Doppler shift in a high speed chase

DOPPLER SHIFT IN A HIGH SPEED CHASE

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 cop car is chasing the fugitive car on a straight long freeway. Consider two cars, cop car [source] moving at 90 mph [144 kph] toward a fugitive car [observer] moving at 65 mph [104 kph] on a long freeway. We’ll determine the apparent frequency as registered by the observer in fugitive car when cop car 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 fugitive car [observer] (m/s)

V_s – Velocity of cop car [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 away from the source while the ‘–’ sign in the denominator indicates that the source is moving toward the observer.

The velocity of fugitive car is V_o = 65 mph

                                                      = 28.88 m/s (Eq. 3)

The velocity of cop car is V_s = 90 mph

                                               = 40 m/s (Eq. 4)

Frequency of cop car siren f₀ = 700 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’ = 700*{[343 – 28.88]/[343 – 40]}

f’ = 725.68 Hz

This is the frequency of sound as registered by the observer [fugitive car] at the freeway when source [cop car] approaches him.

Difference in frequency = f’ – f₀

                                       = 725.68 – 700

                                       = 25.68 Hz

CONCLUSION

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