Changes due to motion of the source and observer (the Doppler effect)

An ambulance with its siren on approaches a pedestrian, passes them and moves away.

1. a) Describe how the pitch perceived by the pedestrian changes as the ambulance approaches and as it moves away (higher or lower pitch).
2. b) Explain the change using the compression and spreading out of the wavefronts, using a diagram in your notebook to help you.
3. c) Does the frequency actually emitted by the siren change? Distinguish between emitted frequency and perceived frequency.

Without making calculations, state in each case whether the perceived frequency is greater than, less than, or equal to the emitted frequency, and justify your answer:

1. a) A train approaching the station sounding its whistle.
2. b) The same train moving away from the station.
3. c) A pedestrian running toward a stationary siren.
4. d) A galaxy moving away from Earth.
5. e) Source and observer in relative rest.

When a source approaches a stationary observer, the observer perceives the frequency $f' = f \times \frac{v}{v - v_f}$ (formula given; speed of sound in air: $v = 340\ \text{m/s}$).

An ambulance emits sound at $500\ \text{Hz}$ and approaches a pedestrian at $20\ \text{m/s}$. Calculate the frequency perceived by the pedestrian.

A car with its alarm sounding at $800\ \text{Hz}$ passes a stationary person while traveling at $20\ \text{m/s}$ ($v = 340\ \text{m/s}$).

1. a) Calculate the frequency perceived as it approaches.
2. b) Calculate the frequency perceived as it moves away.
3. c) Determine the sudden drop in frequency perceived at the exact moment the car passes.

In front of a moving source, the perceived frequency is $f' = f \times \frac{v}{v - v_f}$ (formula given; $v = 340\ \text{m/s}$). The Mach number is the ratio of an aircraft's speed to the speed of sound. The figure shows the wavefronts of a source moving at nearly the speed of sound.

1. a) Calculate $f'$ for an aircraft at Mach $0.9$ ($v_f = 306\ \text{m/s}$) whose engine emits a noise of $1000\ \text{Hz}$.
2. b) Analyze what happens to the denominator of the formula and to $f'$ as $v_f$ approaches $v$ (Mach $1$).
3. c) Relate this to the shock wave and sonic boom: explain why an observer on the ground does not hear a supersonic aircraft approaching until after it has already passed them.

When the observer moves toward a stationary source, the formula is $f' = f \times \frac{v + v_0}{v}$ (formula given; $v = 340\ \text{m/s}$).

A cyclist rides at $10\ \text{m/s}$ toward a loudspeaker emitting a $600\ \text{Hz}$ tone. The figure shows the situation.

Calculate the frequency perceived by the cyclist.

If both the source and the observer move toward each other at the same time, the perceived frequency is $f' = f \times \frac{v + v_0}{v - v_f}$ (formula given; $v = 340\ \text{m/s}$).

A train sounds a $400\ \text{Hz}$ whistle as it approaches at $30\ \text{m/s}$, while a cyclist pedals at $10\ \text{m/s}$ toward it. The figure shows the situation.

Calculate the frequency perceived by the cyclist.

A bat flies at $6\ \text{m/s}$ toward a stationary moth while emitting ultrasound at $80\ \text{kHz}$ ($v = 340\ \text{m/s}$). The diagram shows the emission and the echo.

1. a) Calculate the frequency "received" by the moth, using $f_1 = f \times \frac{v}{v - v_f}$.
2. b) The reflected echo behaves as if the moth were a stationary source of frequency $f_1$ and the bat an observer approaching at $6\ \text{m/s}$: calculate the final frequency heard by the bat, $f_2 = f_1 \times \frac{v + v_0}{v}$.
3. c) Compare $f_2$ with the emitted frequency $f$ and reason why this double Doppler effect makes it easier to detect prey.
4. d) Predict qualitatively how $f_2$ would change if the moth were also moving, either fleeing from the bat or flying toward it.

A traffic radar emits microwaves that reflect off vehicles and return to the device. The diagram shows the emission and the echo.

1. a) Explain the two successive Doppler effects experienced by the wave: the car as an 'observer' receiving the signal and as a 'source' re-emitting the echo.
2. b) Reason why the echo returns with a higher frequency if the car is approaching the radar.
3. c) What wave phenomenon makes it possible for the signal to return to the radar?

A traffic radar emits microwaves at $f = 2.4 \times 10^{10}\ \text{Hz}$. For speeds much smaller than the speed of light, the frequency shift of the echo is $\Delta f \approx \frac{2 \times f \times v}{c}$ (formula given; $c = 3 \times 10^{8}\ \text{m/s}$). An approaching car returns an echo with $\Delta f = 4 \times 10^{3}\ \text{Hz}$.

1. a) Rearrange the formula for the speed of the car and calculate it in $\text{m/s}$.
2. b) Express it in $\text{km/h}$.

In a Doppler ultrasound scan of a blood vessel, the frequency shift of the echo is $\Delta f \approx \frac{2 \times f \times v_{\text{sangre}}}{v}$ (formula given; speed of sound in tissue: $v = 1540\ \text{m/s}$). The probe emits at $5\ \text{MHz}$ and the blood flows at $0.3\ \text{m/s}$.

1. a) Calculate $\Delta f$.
2. b) Compare the result with the range of human hearing ($20\ \text{Hz}$ to $20\ \text{kHz}$) and reason why the equipment converts this shift into an audible tone for the doctor.

A low Earth orbit satellite transmits at $f = 2.2\ \text{GHz}$ and moves at $7.6\ \text{km/s}$ relative to the ground station. For $v \ll c$, $\frac{\Delta f}{f} = \frac{v}{c}$ holds true (formula given; $c = 3 \times 10^{8}\ \text{m/s}$).

1. a) Calculate the maximum shift $\Delta f$, expressed in $\text{kHz}$.
2. b) Determine the frequency range that the station must sweep through during a full pass, with the satellite first approaching and then moving away.
3. c) Explain why satellite receivers must continuously retune during each pass.