Syllabus Edition

First teaching 2023

First exams 2025

|

Doppler Effect for Sound Waves (CIE AS Physics)

Exam Questions

1 hour22 questions
1
Sme Calculator
1 mark

Which of the following statements are true?

  • When a source emitting a sound wave moves away from an observer, the observed frequency increases

  • When a source emitting a sound wave moves away from an observer, the observed wavelength increases

  • When a source emitting a sound wave moves towards an observer, the observed frequency decreases

  • When a source emitting a sound wave moves towards an observer, the observed wavelength increases

Did this page help you?

2
Sme Calculator
1 mark

A source emits a sound wave of speed v and frequency fs. The source moves towards an observer at a speed of vs.

What is the observed frequency of the source?

  • fraction numerator f subscript s v over denominator v space plus space v subscript s end fraction

  • fraction numerator f subscript s v subscript s over denominator v space minus space v subscript s end fraction

  • fraction numerator f subscript s v over denominator v space minus v subscript s end fraction

  • fraction numerator f subscript s v subscript s over denominator v space plus space v subscript s end fraction

Did this page help you?

3
Sme Calculator
1 mark

A train is moving towards a stationary observer. The train sounds its horn. 

What are the correct changes in frequency, wavelength and pitch of the horn as heard by the observer?

 

Frequency

Wavelength

Pitch

A.

higher shorter higher

B.

higher shorter lower

C.

lower shorter lower

D.

lower longer higher

    Did this page help you?

    4
    Sme Calculator
    1 mark

    Which of the following equations is correct for the frequency of a source that is moving toward an observer?

    • f subscript s space equals space fraction numerator f subscript 0 open parentheses v space minus space v subscript s close parentheses over denominator v end fraction

    • f subscript s space equals space fraction numerator f subscript 0 open parentheses v space plus space v subscript s close parentheses over denominator v end fraction

    • f subscript s space equals space fraction numerator f subscript o v over denominator v space minus space v subscript s end fraction

    • f subscript s space equals space fraction numerator f subscript o v over denominator v space plus space v subscript s end fraction

    Did this page help you?

    5
    Sme Calculator
    1 mark

    A car driving at a velocity of 10 m s−1 toward an observer sounds its horn with a frequency of 10 kHz. The speed of sound in air is 330 m s−1.

    What frequency does the observer hear?

    • 8.9 kHz

    • 9.7 kHz

    • 10.1 kHz

    • 10.3 kHz

    Did this page help you?

    1
    Sme Calculator
    1 mark

    A police car siren emits a sound wave with a frequency fs of 440 Hz. The car is travelling away from an observer at a speed of 30 m s−1.

     The speed of sound is 340 m s−1.

     Which of the following is the frequency the observer hears?

    • 478 Hz

    • 460 Hz

    • 440 Hz

    • 404 Hz

    Did this page help you?

    2
    Sme Calculator
    1 mark

    Figure 1.1 shows object P, moving at velocity vs. It emits a sound wave of frequency fs. The speed of sound is v.

    cie-mcq-7-3-q2-cie-ial-physics

    Fig. 1.1

    At which point on the diagram will the observed frequency be fraction numerator space f subscript s v over denominator v plus v subscript s end fraction?

      Did this page help you?

      3
      Sme Calculator
      1 mark

      A source moving at constant velocity vs, emits a sound wave of frequency 450 Hz. A stationary observer records a frequency of 300 Hz.

      What is the value of vs, and is it moving towards or away from the observer?

      Speed of sound, v = 340 m s-1

       

      vs / m s-1

      Towards / Away

      A.

      850

      Away

      B.

      170

      Towards

      C.

      170

      Away

      D.

      850

      Towards

        Did this page help you?

        4
        Sme Calculator
        1 mark

        A car of terrified observers try to get the car to start to drive away from a hungry T. rex, running toward them at a speed of 10 m s−1. They hear the roar of the T. rex at a frequency of 170 Hz. The speed of sound in air is 340 m s−1.

        What is the frequency of the sound as emitted by the T. rex?

        • 175 nm

        • 120 Hz

        • 165 Hz

        • 220 Hz

        Did this page help you?

        5
        Sme Calculator
        1 mark

        A car travels in a straight line towards a stationary observer at a speed of 0.2v, where v is the speed of sound. The car's horn emits a pure sound of frequency fs.

        What is the frequency of the sound of the horn as heard by the observer, f0?

        • f0 = 0.83fs

        • f0 = 1.00fs

        • f0 = 1.20fs

        • f0 = 1.25fs

        Did this page help you?

        1
        Sme Calculator
        1 mark

        The sun emits light of a particular wavelength λs. At any instant, there is a range of wavelengths from less than λs to greater than λs­ that can be observed on Earth. This is due to the Doppler effect.

         Which of the following statements would explain this Doppler effect?

        • the Sun is rotating

        • the Sun is moving toward the Earth

        • the Sun is moving away from the Earth

        • the Sun is moving at right-angles to a line joining the Sun and the Earth

        Did this page help you?

        2
        Sme Calculator
        1 mark

        Two runners, A and B, run past each other in opposite directions, both at speed u. A police car has its siren on. It is traveling at a speed of 10u, in the same direction as Runner A. Runner A and Runner B observe the sound wave from the siren at frequencies of fA and fB respectively. The speed of sound is v.

        What is the ratio of f subscript A over f subscript B?

        • fraction numerator v minus 11 u over denominator v plus 9 u end fraction

        • fraction numerator v over denominator v plus 12 u end fraction

        • fraction numerator v minus 11 u over denominator v minus 9 u end fraction

        • fraction numerator v over denominator v minus 10 u end fraction

        Did this page help you?

        3
        Sme Calculator
        1 mark

        Fig. 1.1 is a graph of the observed wavelength against time of a sound wave emitted from a single source. The emitted sound wave has a frequency f subscript s. The sound travels at a constant velocity directly towards an observer and a frequency of f subscript T is observed. It then travels directly away from the observer and a frequency of f subscript A is observed.

        cie-mcq-hard-7-3-q3-cie-ial-physics

        Fig. 1.1

        What is the value of the ratiof subscript A over f subscript T?

        • 25 over 17

        • 25 over 21

        • 17 over 25

        • 21 over 25

        Did this page help you?

        4
        Sme Calculator
        1 mark

        Fig 1.1 is a diagram of a source that has a mass of 500 g and is connected to a rope of length 20 cm. The rope is attached to point Q and has a tension of 800 N. The source travels around point Q anticlockwise in a circular path with a velocity of vs. It emits a sound wave of frequency fs. When the source is at point P, an observer stands directly in line to the direction of motion at point O.

        cie-mcq-hard-7-3-q4-cie-ial-physics

        Fig 1.1

        What is the value of vs and f subscript o over f subscript s when the source is at point P?

        Speed of sound, v = 340 m s-1

         

        vs / m s-1

        bold italic f subscript bold italic o over bold italic f subscript bold italic s

        A.

        17.9

        1.06

        B.

        17.9

        0.950

        C.

        320

        17.0

        D.

        320

        0.515

          Did this page help you?

          5
          Sme Calculator
          1 mark

          A source moves directly in line with an observer at a constant velocity vs and emits a sound wave with a wavelength of λs. The observed wavelength of the source is λ0 and the ratio lambda subscript 0 over lambda subscript s is equal to 5 over 4.

          What is the value of vs?

          Speed of sound, v = 340 m s-1

          • 68.0 m s-1

          • 85.0 m s-1

          • 612 m s-1

          • 765 m s-1

          Did this page help you?