IB Physics Quiz: Understand Doppler Effect

20 questions · exam conditions

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Understand Doppler EffectQuestion 1 of 20

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?

The pitch is constant but higher than the source pitch.

The pitch is constant but lower than the source pitch.

The pitch continuously decreases.

The pitch continuously increases.

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IB Physics Quiz

IB Physics Quiz: Understand Doppler Effect

Practice Understand Doppler Effect in IB 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 Understand Doppler Effect, giving you a quick way to practice the rules, question types, and explanations that matter most for IB 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

The pitch is constant but higher than the source pitch.
The pitch is constant but lower than the source pitch.
The pitch continuously decreases.
The pitch continuously increases. (correct answer)

Explanation: 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.

Question 2

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?

The star is moving towards the Earth. (correct answer)
The star is moving away from the Earth.
The star is rotating very rapidly.
The star is significantly hotter than the Sun.

Explanation: 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.

Question 3

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 $v$ .

$f \frac{v}{v-u}$ (correct answer)
$f \frac{v+u}{v}$
$f$
$f \sqrt{\frac{v+u}{v-u}}$

Explanation: 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}

Question 4

A car moves towards a stationary observer. The frequency heard by the observer is 5.0% greater than the frequency emitted by the car's horn. What is the speed of the car? The speed of sound in air is 330 m s⁻¹.

15.7 m s⁻¹ (correct answer)
16.5 m s⁻¹
31.4 m s⁻¹
33.0 m s⁻¹

Explanation: The observed frequency is

f' = 1.05 f

. For a source moving towards a stationary observer, the formula is

f' = f \frac{v}{v-v_s}

. Therefore,

1.05 = \frac{330}{330-v_s}

. Rearranging gives

1.05(330-v_s) = 330

, which leads to

346.5 - 1.05 v_s = 330

. Solving for

v_s

gives

1.05 v_s = 16.5

, so

v_s = 16.5 / 1.05 \approx 15.7

m s⁻¹. Distractor B is the result of using the approximation

\Delta f/f \approx v_s/v

, which gives

v_s = 0.05 \times 330 = 16.5

m s⁻¹, a close but incorrect answer.

Question 5

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?

A high constant frequency, which then abruptly drops to a low constant frequency. (correct answer)
A frequency that smoothly and continuously decreases from high to low.
A frequency that is initially high and constant, then drops to zero as the source passes.
A frequency that increases as the source approaches and decreases as it moves away.

Explanation: 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.

Question 6

A star is moving away from the Earth at a speed $v$ . The light from the star is observed to have a redshift corresponding to a fractional wavelength increase of $x = \Delta\lambda/\lambda$ . If the star's recession speed were to double to $2v$ , what would be the new fractional wavelength increase, assuming $2v \ll c$ ?

$x/2$
$x$
$2x$ (correct answer)
$x^2$

Explanation: For electromagnetic waves at non-relativistic speeds (

v \ll c

), the fractional change in wavelength (redshift or blueshift) is directly proportional to the relative speed of the source and observer:

\Delta\lambda/\lambda \approx v/c

. Therefore,

x \propto v

. If the speed

v

is doubled to

2v

, the fractional increase

x

will also double to

2x

Question 7

A police car with its siren on travels at a constant speed along a straight road. A person stands on a sidewalk at some distance from the road. At the instant the car is at its point of closest approach to the person, what is the frequency heard by the person compared to the frequency emitted by the siren, $f_s$ ?

It is higher than $f_s$ .
It is lower than $f_s$ .
It is instantaneously changing from higher to lower than $f_s$ .
It is equal to $f_s$ . (correct answer)

Explanation: The Doppler shift depends on the component of the source's velocity along the line connecting the source and the observer. At the point of closest approach, the car's velocity vector is perpendicular to the line of sight to the observer. Therefore, the component of velocity towards or away from the observer is zero at that specific instant. With no relative velocity along the line of sight, there is no Doppler shift, and the observed frequency is equal to the source frequency

f_s

Question 8

A source of sound and an observer are in relative motion. Which of the following situations will result in the observer hearing a frequency lower than the source frequency?

I. The source and observer are moving away from each other.

II. The source moves towards a stationary observer, but the temperature of the air, and thus the speed of sound, decreases.

III. The observer moves in a circle around the stationary source at a constant speed.

I only (correct answer)
I and II only
III only
I and III only

Explanation: I: When the distance between the source and observer is increasing, the observer perceives a lower frequency (redshift). This is correct. II: A source moving towards an observer causes a frequency increase. A decrease in the speed of sound

v

would modify the result, but the primary effect of approach is an increase in pitch. This statement is incorrect. III: An observer moving in a circle around a stationary source maintains a constant distance from it. The observer's velocity is always perpendicular to the line connecting them to the source. Thus, the radial component of velocity is zero, and there is no Doppler shift. The observed frequency equals the source frequency. Therefore, only statement I is correct.

