Dopple shift during super sonic chase pt2

DOPPLER SHIFT DURING SUPER SONIC CHASE 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 one supersonic military jet [source] is receding away from another supersonic jet [observer]. 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 supersonic military jets, jet1 [source] moving at Mach 2 and jet2 [observer] moving at Mach 1.5. We’ll determine the apparent frequency of source [jet1] noise as registered by the observer in jet2 when jet1 is receding away from him. In other words, jet2 is chasing jet1.

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 [Jet2] (m/s)

V_s – Velocity of Source [Jet1] (m/s)

The Doppler shift equation for this case is,

f’ = f₀*{[V + V_o]/[V + V_s]} (Eqn. 2)

The ‘+’ sign in the numerator of equation (2) indicates that the observer is moving toward the source while the ‘+’ sign in the denominator indicates that the source is moving away from the observer.

The velocity of Jet1 V_s = Mach2

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

              = 2*343

              = 686 m/s (Eqn. 3)

The velocity of Jet2 V_o = Mach1.5

              = 1.5*speed of sound                                     {⸪ Mach1 = speed of sound}

              = 1.5*343

              = 514.5 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 (3), (4), (5) and (6) in equation (2),

f’ = 1000*{[343 + 514.5]/[343 + 686]}

f’ = 833.33 Hz

This is the frequency of sound as registered by the observer in the supersonic military jet when another supersonic source is receding away from him. We observe that the apparent frequency is approximately more than two thirds of the original value. For a stationary observer, it will be only one third of the original value. So when the supersonic jet2 approaches the oncoming sound wave, it experiences a Doppler shift and registers the apparent frequency.

Difference in frequency = f’ – f₀

                                       = 833.33 – 1000

                                       = – 166.67 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 due to Doppler shift and concluded that the observer in jet2 will be able to hear more than two thirds the original value of jet1 sound.

Doppler shift during a super sonic chase

DOPPLER SHIFT DURING A SUPER SONIC 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 one supersonic military jet [source] is chasing another supersonic jet [observer]. 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 supersonic military jets, jet1 [source] moving at Mach 2 and jet2 [observer] moving at Mach 1.5. We’ll determine the apparent frequency of source jet noise as registered by the observer in jet2 when jet1 is chasing jet2.

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 [Jet2] (m/s)

V_s – Velocity of Source [Jet1] (m/s)

The Doppler shift equation for this case is,

f’ = f₀*{[V – V_o]/[V – V_s]} (Eqn. 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 Jet1 V_s = Mach2

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

              = 2*343

              = 686 m/s (Eqn. 3)

The velocity of Jet2 V_o = Mach1.5

              = 1.5*speed of sound                                     {⸪ Mach1 = speed of sound}

              = 1.5*343

              = 514.5 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 (3), (4), (5) and (6) in equation (2),

f’ = 1000*{[343 – 514.5]/[343 – 686]}

f’ = 500 Hz

This is the frequency of sound as registered by the observer in the supersonic jet2 when another supersonic jet1 is chasing jet2. We observe that the apparent frequency is half of the original value. Although it implies that the observer will register only half of the original frequency, this is not the case. This is because sound is traveling slower than the source jet1and observer jet2, thereby it will never catch up with jet1 or jet2.

Difference in frequency = f’ – f₀

                                       = 500 – 1000

                                       = – 500 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 the observer in jet2 will not be able to hear any sound from the chasing jet1.

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.

Super sonic doppler shift in military jet pt5

SUPER SONIC DOPPLER SHIFT IN A MILITARY JET PT5

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 toward 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 toward each other. We’ll determine the apparent frequency of jet noise as registered by the observer in jet2 when jet1 approaches 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 [Jet] (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 toward the source while the ‘–’ sign in the denominator indicates that the source is moving toward 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’ = – 3000 Hz

This is the frequency of sound as registered by the observer in the supersonic military jet when it approaches another supersonic source. We observe that the apparent frequency is negative three times the original value. The negative sign implies that the observer will hear the sound after the source passes away. This means that the observer will register thrice the original frequency but only after the source has passed by. The sound from the source will always lag since it travels slower than the source, so the observer will first see the source and hear it later. However the observer will hear sound of different frequency when he moves away from the source which we’ll discuss in the next post.

Difference in frequency = f’ – f₀

                                       = – 3000 – 1000

                                       = – 4000 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 due to Doppler shift and concluded that the observer will be able to hear thrice the original value but only after the source has passed by.