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
- 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 ± Vo]/[V ± Vs]} (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}
Vo
– Velocity of observer (m/s)
Vs
– Velocity of Source [Train] (m/s)
Since the observer is
stationary,
Vo = 0 (Eq. 2)
Substitute equation (2)
in equation (1),
f’ = f0*{[V]/[V
– Vs]} (Eq. 3)
The ‘–’ sign in the
denominator of equation (3) indicates that the source is approaching the
observer.
The velocity of train Vs
= 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.
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