Super sonic doppler shift in military jet pt2

SUPER SONIC DOPPLER SHIFT IN A MILITARY JET 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 a supersonic military jet is moving away from a stationary observer in a building. 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 a military jet [source] moving at a speed of Mach2 receding away from a stationary observer who is inside a building. We’ll determine the apparent frequency of the jet’s noise as registered by the observer.

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 [Jet] (m/s)

Since the observer is stationary,

V_o = 0  (Eqn. 2)                                                                                                 

Substitute equation (2) in equation (1),

f’ = f₀*{[V]/[V + V_s]} (Eqn. 3)         

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

The velocity of jet V_s = Mach2

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

           = 2*343

           = 686 m/s (Eqn. 4)      

Frequency of jet exhaust 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/[343 + 686]}

f’ = 1000/3

f’ = 333.33 Hz

This is the frequency of sound as registered by the stationary observer in a building when a supersonic military jet moves away from him. We observe that the apparent frequency value drops to one third of the original value. It means the observer will only hear one-third of the original value. The one-third factor is due to the high velocity of the jet, the greater the jet’s velocity the lower is the apparent frequency.

Difference in frequency = f’ – f₀

                                       = [– 1000/3] – 1000

                                       = [– 2000/3] Hz

   = – 666.66 Hz

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

CONCLUSION

We thus determined the apparent frequency as registered by the observer due to Doppler shift and concluded that the observer will be able to hear one third the original frequency of jet noise when the jet is receding away from him.