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.