JEE Advanced · 2009 · 12 Apr 2009 · Paper 2 · Elasticity & SHM

A uniform rod of length

L

and mass

M

is pivoted at the centre. 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

(A) $\frac{1}{2 π} \frac{2 k}{M}$
(B) $\frac{1}{2 π} \frac{k}{M}$
(C) $\frac{1}{2 π} \frac{6 k}{M}$ ✓ Correct
(D) $\frac{1}{2 π} \frac{24 k}{M}$

Answer: (C)

Solution

When the rod is rotated by a small angle

θ

, each end moves by

\frac{L}{2} θ

. The restoring torque is

τ = - 2 \times k (\frac{L}{2} θ) \times \frac{L}{2} = - \frac{k L ^{2}}{2} θ

. The moment of inertia of the rod about its center is

I = \frac{M L ^{2}}{12}

. Using

τ = I α

, we get

\frac{d ^{2} θ}{d t ^{2}} = - \frac{k L ^{2} /2}{M L ^{2} /12} θ = - \frac{6 k}{M} θ

. Thus

ω = \frac{6 k}{M}

and frequency

f = \frac{1}{2 π} \frac{6 k}{M}

JEE Advanced · 2009 · 12 Apr 2009 · Paper 2 · Single correct option

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JEE Advanced · 2009 · 12 Apr 2009 · Paper 2 · Elasticity & SHM

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