Question 9

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 $c = 3.0 \times 10^8 \, \text{m s}^{-1}$ .

643 nm
656 nm
669 nm (correct answer)
682 nm

Explanation: 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}

Question 10

This question is intended for Higher Level (HL) students.

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.

471 Hz
515 Hz
516 Hz (correct answer)
531 Hz

Explanation: 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.

Question 11

A train whistle emits a sound of frequency 500 Hz and wavelength 0.68 m. The train moves away from a stationary observer at 34 m s⁻¹. What are the frequency and wavelength of the sound as measured by the observer? The speed of sound is 340 m s⁻¹.

Frequency = 455 Hz, Wavelength = 0.68 m
Frequency = 455 Hz, Wavelength = 0.75 m (correct answer)
Frequency = 500 Hz, Wavelength = 0.75 m
Frequency = 556 Hz, Wavelength = 0.61 m

Explanation: First, calculate the observed frequency for a source moving away:

f' = f \left( \frac{v}{v+v_s} \right) = 500 \left( \frac{340}{340+34} \right) = 500 \left( \frac{340}{374} \right) \approx 455

Hz. Next, calculate the observed wavelength. The speed of the sound waves in the air,

v

, remains constant at 340 m s⁻¹. The new wavelength is

\lambda' = v/f' = 340 / 455 \approx 0.747

m, which is 0.75 m. The wavelength increases because the source moves a certain distance between emitting each wave crest, effectively stretching the waves.

Question 12

Light from a star is analysed. A spectral line with a laboratory wavelength of 400 nm is observed at 402 nm. Six months later, the same line from the same star is observed at 399 nm. What is a plausible explanation for this change?

The star reversed its direction of motion relative to the Sun.
The Earth moved from one side of its orbit to the other relative to the star. (correct answer)
The star's temperature significantly increased, then decreased.
The light passed through a different interstellar dust cloud for the second measurement.

Explanation: The first measurement (400 nm -> 402 nm) is a redshift, indicating the net velocity is away from Earth. The second measurement (400 nm -> 399 nm) is a blueshift, indicating the net velocity is towards Earth. The Earth's orbital velocity around the Sun is approximately 30 km/s. Over six months, the Earth moves to the opposite side of its orbit, reversing its velocity vector relative to a distant star. This periodic change in Earth's velocity accounts for the observed shift from red to blue for many stars.

Question 13

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.

0 Hz
12 Hz
48 Hz
24 Hz (correct answer)

Explanation: 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.

Question 14

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?

$f_d = f_0 \left( \frac{v_s - v_i}{v_s + v_i} \right)$ (correct answer)
$f_d = f_0 \left( \frac{v_s}{v_s + v_i} \right)$
$f_d = f_0 \left( 1 - \frac{2v_i}{v_s} \right)$
$f_d = f_0 \left( \frac{v_s + v_i}{v_s - v_i} \right)$

Explanation: 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)

Question 15

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?

The signals lose energy as they travel through space, which reduces their frequency over vast distances.
The probe is moving away from Earth, causing a Doppler redshift in the radio signals. (correct answer)
Gravitational fields between the probe and Earth stretch the wavelength of the signals, causing a gravitational redshift.
The Earth's atmosphere preferentially absorbs higher-frequency components of the radio signal.

Explanation: 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.

Question 16

This question is intended for Higher Level (HL) students.

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?

$f_O = f_S$
$f_O > f_S$
The relationship depends on whether $u$ is greater or less than $v/2$ .
$f_O < f_S$ (correct answer)

Explanation: 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).

Question 17

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?

$863$ Hz
$741$ Hz
$930$ Hz (correct answer)
$800$ Hz

Explanation: 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.

Question 18

Two cars approach each other on a straight road. Car A has a horn with frequency $400$ Hz and travels at $20$ m/s. Car B travels at $15$ m/s toward Car A. If the speed of sound is $340$ m/s, what frequency does the driver of Car B hear?

$442$ Hz (correct answer)
$418$ Hz
$383$ Hz
$358$ Hz

Explanation: Both cars are moving toward each other, so both source and observer motions increase the observed frequency. Using

f' = f_0 \frac{v + v_o}{v - v_s}

where

v_o = 15

m/s (observer velocity) and

v_s = 20

m/s (source velocity):

f' = 400 \frac{340 + 15}{340 - 20} = 400 \frac{355}{320} = 442.5

Hz ≈ 442 Hz. Choice B only accounts for the moving source. Choice C incorrectly uses subtraction for the observer term. Choice D uses incorrect signs for both velocities.

Question 19

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?

$200$ Hz
$212$ Hz (correct answer)
$424$ Hz
$106$ Hz

Explanation: 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.

Question 20

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?

$150$ Hz
$300$ Hz (correct answer)
$450$ Hz
$600$ Hz

Explanation: 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.