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.

Doppler effect of Sound

DOPPLER EFFECT OF SOUND

INTRODUCTION

The Doppler Effect or Doppler Shift named after Austrian physicist Christian Doppler is a phenomenon of change in frequency or wavelength of a wave for an observer moving relative to the source. When the source [body emitting sound] is in motion toward the observer, waves [ex: sound, light] are compressed leading to shorter wavelengths or greater frequency. Similarly when the source moves away from the observer, the waves are spread apart leading to longer wavelengths or smaller frequency. We’ll discuss the effect of Doppler shift in everyday life particularly two cases where Doppler shift doesn’t occur.

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    2. The effect of humidity on sound is negligible

3    3. The amplitude of sound is negligible

4    4. The air molecules do not move with respect to source and observer

CALCULATION

The Doppler shift arises when two observers in different frames register two different values of frequency while there is only one original frequency which is the source frequency.

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 (m/s)

There are two cases where the apparent and original frequency will be the same.

CASE # 1

If both source and observer are stationary, no matter how far or near they are, both will register the same frequency. When both source and observer are stationary,

V_o = V_s = 0 (Eq. 2)

Substitute equation (2) in equation (1),

f’ = f₀*{[V ± 0]/[V ± 0]}

f’ = f₀

CASE # 2

If both source and observer are moving in same direction with the same velocity, they both will register the same frequency. In this case,

V_o = V_{s  (Eq. 3)}

Substitute equation (3) in equation (1),

f’ = f₀*{[V ± V_s]/[V ± V_s]}

f’ = f₀

CONCLUSION

Thus the Doppler shift was explained and the two cases where the shift doesn’t occur were also explained.

Velocity Time dilation on International Space Station

VELOCITY TIME DILATION ON THE INTERNATIONAL SPACE STATION

INTRODUCTION

International Space Station [ISS] is a satellite orbiting the Earth at a height of 250 miles from the surface. It completes one revolution around the Earth in approximately 90 Earth minutes. Consider the ISS which revolves at a certain velocity, an observer [observer 1] on Earth’s surface who also performs circular motion since Earth is rotating and another observer [observer 2] at the center of Earth who will not perform circular motion. According to the Special Theory of Relativity, a clock on ISS would run slower relative to the stationary observer at the center of Earth. We’ll find the time gained by the ISS and observer 1 with respect to observer 2.

ASSUMPTIONS

1. The revolution path is exactly circular.
2. Earth is a perfect homogenous sphere.
3. The effect of Gravitational time dilation on ISS is negligible.

CALCULATION

The Angular velocity of Earth

ω_e = 2π/T_e [rad/s]

T_e – Rotational time period of Earth [T ≈ 24 hours = 86400s]

ω_e = 2*3.14/86400

ω_e = 7.2722*10^-5 rad/s

The angular velocity is same at all points on earth since it doesn’t exhibit differential rotation. But the tangential velocity on surface varies with the distance from the center. The greater the distance, higher is the velocity. It is important to note that since ground level observer is also performing a circular motion, he will also experience time dilation relative to observer at the center of Earth.

The Angular velocity of ISS

ω_i = 2π/T_i [rad/s]

T_i – Revolution time period [T ≈ 90 minutes = 5400s]

ω_i = (2*3.14)/5400

ω_i = 1.1629*10^-3 rad/s

The tangential velocity of observer at center of Earth is given by,

V₀ = [R]*ω_e [m/s]

R – Average radius of earth [R = 0 m]

V₀ = 0 m/s

The tangential velocity of ground observer is given by,

V_e = [R]*ω_e [m/s]

R – Average radius of earth [R = 6371000m]

V_e = 6371000*7.2722*10^-5

V_e = 463.3118 m/s

The tangential velocity of ISS is given by,

V_i = [R + h]*ω_i [m/s]

R – Average radius of earth [R = 6371000m]

h – Height of ISS from ground [h = 400 km]

V_i = 6771*1000*1.1629*10^-3

V_i = 7873.995 m/s

According to the Special Theory of Relativity, the time dilation equation is,

t’ = t/γ [s]

t’– ISS time. [s]

t₁ – Ground level observer’s time. [s]

t₂ - Proper time or center of earth observer’s time. [s]

γ – Relativistic gamma factor, γ = 1/√ [1-(v/c) ²]

c - Velocity of light [c = 3*10⁸ m/s]

t’ = t*√ [1-(v/c) ²]

Center of Earth observer time

t₂ = t since v = 0

Ground level observer time

t₁ = t*√ [1-2.3850*10^-12]

t₁ = t*√ [0.999999999997615]

t₁ = t* 0.999999999998808

ISS time

t’ = t*√ [1-6.4448*10^-10]

t’ = t*√ [0.99999999935552]

t’ = t* 0.99999999967776

CONCLUSION

We can observe that proper and actual time isn’t the same which proves that time dilates on revolving ISS relative to the stationary one. We’ll consider 5 different t’ values and calculate t value. The larger the t₂ the more is the difference between t₂ and t’. Thus ISS will gain time over both observer1 and observer2. In fact in one year it gains 0.01 second with respect to observer2.

Time	t2 [Center of Earth observer] (s)	t1 [Ground level observer] (s)	t’ [Revolving ISS] (s)
1 minute	60	59.9999999999285	59.9999999806656
1 hour	3600	3599.99999999571	3599.99999883994
1 day	86400	86399.999999897	86399.9999721585
1 month	2592000	2591999.99999691	2591999.99916475
1 year	31536000	31535999.9999624	31535999.9898378