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