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JEE Main 11 Jan 2019 Shift 2 Solved Paper
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© examsnet.com
Question : 5
Total: 90
A simple pendulum of length 1 m is oscillating with an angular frequency 10 rad/s. The support of the pendulum starts oscillating up and down with a small angular frequency of 1 rad/s and an amplitude of
10
–
2
m. The relative change in the angular frequency of the pendulum is best given by :-
10
–
3
rad/s
10
–
1
rad/s
1 rad/s
10
–
5
rad/s
Validate
Solution:
Angular frequency of pendulum
ω =
√
g
eff
l
∴
Δ
ω
ω
=
1
2
Δ
g
eff
g
eff
Δω =
1
2
Δ
g
g
× ω
[
ω
s
= angular frequency of support]
Δω =
1
2
×
2
A
ω
s
2
100
× 100
Δω =
10
–
3
rad/s
© examsnet.com
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