JEE Main · 2019 · 12 Jan 2019 · Shift 1 · Elasticity & SHM

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Two light identical springs of spring constant k are attached horizontally at the two ends of a uniform horizontal rod AB of length I and mass m . the rod is pivoted at its centre '

O

' and can rotate freely in horizontal plane. The other ends of the two springs are fixed to rigid supports as shown in figure. The rod is gently pushed through a small angle and released. The frequency of resulting oscillation is:

(A) $\frac{1}{2 π} \frac{3 k}{m}$
(B) $\frac{1}{2 π} \frac{2 k}{m}$
(C) $\frac{1}{2 π} \frac{6 k}{m}$ ✓ Correct
(D) $\frac{1}{2 π} \frac{k}{m}$

Answer: (C)

Solution

When the rod turns through a small angle

θ

about its centre, each end moves through

x = \frac{L}{2} θ

and stretches its spring, so each spring pushes back with force

k x

. Both springs give a restoring torque in the same sense, each with moment arm

\frac{L}{2}

, so the total restoring torque is

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

. The rod's moment of inertia about its centre is

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

. Writing

τ = I \ddot{θ}

gives

\ddot{θ} = - \frac{6 k}{m} θ

, so the motion is simple harmonic with

ω = \frac{6 k}{m}

and frequency

f = \frac{1}{2 π} \frac{6 k}{m}

JEE Main · 2019 · 12 Jan 2019 · Shift 1 · Single correct option

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JEE Main · 2019 · 12 Jan 2019 · Shift 1 · Elasticity & SHM

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