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TS EAMCET 9-Sep-2020 Shift 2 Solved Paper
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© examsnet.com
Question : 101
Total: 160
An engine sounding a whistle of frequency 2000 Hz is receding from the stationary observer at 72 km/h. What is the apparent frequency of the observer? The velocity of sound in air is 340 m/s
1889 Hz
2889 Hz
3889 Hz
4889 Hz
Validate
Solution:
Here a source is moving away from a stationary observer.
So, frequency recorded at source will be lower than true frequency.
Apparent frequency at observer,
f
′
=
f
(
v
v
+
v
s
)
where,
v
=
speed of sound
=
340
m
/
s
and
v
s
=
72
k
m
/
h
=
72
×
5
18
=
20
m
s
−
1
∴
f
′
=
2000
(
340
340
+
20
)
© examsnet.com
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