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AP EAMCET Engineering 21 Apr 2019 Shift 1 Paper
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© examsnet.com
Question : 101
Total: 160
A source of sound "S' in the form of a block kept oil a smooth horizontal surface is connected to a spring, as shown in the figure. If the spring oscillates with an amplitude of 50 cm along horizontal between the wall and the observer O. the maximum frequency heard by the observer is 12.5% more than the minimum frequency heard by him. If the mass of the source of sound is 100 g. the force constant of the spring is (Speed of sound in air is 340
m
s
−
1
)
40 N
m
−
1
80 N
m
−
1
160 N
m
−
1
320 N
m
−
1
Validate
Solution:
The expression for the apparent frequency for the moving source is,
n
=
n
0
v
v
−
v
s
The condition for
n
max
,
v
s
=
v
s
max
and for
n
min
,
v
s
=
−
v
s
max
The expression for
n
max
is given by
n
max
=
n
0
v
v
−
v
smax
.
.
.
(
1
)
The expression for
n
min
is given by
n
min
=
n
0
v
v
+
v
s
m
a
x
.
.
.
(
2
)
Divide equation (1) by (2)
1.125
n
min
n
min
=
v
+
v
s
max
v
−
v
s
max
1.125
(
v
−
v
s
max
)
=
v
+
v
s
max
v
(
0.125
)
=
2.125
v
s
max
v
s
max
=
340
×
0.125
2.125
=
20
m
s
−
1
The expression for the
v
s
max
is given by
v
s
m
a
x
=
A
√
k
m
k
=
v
2
s
max
m
A
2
Substitute the given values in the above equation.
k
=
v
2
s
max
m
A
2
=
20
×
20
×
0.10
0.50
×
0.50
=
160
N
m
−
1
© examsnet.com
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