Super sonic doppler shift in military jet pt6

SUPER SONIC DOPPLER SHIFT IN A 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 supersonic military jets are moving away from each other. A supersonic military jet is a jet that moves faster than the speed of sound thereby leading to a sonic boom. A sonic boom is an explosion that occurs when any object travels faster than sound. Consider two military jets, jet1 [source] and jet2 [observer] moving at Mach2 away from each other. We’ll determine the apparent frequency of jet noise as registered by the observer in jet2 when jet1 recedes away from him.

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]} (Eqn. 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 (m/s)

V_s – Velocity of source (m/s)

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

V_o = V_s (Eqn. 2)

Substitute equation (2) in equation (1),

f’ = f₀*{[V – V_o]/[V + V_s]} (Eqn. 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 = Mach2

                    = 2*speed of sound                                  {⸪ Mach1 = speed of sound}

                    = 2*343

                    = 686 m/s (Eqn. 4)

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

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

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

f’ = 1000*{[343 – 686]/[343 + 686]}

f’ = – 1000/3 Hz

This is the frequency of sound as registered by the observer in the supersonic military jet when it recedes away from a supersonic source. We observe that the apparent frequency is negative one third the original value. The negative sign implies that the observer in jet2 is traveling faster than the speed of sound. This means that the observer in jet2 will not register any sound since sound waves cannot catch up with the supersonic observer.

Difference in frequency = f’ – f₀

                                       = – 1000/3 – 1000

                                       = – (4/3)*1000 Hz

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

CONCLUSION

We thus determined the apparent frequency as registered by the observer in jet2 due to Doppler shift and concluded that this observer will not be able to hear sound of source jet1.