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JEE Main Physics Class 11 Oscillations Part 1 Questions
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© examsnet.com
Question : 30
Total: 100
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
r
a
d
∕
s
and an amplitude of
10
−
2
m
.
The relative change in the angular frequency of the pendulum is best given by:
[11 Jan 2019, II]
10
−
3
rad
∕
s
1
rad
∕
s
10
−
1
rad
∕
s
10
−
5
rad
∕
s
Validate
Solution:
Angular frequency of pendulum
ω
=
√
g
l
∴
relative change in angular frequency
Δ
ω
ω
=
1
2
Δ
g
g
[
as length remains constant
]
Δ
g
=
2
A
ω
s
2
[
ω
s
=
angular frequency of support and,
A
=
amplitude]
Δ
ω
ω
=
1
2
×
2
A
ω
S
2
g
Δ
ω
=
1
2
×
2
×
1
2
×
10
−
2
10
=
10
−
3
rad
∕
sec
© examsnet.com
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