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AP EAMCET Engineering 2018 Apr 22 Shift 2 Paper
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© examsnet.com
Question : 101
Total: 160
An observer and a source emitting sound of frequency 120 Hz are on the X-axis. The observer is stationary while the source of sound is in motion given by the equation
x
=
3
sin
ω
t
( x inmeters and t in second). If the difference between the maximum and minimum frequencies of the sound observed by the observer is 22 Hz, then the value of ω is (Speed of sound in air = 330 ms
−
1
)
33 rads
−
1
36 rads
−
1
20 rads
−
1
10 rads
−
1
Validate
Solution:
The frequency of source is f =120 Hz .
The following figure represents the required diagram.
Write the expression of the instantaneous speed of the source.
v
=
d
x
d
t
=
3
ω
sin
ω
t
Here, the difference between the maximum and minimum frequency is 22 Hz.
f
max
−
f
min
=
f
(
v
v
−
v
s
)
−
f
(
v
v
+
v
s
)
f
max
−
f
min
=
f
(
v
v
−
v
s
−
v
v
+
v
s
)
f
(
v
2
+
v
v
s
−
v
2
+
v
v
s
v
2
−
v
s
2
)
=
f
max
−
f
min
120
(
2
v
v
s
v
2
−
v
s
2
)
=
22
2
(
330
)
(
3
ω
)
(
330
)
2
−
(
3
ω
)
2
=
22
120
ω
=
10
rad.s
−
1
© examsnet.com
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