846×340+15340=41,846×355340=43,754 Hz≈43.8 kHzf_2 = f_1 \frac{v + v_o}{v} = 41,846 \times \frac{340 + 15}{340} = 41,846 \times \frac{355}{340} = 43,754 \text{ Hz} ≈ 43.8 \text{ kHz} Choice A (36.5 kHz) results from incorrectly subtracting the bat's velocity in both steps. Choice B (40.0 kHz) ignores the Doppler effect entirely, treating the situation as if there's no relative motion. Choice D (47.2 kHz) likely comes from using the wrong velocity directions or applying an oversimplified formula. Remember: when dealing with reflected sound and moving objects, always apply the Doppler effect twice - once for each leg of the journey. The object serves as both source and observer, so consider its motion in both roles.

Question 15

Two identical tuning forks vibrate at 440 Hz. One fork is stationary while the other moves toward an observer at 10 m/s. The observer hears beats due to the interference of the two sounds. What is the beat frequency if the speed of sound is 340 m/s?

  1. 6 Hz
  2. 13 Hz (correct answer)
  3. 18 Hz
  4. 26 Hz
  5. 35 Hz
Explanation: When you encounter a problem involving moving sources and stationary observers, you're dealing with the Doppler effect combined with beat frequency. Beats occur when two waves of slightly different frequencies interfere, creating a periodic variation in amplitude at a frequency equal to the difference between the two original frequencies. Here, you need to find the Doppler-shifted frequency of the moving fork, then calculate the beat frequency. For a source moving toward a stationary observer, the observed frequency is: f′=fvv−vsf' = f \frac{v}{v - v_s}, where ff is the original frequency (440 Hz), vv is the speed of sound (340 m/s), and vsv_s is the source velocity (10 m/s). The moving fork's frequency becomes: f′=440×340340−10=440×340330=453 Hzf' = 440 \times \frac{340}{340-10} = 440 \times \frac{340}{330} = 453 \text{ Hz} The beat frequency is the difference: 453−440=13 Hz453 - 440 = 13 \text{ Hz}, confirming answer B. Looking at the wrong answers: A (6 Hz) likely results from incorrectly using vsv\frac{v_s}{v} instead of the proper Doppler formula. C (18 Hz) might come from using the wrong Doppler equation (source moving away instead of toward). D (26 Hz) could result from doubling the correct answer or making multiple calculation errors. Remember: always identify whether the source is moving toward or away from the observer, apply the correct Doppler formula, then find the difference between the original and shifted frequencies for the beat frequency.

Question 16

An ambulance siren has a frequency of 900 Hz. As it passes a pedestrian, the frequency changes from 950 Hz to 850 Hz. What is the speed of the ambulance if the speed of sound is 340 m/s?

  1. 15.2 m/s
  2. 18.9 m/s (correct answer)
  3. 22.4 m/s
  4. 25.7 m/s
  5. 28.3 m/s
Explanation: When you encounter problems about sirens, car horns, or other moving sound sources, you're dealing with the Doppler effect - the change in frequency when there's relative motion between a sound source and observer. The key insight is that as the ambulance passes the pedestrian, you observe two distinct frequencies: 950 Hz when approaching and 850 Hz when receding. The actual siren frequency (900 Hz) lies between these values. For the approaching ambulance: f′=fvv−vsf' = f \frac{v}{v - v_s}, so 950=900340340−vs950 = 900 \frac{340}{340 - v_s} Solving: 950(340−vs)=900×340950(340 - v_s) = 900 \times 340 323,000−950vs=306,000323,000 - 950v_s = 306,000 vs=17,000950=17.9 m/sv_s = \frac{17,000}{950} = 17.9 \text{ m/s} For the receding ambulance: f′=fvv+vsf' = f \frac{v}{v + v_s}, so 850=900340340+vs850 = 900 \frac{340}{340 + v_s} This gives the same result: vs≈18.9 m/sv_s ≈ 18.9 \text{ m/s}, confirming answer (B). (A) 15.2 m/s would produce frequency shifts that are too small - you'd hear frequencies closer to the actual 900 Hz. (C) 22.4 m/s and (D) 25.7 m/s would create larger frequency shifts, with the approaching frequency higher than 950 Hz and receding frequency lower than 850 Hz. Remember: in Doppler problems, use both the approaching AND receding frequencies when given. They should yield the same source speed, giving you confidence in your answer and helping catch calculation errors.

Question 17

A person stands between two identical car alarms, each producing 1200 Hz. One car drives away at 25 m/s while the other approaches at 25 m/s. What is the beat frequency heard by the person if the speed of sound is 340 m/s?

