Doppler shift on a freeway Pt2

DOPPLER SHIFT ON A FREEWAY PT2

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 two cars are moving away from each other on a straight long freeway. Consider two cars, car1 [source] and car2 [observer] moving at a speed of 65 mph [104 kmph] away from each other on a divided freeway. We’ll determine the apparent frequency as registered by the observer in car2 when car1 sounds the horn.

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 car2 [observer] (m/s)

V_s – Velocity of car1 [source] (m/s)

The observer and source are moving at the same velocity hence,

V_o = V_{s (Eq. 2)}

Substitute equation (2) in equation (1),

f’ = f₀*{[V – V_o]/[V + V_o]} (Eq. 3)

The ‘–’ sign in the numerator of equation (3) indicates that the observer is moving away from the source while the ‘+’ sign in the denominator indicates that the source is moving away from the observer.

The velocity of car1 and car2 V_o = 65 mph

                                                     = 28.88 m/s (Eq. 4)

Frequency of car horn f₀ = 400 Hz (Eq. 5)

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

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

f’ = 400*{[343 – 28.88]/[343 + 28.88]}

f’ = 337.87 Hz

This is the frequency of sound as registered by the observer [car2] at the freeway when source [car1] is receding away from him.

Difference in frequency = f’ – f₀

                                       = 337.87 – 400

                                       = – 62.13 Hz

Negative sign indicates that the apparent frequency is less than the original one but the magnitude of the frequency is always positive.

CONCLUSION

We thus determined the apparent frequency as registered by the observer due to Doppler shift. In fact if both cars honk then they both will experience Doppler shift since they act as source and observer at the same time for each other.