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JEE Main Physics Class 11 Oscillations Part 1 Questions
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© examsnet.com
Question : 32
Total: 100
A particle executes simple harmonic motion with an amplitude of
5
c
m
. When the particle is at
4
c
m
from the mean position, the magnitude of its velocity in SI units is equal to that of its acceleration. Then, its periodic time in seconds is:
[10 Jan 2019, II]
4
π
3
3
8
π
8
π
3
7
3
π
Validate
Solution:
Velocity,
v
=
ω
√
A
2
−
x
2
........(i)
acceleration,
a
=
−
ω
2
x
........(ii)
and according to question,
|
v
|
=
|
a
|
⇒
ω
√
A
2
−
x
2
=
ω
2
x
⇒
A
2
−
x
2
=
ω
2
x
2
⇒
5
2
−
4
2
=
ω
2
(
4
2
)
3
=
ω
×
4
⇒
ω
=
3
4
∴
T
=
2
π
∕
ω
=
2
π
3
∕
4
=
8
π
3
© examsnet.com
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