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DOPPLER EFFECT QUESTIONS AND ANSWERS

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DOPPLER EFFECT QUESTIONS AND ANSWERS Have you ever noticed how the sound of an ambulance changes as it passes by? Or observed how the pitch of a siren alters as a police car speeds past you? These phenomena are not merely auditory curiosities; they are manifestations of a fundamental principle in physics known as the Doppler Effect. Whether you’re a seasoned physicist or just someone curious about the world around you, understanding the Doppler Effect can enrich your appreciation of everyday experiences and deepen your understanding of the universe. In this Page, we’ll explore the Doppler Effect, answer common questions, and delve into its applications across various fields.

Activity 1

  1. A sound source approaches a stationary (not moving) observer at constant velocity. Which ONE of the following describes how the observed frequency and wavelength differ from that of the sound source?
    Observed WavelengthObserved Frequency
     A. Greater than Greater than
     B. Less than Less than
     C. Greater than Less than
     D. Less than Greater than
    1. Greater than
    2. Less than
    3. Greater than Less than
    4. Less than Greater than (2)
  2. Which one of the following is the main principle applied when using the rate of blood flow or the heartbeat of a foetus in the womb?
    1. Doppler Effect.
    2. Photoelectric effect
    3. Huygens principle
    4. Diffraction (2)
  3. An ambulance approaches an accident scene at constant velocity.
    The siren of the ambulance emits sound waves at a frequency of 980 Hz. A detector at the scene measures the frequency of the emitted sound waves as 1 050 Hz.

    1. Calculate the speed at which the ambulance approaches the accident scene. Use the speed of sound in air as 340 m·s–1 (6)
    2. Explain why the measured frequency is higher than the frequency of the source. (2)
    3. The principle of the Doppler Effect is applied in the Doppler flow meter. State ONE positive impact of the use of the Doppler flow meter on humans. (2)
  4. The siren of a stationary (not moving) ambulance emits sound waves at a frequency of 850 Hz.
    An observer (person witnessing this) who is travelling in a car at a constant speed in a straight line, begins measuring the frequency of the sound waves emitted by the siren when he is at a distance x from the ambulance. The observer continues measuring the frequency as he approaches, passes, and moves away from the ambulance. The results obtained are shown in the graph below.
    7

    1. The observed frequency suddenly changes at t = 6s. Give a reason for this sudden change in frequency. (1)
    2. Calculate
      b(1) The speed of the car (Take the speed of sound in air as 340 m·s–1) (5)
      b(2) Distance x between the car and the ambulance when the observer BEGINS measuring the frequency. (3)
      [23]
Solutions

  1. D (2)
  2. A (2)
  3. fL = V ± VL (fs) OR fL = V (fs)
    V ± VL                (V – Vs)
    ∴ 1050 3 – (340 – 0) / (340 – Vs) × 980
    ∴ Vs = 22,67 m/s (6)

    1. Waves in front of the moving source are compressed. The observed wavelength decreases (3). For the same speed of sound (3), a higher frequency will be observed. (2)
    2. Determine whether arteries are clogged or narrowed (3), so that precautions can be taken to prevent heart attack or stroke (3), OR Determine the heartbeat (3) of a foetus to assure that the child is alive or does not have a heart defect (3). (2)
  4. The approaching observer (higher f) passes the source at t = 6 s and moves away (lower f) from the source. (3) (1)
    1. (1)
      Option 1. Approaching observer:
      fL = V ± VL (fs)
      V ± Vs
      OR
      fL = V + VL (fs)
      V
      ∴ 900 3 = (340 + VL) (850) / (340)
      ∴ VL = 20 m/s (5)
      Option 2. Observer moving away:
      fL = V ± VL (fs)
      V ± Vs
      OR
      fL = V – VL (fs)
      V
      ∴ 800 = (340 – VL) (850) / (340)
      ∴ VL = 20 m/s (5)
      Notes: any other correct Doppler Effect formula gets maximum 3/4 marks.
    2. (2)
      Option 1.
      Δx = v1Δt + ½ aΔt2
      = (20)(6) + ½ (0)Δt2
      = 120 m
      Option 2.
      Δx = vΔt
      = (20)(6)
      = 120 m (3)
      Note: accept s = ut or s = vt, as well as s = ut + ½at2, as well as Δy = v1Δt + ½aΔt2
      [23]

Activity 2
An ambulance moving at 40 m·s–1 approaches a traffic light where a blind man and his dog wait to cross a road. The siren of the ambulance (source) emits sound waves at a frequency of 350 Hz (fs). The pitch of the sound that the man hears increases as the ambulance moves towards him and decreases as the ambulance passes him and moves away.

  1. If the speed of sound in air is 340 m·s–1, determine the frequency (fL) of the sound waves that the man hears while the ambulance approaches him. (3)
  2. Explain how this effect can help a blind person waiting to cross the road. (2)
    [5] 
Solutions

  1. fL =   (V±VL)
    (V±VS)fS
    fL =         (340+0)
    (340 – 40) x 350Hz
    fL = 396,67 Hz (3)
  2. When crossing a street, a blind person can determine whether a car is moving towards or away from him. If the pitch of a vehicle decreases, the person knows that the vehicle is moving away from him, and vice versa. (2)
    [5] 

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