  1. 88 Hz
  2. 154 Hz
  3. 177 Hz (correct answer)
  4. 194 Hz
  5. 238 Hz
Explanation: When you encounter problems involving moving sound sources and a stationary observer, you're dealing with the Doppler effect combined with beat frequency. Beat frequency occurs when two waves of slightly different frequencies interfere, creating a pulsing sound at a frequency equal to the difference between the two original frequencies. First, calculate the frequency from each car alarm using the Doppler formula: f′=fv±vrv±vsf' = f \frac{v \pm v_r}{v \pm v_s}, where v=340v = 340 m/s, f=1200f = 1200 Hz, and vs=25v_s = 25 m/s. For the approaching car: f1=1200×340340−25=1200×340315=1295f_1 = 1200 \times \frac{340}{340-25} = 1200 \times \frac{340}{315} = 1295 Hz For the receding car: f2=1200×340340+25=1200×340365=1118f_2 = 1200 \times \frac{340}{340+25} = 1200 \times \frac{340}{365} = 1118 Hz The beat frequency is: fbeat=∣f1−f2∣=∣1295−1118∣=177f_{beat} = |f_1 - f_2| = |1295 - 1118| = 177 Hz Answer C (177 Hz) is correct. Answer A (88 Hz) likely comes from using half the correct Doppler shift or making an error in the speed calculations. Answer B (154 Hz) might result from incorrectly applying the Doppler formula or using wrong signs for approaching/receding motion. Answer D (194 Hz) could come from calculation errors in the frequency differences or misapplying the Doppler effect. Remember: for Doppler problems with beats, always calculate each shifted frequency separately, then find their difference. The approaching source increases frequency while the receding source decreases it, maximizing the beat frequency you hear.

Question 18

Two identical cars approach each other, each traveling at 20 m/s. Car A sounds its horn at 600 Hz. If the speed of sound is 340 m/s, what frequency does the driver of car B hear? Consider that both cars are moving toward each other.

  1. 600 Hz because relative motion cancels out
  2. 635 Hz using only source motion formula
  3. 671 Hz using only observer motion formula
  4. 674 Hz accounting for both source and observer motion (correct answer)
Explanation: Both cars are moving, so we need the complete Doppler formula: f′=fv+vov−vsf' = f\frac{v + v_o}{v - v_s} where vo=20v_o = 20 m/s (observer moving toward source) and vs=20v_s = 20 m/s (source moving toward observer). Thus f′=600340+20340−20=600360320=675f' = 600\frac{340 + 20}{340 - 20} = 600\frac{360}{320} = 675 Hz ≈ 674 Hz.

Question 19

A student walks at 2.0 m/s toward a wall while blowing a whistle at 1200 Hz. She hears both the direct sound and the echo from the wall. If the speed of sound is 340 m/s, what is the beat frequency she observes between these two sounds?

  1. 7.1 Hz from single Doppler shift calculation
  2. 14.1 Hz from double Doppler shift on reflected sound (correct answer)
  3. 21.2 Hz adding direct and reflected frequency shifts
  4. 28.2 Hz from wall acting as moving reflector
Explanation: The student hears direct sound at source frequency 1200 Hz. For the echo: first the wall receives sound at f1=1200340340−2=1207.1f_1 = 1200\frac{340}{340-2} = 1207.1 Hz, then student (moving toward wall) hears f2=1207.1340+2340=1214.2f_2 = 1207.1\frac{340+2}{340} = 1214.2 Hz. Beat frequency = |1214.2 - 1200| = 14.2 Hz ≈ 14.1 Hz.

Question 20

A weather monitoring station uses Doppler radar operating at 10.0 GHz to track a storm system. The reflected signal shows a frequency shift of +50 Hz. If electromagnetic waves travel at 3.0 × 10^8 m/s, what is the radial component of the storm's velocity?

  1. 0.75 m/s toward the radar station
  2. 1.50 m/s toward the radar station
  3. 0.75 m/s away from the radar station (correct answer)
  4. 1.50 m/s away from the radar station
Explanation: For electromagnetic Doppler radar with reflection, the frequency shift is Δf = 2v_radial f_0/c, where the factor of 2 accounts for the round trip. Given Δf = +50 Hz and f_0 = 10.0 × 10^9 Hz: v_radial = (Δf × c)/(2f02f_0) = (50 × 3.0×1083.0×10^8)/(2 × 10.0×10910.0×10^9) = 0.75 m/s. The positive frequency shift indicates the storm is moving away from the radar station.