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JEE Main Physics Class 11 Oscillations Part 1 Questions
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© examsnet.com
Question : 43
Total: 100
A block of mass
0.1
k
g
is connected to an elastic spring of spring constant
640
N
m
−
1
and oscillates in a medium of constant
10
−
2
k
g
s
−
1
. The system dissipates its energy gradually. The time taken for its mechanical energy of vibration to drop to half of its initial value, is closest to :
[Online April 9,2017]
2
s
3.5
s
5
s
7
s
Validate
Solution:
Since system dissipates its energy gradually, and hence amplitude will also decreases with time according to
a
=
a
0
e
−
b
t
∕
m
......(i)
∵
Energy of vibration drop to half of its initial value
(
E
0
)
,
as
E
∝
a
2
⇒
a
∝
√
E
a
=
a
0
√
2
⇒
b
t
m
=
10
−
2
t
0.1
=
t
10
From eq
n
(i),
a
0
√
2
=
a
0
e
−
t
∕
10
1
√
2
=
e
−
t
∕
10
or
√
2
=
e
t
10
ln
√
2
=
t
10
∴
t
=
3.5
seconds
© examsnet.com
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