Doppler shift in a high speed chase

DOPPLER SHIFT IN A HIGH SPEED 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 a cop car is chasing the fugitive car on a straight long freeway. Consider two cars, cop car [source] moving at 90 mph [144 kph] toward a fugitive car [observer] moving at 65 mph [104 kph] on a long freeway. We’ll determine the apparent frequency as registered by the observer in fugitive car when cop car sounds the siren.

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 fugitive car [observer] (m/s)

V_s – Velocity of cop car [source] (m/s)

The Doppler shift equation for this case is,

f’ = f₀*{[V – V_o]/[V – V_s]} (Eq.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 fugitive car is V_o = 65 mph

                                                      = 28.88 m/s (Eq. 3)

The velocity of cop car is V_s = 90 mph

                                               = 40 m/s (Eq. 4)

Frequency of cop car siren f₀ = 700 Hz (Eq. 5)

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

Substitute equations (3), (4), (5) and (6) in equation (2),

f’ = 700*{[343 – 28.88]/[343 – 40]}

f’ = 725.68 Hz

This is the frequency of sound as registered by the observer [fugitive car] at the freeway when source [cop car] approaches him.

Difference in frequency = f’ – f₀

                                       = 725.68 – 700

                                       = 25.68 Hz

CONCLUSION

We thus determined the apparent frequency as registered by the observer due to Doppler shift. 

Doppler shift near church pt2

DOPPLER SHIFT NEAR CHURCH 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 car is moving away from a stationary sound source. Consider a car [observer] moving at a speed of 35 mph [56 kph] moving away from a Church [source] which continuously emits sound waves by ringing the bell. We’ll determine the apparent frequency as registered by the observer in car.

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 observer [car] (m/s)

V_s – Velocity of Source [Church bell] (m/s)

Since the source is stationary,

V_s = 0 (Eq. 2)

Substitute equation (2) in equation (1),

f’ = f₀*{[V – V_o]/[V]} (Eq. 3)

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

The velocity of car V_o = 35 mph

                                    = 15.55 m/s (Eq. 4)

Frequency of Church bell 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*{[343–15.55]/[343]}

f’ = 954.66 Hz

This is the frequency of sound as registered by the observer in car when he recedes away from a church that rings the bell.

Difference in frequency = f’ – f₀

                                       = 954.66 – 1000

                                       = –45.34 Hz

The negative value 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.

Doppler effect near Church

DOPPLER SHIFT NEAR CHURCH

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 car is moving toward a stationary sound source. Consider a car [observer] moving at a speed of 35 mph [56 kmph] approaching a Church [source] which continuously emits sound waves by ringing the bell. We’ll determine the apparent frequency as registered by the observer in car.

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 observer [car] (m/s)

V_s – Velocity of Source [Church bell] (m/s)

Since the source is stationary,

V_s = 0 (Eq. 2)

Substitute equation (2) in equation (1),

f’ = f₀*{[V + V_o]/[V]} (Eq. 3)

The ‘+’ sign in the numerator of equation (3) indicates that the observer is approaching the source.

The velocity of car V_o = 35 mph

                                    = 15.55 m/s (Eq. 4)

Frequency of Church bell 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+15.55]/[343]}

f’ = 1045.33 Hz

This is the frequency of sound as registered by the observer in car when he approaches a church that rings the bell.

Difference in frequency = f’ – f₀

                                       = 1045.33 – 1000

                                       = 45.33 Hz

CONCLUSION

We thus determined the apparent frequency as registered by the observer due to Doppler shift.

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.

Doppler shift on a Freeway Pt1

DOPPLER SHIFT ON A FREEWAY PT1

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 toward each other on a straight long freeway. Consider two cars, car1 [source] and car2 [observer] moving at a speed of 65 mph [104 kmph] toward 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 toward the source while the ‘–’ sign in the denominator indicates that the source is moving toward 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’ = 473.55 Hz

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

Difference in frequency = f’ – f₀

                                       = 473.55 – 400

                                       = 73.55 Hz

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.

Doppler shift at Railroad Crossing Pt2

DOPPLER SHIFT AT RAILROAD CROSSING PT 2

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 train moves away from a stationary observer at a railroad crossing. Consider a train emitting sound moving at a speed of 70 mph [112kmph] receding away from an observer at an unmanned railroad crossing. We’ll determine the apparent frequency as registered by the observer.

ASSUMPTIONS

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

The effect of humidity on sound is negligible

The amplitude of sound is unity

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 (m/s)

V_s – Velocity of Source [Train] (m/s)

Since the observer is stationary,

 V_o = 0 (Eq. 2)

Substitute equation (2) in equation (1),

f’ = f₀*{[V]/[V + V_s]} (Eq. 3)

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

The velocity of train V_s = 70 mph

                                       = 31.11 m/s (Eq. 4)

Frequency of train horn f₀ = 500 Hz (Eq. 5)

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

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

f’ = 500*{343/[343 + 31.11]}

f’ = 458.42 Hz

This is the frequency of sound as registered by the stationary observer at the unmanned railroad crossing when a train sounding horn recedes away from him.

Difference in frequency = f’ – f₀

                                       = 458.42 – 500

                                       = – 41.58 Hz

The answer is negative since apparent frequency is less than original one but the magnitude is always positive.

CONCLUSION

We thus determined the apparent frequency as registered by the observer due to Doppler shift.

Doppler Shift at Railroad Crossing

DOPPLER SHIFT AT RAILROAD CROSSING 

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 train is moving toward a stationary observer at a railroad crossing. Consider a train emitting sound moving at a speed of 70 mph [112 kmph] approaching an observer at an unmanned railroad crossing. We’ll determine the apparent frequency 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]} (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 (m/s)

V_s – Velocity of Source [Train] (m/s)

Since the observer is stationary,

V_o = 0 (Eq. 2)

Substitute equation (2) in equation (1),

f’ = f₀*{[V]/[V – V_s]} (Eq. 3)

The ‘–’ sign in the denominator of equation (3) indicates that the source is approaching the observer.

The velocity of train V_s = 70 mph

                                       = 31.11 m/s (Eq. 4)

Frequency of train horn f₀ = 500 Hz (Eq. 5)

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

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

f’ = 500*{343/[343 – 31.11]}

f’ = 549.87 Hz

This is the frequency of sound as registered by the stationary observer at the unmanned railroad crossing when a train sounding horn approaches him.

Difference in frequency = f’ – f₀

                                       = 549.87 – 500

                                       = 49.87 Hz

CONCLUSION

We thus determined the apparent frequency as registered by the observer due to Doppler shift.