Sonic Doppler Shift in Military Jet Pt6

SONIC DOPPLER SHIFT IN MILITARY JET PT6

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 sonic military jets are moving away from each other. A sonic military jet is a jet that moves at the speed of sound. Consider two military jets, jet1 [source] and jet2 [observer] moving at a speed of 343 m/s away from each other. We’ll determine the apparent frequency of jet noise as registered by the observer in jet2 when jet1 moves away from him.

ASSUMPTIONS

1    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    2. The effect of humidity on sound is negligible

3    3. The amplitude of sound is unity

4    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 observer [Jet2] (m/s)

V_s – Velocity of Source [Jet1] (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_s]} (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 jet V_o = V_s = 343 m/s (Eq. 4)                                                              

Frequency of jet noise f₀ = 1000 Hz (Eq. 5)                                                                   

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

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

f’ = 1000*{[343 – 343]/[343 + 343]}

f’ = 0 Hz

This is the frequency of source noise as registered by observer [Jet2] in the military jet when moving away from the source [Jet1]. We can observe that frequency is zero which means the observer will not be able to hear anything.

Difference in frequency = f’ – f₀

                                       = 0 – 1000

                                       = – 1000 Hz

Negative value indicates that apparent frequency is less than the original but magnitude is always positive.

CONCLUSION

We thus determined the apparent frequency of source noise as registered by the observer due to Doppler shift and concluded that the observer will be able to hear nothing when he is moving away from the source. In fact both source and observer will each experience the exact same Doppler shift since they both are source and observer for each other.