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Describe the Doppler Effect and calculate the period and frequency of a detected wave from a given graph.

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3 parts
3.2

1.40×10−3 s\text{1.40} \times 10^{-3} \text{ s}

3.3

714 Hz\text{714 Hz}

3.4

54.4 m/s\text{54.4 m/s}

Step-by-step solution

8 steps

Here are the solutions to Question 3.

Part 3.1

State the Doppler Effect in words.

The Doppler Effect is the apparent change in the frequency (and thus pitch) of a wave observed when the source of the wave and the observer are moving relative to each other. The frequency increases when they move closer and decreases when they move apart.

Part 3.2

1.40×10−3 s\text{1.40} \times 10^{-3} \text{ s}

Determine the period of the detected wave.

  1. Identify the time for half a period from the graph.

    The graph shows a peak at t=3.5×10−4t = 3.5 \times 10^{-4} s and a trough at t=10.5×10−4t = 10.5 \times 10^{-4} s. The time difference between a peak and a trough is half a period (T/2T/2).

    T2=(10.5−3.5)×10−4 s\frac{T}{2} = (10.5 - 3.5) \times 10^{-4} \text{ s} T2=7.0×10−4 s\frac{T}{2} = 7.0 \times 10^{-4} \text{ s}

  2. Calculate the full period (TT).

    T=2×7.0×10−4 sT = 2 \times 7.0 \times 10^{-4} \text{ s} T=14.0×10−4 sT = 14.0 \times 10^{-4} \text{ s} T=0.0014 sT = 0.0014 \text{ s}

    The period of the detected wave is 1.40×10−3 s\boxed{\text{1.40} \times 10^{-3} \text{ s}}.

Part 3.3

714 Hz\text{714 Hz}

Calculate the frequency of the detected sound wave.

  1. Use the relationship between frequency (ff) and period (TT).

    f=1Tf = \frac{1}{T}

  2. Substitute the period calculated in 3.2.

    f=114.0×10−4 sf = \frac{1}{14.0 \times 10^{-4} \text{ s}} f=1000014 Hzf = \frac{10000}{14} \text{ Hz} f=50007 Hzf = \frac{5000}{7} \text{ Hz} f≈714.29 Hzf \approx 714.29 \text{ Hz}

    The frequency of the detected sound wave is 714 Hz\boxed{\text{714 Hz}}.

Part 3.4

54.4 m/s\text{54.4 m/s}

Calculate the speed of the moving car.

  1. Identify the given values.

    Source frequency, fs=600f_s = 600 Hz
    Speed of sound in air, v=340v = 340 m/s
    Detected frequency, fL=714.29f_L = 714.29 Hz (using the more precise fraction 50007\frac{5000}{7} Hz for calculation)

  2. Determine the appropriate Doppler Effect formula.

    Since the detected frequency (fL=714.29f_L = 714.29 Hz) is higher than the source frequency (fs=600f_s = 600 Hz), the car (source) must be moving towards the stationary listener. The formula for a moving source and stationary listener approaching each other is:

    fL=fs(vv−vs)f_L = f_s \left( \frac{v}{v - v_s} \right)

    where vsv_s is the speed of the source (car).

  3. Rearrange the formula to solve for vsv_s.

    fLfs=vv−vs\frac{f_L}{f_s} = \frac{v}{v - v_s} fL(v−vs)=fsvf_L (v - v_s) = f_s v fLv−fLvs=fsvf_L v - f_L v_s = f_s v fLvs=fLv−fsvf_L v_s = f_L v - f_s v vs=v(fL−fsfL)v_s = v \left( \frac{f_L - f_s}{f_L} \right)

  4. Substitute the values and calculate vsv_s.

    vs=340 m/s(50007 Hz−600 Hz50007 Hz)v_s = 340 \text{ m/s} \left( \frac{\frac{5000}{7} \text{ Hz} - 600 \text{ Hz}}{\frac{5000}{7} \text{ Hz}} \right) vs=340 m/s(5000−4200750007)v_s = 340 \text{ m/s} \left( \frac{\frac{5000 - 4200}{7}}{\frac{5000}{7}} \right) vs=340 m/s(800750007)v_s = 340 \text{ m/s} \left( \frac{\frac{800}{7}}{\frac{5000}{7}} \right) vs=340 m/s(8005000)v_s = 340 \text{ m/s} \left( \frac{800}{5000} \right) vs=340 m/s(850)v_s = 340 \text{ m/s} \left( \frac{8}{50} \right) vs=340 m/s(425)v_s = 340 \text{ m/s} \left( \frac{4}{25} \right) vs=136025 m/sv_s = \frac{1360}{25} \text{ m/s} vs=54.4 m/sv_s = 54.4 \text{ m/s}

    The speed of the moving car is 54.4 m/s\boxed{\text{54.4 m/s}}.

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