Doppler shift in emergency vehicle

DOPPLER SHIFT IN EMERGENCY VEHICLE

INTRODUCTION

We know that Doppler Effect or Doppler shift occurs between a source and observer when they are in relative motion with respect to each other. In this case we’ll determine the Doppler shift that occurs when a passenger car slows down while an emergency vehicle [Ambulance] passes beside at a higher speed. Consider two vehicles, an Ambulance [source] moving at 35 mph [56 kmph] away from a passenger car [observer] moving at 5 mph [8 kmph] on a long street. We’ll determine the apparent frequency as registered by the observer in passenger car when Ambulance sounds the siren.

ASSUMPTIONS

1. The atmospheric air has standard temperature and pressure conditions

·         Temperature T = 298 K or 25°C or 77°F

·         Pressure = 1 bar = 10⁵ N/m²

2. The effect of humidity on sound is negligible

3. The amplitude of sound is unity

4. The air molecules do not move with respect to source and observer

CALCULATION

The equation for Doppler shift is given by,

f’ = f₀*{[V ± V_o]/[V ± V_s]} (Eq. 1)

f₀ – Original frequency (Hz)

f’ – Apparent or observed frequency (Hz)

V – Velocity of Sound in air at standard temperature and pressure conditions (m/s) {V = 343 m/s}

V_o – Velocity of passenger car [observer] (m/s)

V_s – Velocity of Ambulance [source] (m/s)

The Doppler shift equation for this case is,

f’ = f₀*{[V + V_o]/[V + V_s]} (Eq. 2)   

The ‘+’ sign in the numerator of equation (2) indicates that the observer is moving toward the source while the ‘+’ sign in the denominator indicates that the source is moving away from the observer.

The velocity of passenger car is V_o = 5 mph

                                                         = 2.22 m/s (Eq. 3)

The velocity of Ambulance is V_s = 35 mph

                                                      = 15.55 m/s (Eq. 4)

Frequency of Ambulance siren f₀ = 800 Hz (Eq. 5)

Speed of sound in air V = 343 m/s (Eq. 6)

Substitute equations (3), (4), (5) and (6) in equation (2),

f’ = 800*{[343 + 2.22]/[343 + 15.55]}

f’ = 770.25 Hz

This is the frequency of sound as registered by the observer [passenger car] in the street when source [Ambulance] recedes away from him.

Difference in frequency = f’ – f₀

                                       = 770.25 – 800

                                       = –29.75 Hz

The negative sign indicates that the apparent frequency is less than the original but the magnitude is always the same.

CONCLUSION

We thus determined the apparent frequency as registered by the observer due to Doppler shift.