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IIT JEE Advanced 2009 Paper 2
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© examsnet.com
Question : 4
Total: 57
A uniform rod of length L and mass M is pivoted at thecentre.Its two ends are attached to two springs of equal spring constants k. The springs are fixed to rigid supports as shown in the figure, and the rod is free to oscillate in the horizontal plane. The rod is gently pushed through a small angle θ in one direction and released. The frequency of oscillation is
1
2
π
√
2
k
M
1
2
π
√
k
M
1
2
π
√
6
k
M
1
2
π
√
24
k
M
Validate
Solution:
The restoring torque is
J = - 2 × kx
(
L
2
)
cos θ = I
(
d
2
θ
d
t
2
)
Now, x =
L
2
sin θ
Therefore,
J = - k
(
L
2
)
cos θ = I
(
d
2
θ
d
t
2
)
⇒
(
−
k
L
2
4
)
sin 2θ = I
(
d
2
θ
d
t
2
)
For small θ, sin 2θ = 2θ. Therefore,
−
k
L
2
θ
2
= I
(
d
2
θ
d
t
2
)
where I =
M
L
2
12
. Therefore,
d
2
θ
d
t
2
=
(
−
6
k
M
)
θ =
−
ω
2
θ
(SHM)
⇒ ω =
√
6
k
M
Hence, the frequency of oscillation is
ω
2
π
=
1
2
π
√
6
k
M
© examsnet.com
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